To show $v^G = z$ we show $v^G \subseteq z$ and $z \subseteq v^G$. Let $\sigma^G \in v^G$, therefore $(\sigma, p) \in \tau$ and $p \Vdash \phi(\sigma)$, therefore $\phi(\sigma^G)$, which implies that $\sigma^G \in z$.

Conversely, let $\sigma^G \in z$ therefore $\sigma^G \in \tau^G$ and $\phi(\sigma^G)$. Then, $(\sigma, p) \in \tau$ for some $p \in G$. Moreover, by FTF:Truth there is some $q$ that forces $\phi(\sigma^G)$. Let $r \in G$ be a common extension of $p, q$. Then $(\sigma, r) \in \tau$ and by FTF:Coherence $\tau \Vdash \phi(\sigma^G)$ so by definition of $v$, $(\sigma, r) \in v$ and therefore $\sigma^G \in v^G$. $\square$

**Replacement Schema.** Given any formula $\phi(x, y, z)$ relativized to $M[G]$ and set $\pi^G \in M[G]$ (the domain) such that for any $\sigma^G \in \pi^G$ there is a unique $\tau^G$ such that $\phi(\sigma^G, \tau^G, \pi^G)$, we need to show that there is a set $\rho^G \in M[G]$ (the image of the domain under the class function $\phi$) that contains all elements $\tau^G$ that are making $\phi(\sigma^G, \tau^G, \pi^G)$ true for some $\sigma^G \in \pi^G$.

Fix $\mathbb{P}$-name $\pi$ in $M$. For every $(\sigma, p) \in \pi$, let $\alpha_{\sigma,p}$ be the least ordinal in $M$ such that there exists $\tau$ with $(p, \sigma, \tau, \pi) \in \mathcal{F}_{\alpha, \sigma, p}^{\phi}$, i.e., with $p \Vdash \phi(\sigma, \tau, \pi)$, if any such $p$ exists, and let $\alpha_{\sigma,p} = \emptyset$ if not. By Replacement in $M$, the set of these ordinals is in $M$. To see this, consider the formula $\text{Ord}(\alpha) \wedge (\exists p \exists \tau (p, \sigma, \tau, \pi) \in \mathcal{F}_{\alpha}^{\phi} \lor \alpha = \emptyset) \wedge \forall \beta \in \alpha \neg(\exists p \exists \tau (p, \sigma, \tau, \pi) \in \mathcal{F}_{\beta}^{\phi})$. For any set $\pi$ and any $(\sigma, p) \in \pi$ there is a unique $\alpha$ that makes this formula true, therefore by Replacement we can collect into a set all ordinals $\alpha$ that make the formula true for elements of $\pi$'s domain.

Therefore, the supremum of $\alpha_{\sigma,p}$ for any $(\sigma, p)$, call it $\alpha^*$ is also in $M$.

Now define $V_\alpha := \{\sigma^G \mid \sigma \in N_\alpha\} \in M[G]$ for any $\alpha \in M$. If we can show that any $\tau^G$ such that $\phi(\sigma^G, \tau^G, \pi^G)$ for $\sigma^G \in \pi^G$ has to be in $V_{\alpha^*}$, then we can define the set of all of such $\tau^G$ using Comprehension$^{M[G]}$, which we just proved. That is to say, we can define $\{y \in V_{\alpha^*} \mid \exists x \in \pi^G \phi(x, y, \pi^G)\} \in M[G]$ by Comprehension$^{M[G]}$. Therefore, all we need to show is that $\text{nr}(\tau) \le \alpha^*$ for any $\tau$ such that $\phi(\sigma^G, \tau^G, \pi^G)$.

Let $\sigma^G \in \pi^G$ be arbitrary. We know $(\sigma, p) \in \pi$ for some $p \in G$. Also, there is some $\tau^G$ making $\phi$ true and by FTF.Truth there is some $q \in G$ that forces $\phi(\sigma, \tau, \pi)$. Taking $r \ge p, q$ as a common extension, we know $(\sigma, r) \in \pi$ as well as $r \Vdash \phi(\sigma, \tau, \pi)$. Because $\alpha^*$ is the supremum of the least ordinals $\alpha_{\sigma,p}$ where $p$ forces $\phi(\sigma, \tau', \pi)$, there must be some $\tau'$ such that $\text{nr}(\tau') \le \alpha^*$ as well as $r \Vdash \phi(\sigma, \tau', \pi)$. However, $\phi$ maps unique $\sigma$ to unique $\tau$, therefore $\tau = \tau'$ and consequently $\text{nr}(\tau) \le \alpha^*$. $\square$

**Axiom of Choice.** (Sketch.) AC is equivalent to the well-ordering principle. Let $\tau^G \in M[G]$. We need to show that $\tau^G$ can be well ordered. Starting with $\tau \subset N_\alpha \times \mathbb{P}$ for $\alpha = \text{nr}(\tau)$, we can well order $\tau$ using AC in M: $\tau = \{(\sigma_\beta, p_\beta) \mid 0 \le \beta < \gamma\}$, whereby the function $f: \tau \to \gamma$ mapping $(\sigma_\beta, p_\beta)$ to $\beta$ is in M. We can lift it up to $M[G]$ with its canonical $\mathbb{P}$-name $\tilde{f}$ and it will easily do the job of well ordering $\tau^G$. Details are left to the reader. $\square$

Therefore, given any axiom $\phi$ of ZFC and any separative poset $\mathbb{P}$ defined in $M$, it is a ZFC$+$ theorem that a generic $G$ over $\mathbb{P}$ exists and $M[G]$ is a set containing all elements of $M$ as well as $G$ and moreover $\phi^{M[G]}$. Given any finite list of axioms $\phi_1, \dots, \phi_k$, $(\phi_1 \land \dots \land \phi_k)^{M[G]}$ can be derived in ZFC$^+$. Those axioms can then be used as explained in the next section to prove in ZFC$^+$ that $\neg\text{CH}^{M[G]}$. Then, proof of CH in ZFC would lead to contradiction by relativizing the proof within $M[G]$ (adding more axioms to the $\phi_i$'s as needed). We just finished outlining the proof of the following theorem, which is a numerical statement about syntactic properties of ZFC$^+$ and can be formalized in Peano Arithmetic:

**Metatheorem.** If $\mathbb{P}$ is a separative poset in $M$ and $\phi$ is any theorem of ZFC, then there is a ZFC$^+$ proof of $\text{"}(G$ is a generic ideal over $\mathbb{P}$)$\text{ and (the set of }\mathbb{P}\text{-names }M^\mathbb{P} \in M\text{)}\text{ and (the set }M[G]\text{ of }G\text{-evaluations of elements of }M^\mathbb{P}\text{ is countable and transitive)}\text{ and }M \subsetneq M[G]\text{ and }G \in M[G]\text{ and }\phi^{M[G]}\text{"}$.

## 4.6 Forcing $\neg\text{CH}$ to hold in $M[G]$

To recap, $M[G]$ contains $G$ together with $\aleph_2^M$-many new subsets of $\aleph_0$ and simultaneously the proof of $CH^{M[G]}$ goes through. Are we done?

Not yet: we know that $\bigcup G : \aleph_2^M \times \aleph_0 \to \{0, 1\}$ is in $M[G]$ and consequently $|(2^{\aleph_0})^{M[G]}| \ge \aleph_2^M$, but we haven't shown that $\aleph_2^{M[G]} = \aleph_2^M$. Perhaps among its many new sets $M[G]$ also contains a function $\aleph_1^{M[G]} \to \aleph_2^M$. Then, no contradiction with $CH^{M[G]}$ is shown.
**Cardinal collapse.** We need to show that introducing $G$ through forcing does not create new functions from a smaller infinite cardinal to a larger infinite cardinal. Such a function would collapse the larger cardinal to be equal to the smaller one. Here is a simple way to achieve that through forcing. At the ground countable model $M$ define the forcing notion $\mathbb{P}'$ to consist of all finite partial bijections between $\aleph_\alpha^M$ and $\aleph_\beta^M$ with $\beta > \alpha$. Then $M[G]$ will contain $\bigcup G$, which is a complete bijection between $\aleph_\alpha^M$ and $\aleph_\beta^M$. Recall that both of these are countable ordinals. $\bigcup G$ would be a complete bijection because the set $D_\gamma := \{p \in \mathbb{P}' | \gamma \text{ is in the range of } p\}$ is a dense subset of $\mathbb{P}'$ for all $\gamma \in \aleph_\beta^M$ and $D_\gamma \in M$, therefore $G$ has to intersect it, i.e., $\bigcup G$ has to include $\gamma$ in its range. Similarly for its domain. Therefore $\aleph_\beta^M$ is of the same cardinality as $\aleph_\alpha^M$ in $M[G]$. One may ask now, what about $\aleph_\beta^{M[G]}$? Well, this is some countable ordinal larger than $\aleph_\beta^M$ that $M[G]$ "thinks" is the $\beta$th infinite cardinal. Could new cardinals be introduced in $M[G]$? No: if a bijection already exists in $M$ between two ordinals, the same function still exists in $M[G]$. New cardinals can never appear, but cardinals of the ground model $M$ can collapse. Therefore, to conclude the proof we need to show that forcing with $\mathbb{P}$ does not collapse cardinals.

The main idea is to examine the new functions $f^G \in M[G]$ created from the “perspective” of $M$ and observe that their range cannot be “too big” because each $\mathbb{P}$-name $\sigma$ of the domain has at most countably $M$ many potential $\mathbb{P}$-name partners $\tau$ in the range. A potential partner $\tau$ in the range is one for which there is any $q_\tau$ forcing $f(\sigma) = \tau$. If each $\sigma$ in the domain has at most countably $M$ many potential partners in the range, then the range is at most $\aleph_0$ times the domain in size from the perspective of $M$, and for infinite domain that implies same cardinality from the perspective of $M$.

More specifically, any function $f^G \in M[G]$ between cardinals $\kappa^M < \lambda^M$ has an associated $\mathbb{P}$-name $f$ and some $p \in G$ that forces “$f$ is a function $\kappa^M \to \lambda^M$”. Then, for every $(\sigma^G, \tau^G) \in f^G$ (i.e., $f(\sigma^G) = \tau^G$) there has to be some $q \in G, q \ge p$ that forces $(\sigma, \tau) \in f$. Fixing $p$ and $\sigma$, the potential distinct partners $\tau$ of $\sigma$ correspond to distinct $q_\tau \ge p$ that are pairwise incompatible. The key property to prove now is that there are at most countably $M$ many pairwise incompatible elements in $\mathbb{P}$. If so, then there are at most countably $M$ many potential partners $\tau$ for each $\sigma$.

**Definition.** An antichain over a poset $\mathbb{P}$ is a set of pairwise incompatible elements. $\mathbb{P}$ satisfies the countable chain condition (ccc) if there is no uncountable antichain over $\mathbb{P}$.

Because $M$ is countable, $\mathbb{P}$ is trivially ccc. The key will be to prove $(\mathbb{P}$ is ccc)$^M$. Once this is done, the following lemma establishes that cardinals in $M$ remain cardinals in $M[G]$. The idea is simple: given a function $f^G$, any assignment $f^G(\sigma^G) = \tau^G$ is forced by some $q \in G$. If $(\mathbb{P}$ is ccc)$^M$, then $M$ “knows” that for any given $\sigma$, there are at most countably many distinct $\tau$ forced by distinct and pairwise incompatible $q_\tau \models f(\sigma) = \tau$. Therefore, the cardinality of the range of $f$ is forced to be at most that of the domain of $f$ times $\aleph_0$, therefore $f$ cannot map an infinite cardinal to a larger infinite cardinal. All of these statements are relativized to $M$.

**Lemma.** Let $\mathbb{P}$ be a poset in $M$ such that $(\mathbb{P}$ is ccc)$^M$, $G$ a generic ideal, and $f^G : \eta \to \theta$ be a function in $M[G]$ between ordinals $\eta < \theta$. Then:

1. There exists a function $g \in M$ from $\eta$ to $2^\theta$ such that $\forall \alpha < \eta f^G(\alpha) \in g(\alpha)$, and moreover $(g(\alpha)$ is countable)$^M$.

2. Infinite ordinals that are cardinals in $M$ remain cardinals in $M[G]$.

**Proof.** (1) Let $f$ be the $\mathbb{P}$-name for $f^G$ and $p \in G$ force “$f : \tilde{\eta} \to \tilde{\theta}$ is a function”. For each $\alpha \in \eta$ let $g(\alpha) = \{\beta \in \theta \mid \exists q \ge p \text{ such that } q \models f(\tilde{\alpha}) = \tilde{\beta}\}$. This is a set in $M$ by Comprehension and FTF. Definability. Moreover, $f^G(\alpha) \in g(\alpha)$. Finally, if for each $\beta$ we pick one element $q_\beta$ that forces $f(\tilde{\alpha}) = \tilde{\beta}$, then $\{q_\beta \mid \beta \in g(\alpha)\}$ is a set of pairwise incompatible elements and therefore at most countable$^M$ by $(\mathbb{P}$ is ccc)$^M$, and is in bijection with $g(\alpha)$ by $\beta \to q_\beta$. Finally, because the latter function is in $M$, we have $(g(\alpha)$ is countable)$^M$.

(2) Let $\theta$ be an infinite ordinal in $M$ such that $(\theta$ is not a cardinal)$^{M[G]}$. Then let $\eta < \theta$ be an infinite ordinal and $f^G : \eta \to \theta$ a surjection in $M[G]$. Let $g : \eta \to 2^\theta$ be a function in $M$ as in part (1). Notice that $\theta = \bigcup_{\alpha \in \eta} g(\alpha)^{63}$ and now we get $(|\theta| = |\bigcup_{\alpha \in \eta} g(\alpha)| \le |\eta \cdot \aleph_0| = |\eta|)^M$. Therefore $\theta$ is not a cardinal in $M$ either. In conclusion, a cardinal in $M$ remains a cardinal in $M[G]$. $\square$

<footnote>$^{63}$Because $f^G$ is a surjection $\eta \to \theta$ and $f^G(\alpha) \in g(\alpha)$ for all $\alpha \in \eta$.</footnote>
The key is now to prove that $(P \text{ is ccc})^M$. Equivalently, that any uncountable$^M$ subset of $\mathbb{P}$ contains some pair of compatible elements $p, q$. Recalling that all $p \in \mathbb{P}$ are finite partial functions, this will boil down to a basic result in infinite combinatorics, the $\Delta$-system lemma.

**$\Delta$-system lemma.** Let $X$ be an uncountable family of finite sets. Then there is an uncountable $Y \subseteq X$ and a finite set $r$, the *root*, such that every pair of sets in $Y$ intersect in precisely $r$:
$$ \forall x, y \in Y \ x \cap y = r. $$

**Proof sketch.** This is a basic result with many excellent expositions of the proof available online. Briefly, let $X_n$ be the sets of size $n$ among sets in $X$. At least one of these, say $X_k$ has to be uncountable otherwise $X$ would be a countable union of countable sets. Let $m$ be the maximum size of a set $r$ such that $r \subseteq x$ for uncountably many $x \in X_k$. $\emptyset \subset x$ for every $x \in X_k$, therefore $0 \le m \le k$. So let $r$ be a set of size $m$ and $Y \subseteq X_k$ the uncountable collection of all $x \supseteq r$. Any element $a$ of any $x \in Y$ is either $\in r$ or in at most countably many sets in $Y$.

Form a graph whose vertices are the sets $x \in Y$ and edges connect any sets $x, y$ whose intersection contains elements not in $r$. Any vertex in the graph has at most a finite number of elements not in $r$, and consequently at most countably many neighbors. So the graph has uncountably many vertices, each of which has at most countably many neighbors. We show that there is an uncountable *independent* set of vertices, where no two vertices are neighbors. Such a set can be constructed with transfinite induction. At stage 0 pick any vertex $x_0$. At stage $\alpha \le \aleph_1$, we have selected a countable family of vertices $\{x_\beta | \beta < \alpha\}$ each of which has at most countably many neighbors outside this collection, so there are still uncountably many vertices from which to pick some $x_\alpha$ that is not a neighbor of any $x_\beta$, $\beta < \alpha$.

Therefore, pick an uncountable independent subset of $Y$. This is an uncountable collection in which any $x \cap y = r$. $\square$

This lemma will prove useful in showing that $(P \text{ is ccc})^M$. Given an uncountable$^M$ collection of finite partial functions in $\mathbb{P}$, we will find an uncountable$^M$ subcollection whose domains pairwise intersect precisely in some finite root domain $r$. Then, because each finite domain can only generate finitely many distinct functions whose range is $\{0, 1\}$, there has to be some functions that are identical within $r$ and whose remaining distinct domains can be patched together. That means they are compatible, showing that there is no uncountable$^M$ incompatible collection of partial functions in $\mathbb{P}$. More formally:

**Lemma.** $(P \text{ is ccc})^M$.

**Proof.** Let $S$ be an uncountable$^M$ collection of $p \in \mathbb{P}$ and let $X = \{\text{Dom}(p) | p \in S\}$. Any specific $\text{Dom}(p)$ is a finite set, which can support at most finitely many distinct functions to $\{0, 1\}$. Therefore $X$ is uncountable$^M$. Now prove the $\Delta$-system lemma relativized to $M$ and find an uncountable$^M$ $Y \subseteq X$ and a finite root $r$ such that $x \cap y = r$ for every $x, y \in Y$. For every $x \in Y$ pick some $p_x \in S$ such that $x = \text{Dom}(p_x)$. The domains of functions $\{p_x | x \in Y\}$ all intersect in precisely $r$. For any two functions to be incompatible, they should differ in how they map $r$ to $\{0, 1\}$. Because there are finitely many ways to map $r$ to $\{0, 1\}$, there are distinct $p_x, p_y$ such that $p_x|_r = p_y|_r$, i.e., which are compatible. $\square$

**Theorem.** $(\neg\text{CH})^{M[G]}$ is a ZFC$+$ theorem.

**Proof.** In ZFC$+$, because $(P \text{ is ccc})^M$ we get $\aleph_2^{M[G]} = \aleph_2^M$. Also, $(\bigcup G : \aleph_2 \times \aleph_0 \to \{0, 1\})$ is a function)$^{M[G]}$. Because for each $(\alpha, n) \in \aleph_2^M \times \aleph_0$ the set of $p \in \mathbb{P}$ containing $(\alpha, n)$ in their domain is a dense set in $M$, $G$ must intersect that set and therefore all of $\aleph_2^M \times \aleph_0$ is in the domain of $\bigcup G$, i.e., $(\bigcup G \text{ is a complete function})^{M[G]}$. Moreover, for any distinct $\alpha, \beta \in \aleph_2^M$ the set of all $p \in \mathbb{P}$ that contain $(\alpha, n), (\beta, n)$ mapping to distinct values for some $n$ is a dense set in $M$ and therefore $G$ must intersect that set. Therefore each “row” of $\bigcup G$ is distinct. By a similar argument each row of $\bigcup G$ is a subset of $\aleph_0$ not in $M$. In conclusion, (there are $\aleph_2$ distinct subsets of $\aleph_0$)$^{M[G]}$ is a ZFC$+$ theorem. $\square$

Therefore, if ZFC is consistent, then so is ZFC + $\neg$CH.

### 4.6.1 What values can $2^{\aleph_0}$ take?

The above proof shows that $|(2^{\aleph_0})^M| \ge \aleph_2^M$ but no exact value for the cardinality of $2^{\aleph_0}$ is given. Forcing $|2^{\aleph_0}| = \aleph_2$ is possible but will not be covered here. However, notice that the value $\aleph_2$ plays no role in the proof. Can the continuum be forced to take other values?
In fact, it can take almost any value! It can be ℵ₃, ℵ₂₀₂₂, ℵ<sub>ℝ</sub>, ℵ<sub>ℝ₂</sub>, ℵ<sub>ℝ₂₀₂₂</sub>, .... There is only one restriction that comes from a well known and useful result, König's theorem:

**König's theorem.** Let $κ_i, λ_i, i \in I$ be two infinite sequences of cardinals with $κ_i \le λ_i$, $\forall i \in I$, where $I$ is an index set of any cardinality. Then $\sum κ_i < \Pi λ_i$.

Note the strict inequality in the above theorem, as well as the fact that the $κ_i$ and $λ_i$ can be finite or infinite. Proof of this theorem requires the Axiom of Choice.

A constraint on the value of the continuum follows. To describe it, the notion of cofinality needs to be introduced.

**Definition: cofinality.** Let $κ$ be any ordinal. A sequence of ordinals $\langle λ_ξ < κ | ξ < β \rangle$ is cofinal to $κ$ if for every $α < κ$ there is a $ξ < β$ such that $λ_ξ > α$. The cofinality of $κ$, denoted $cf(κ)$, is the least $β$ for which such a sequence exists.

**Examples.** The cofinality of the ordinal 2022 is 1: the sequence $\langle 2021 \rangle$ suffices. The cofinality of $ω$ is $ω$: no finite sequence of finite numbers has $ω$ as its limit. A cardinal $κ$ is called **regular** when it has this precise property of $cf(κ) = κ$; otherwise it is called **singular**. The cofinality of $ℝ_0$ is $ω$: the sequence $\langle ℝ_i | i < ℝ_0 \rangle$ is cofinal; therefore $ℝ_0$ is singular.

An important corollary of König's theorem involves the cofinality of an infinite cardinal:

**Corollary.** If $λ ≥ ℵ_0$ then $λ < λ^{cf(λ)}$.

**Proof.** Let $\langle κ_ξ < λ | ξ < cf(λ) \rangle$ be a cofinal sequence to $λ$. Use König's theorem with $λ_i = λ$ for $i \in cf(λ)$ to obtain $\sum_{ξ<cf(κ)} κ_ξ < λ^{cf(λ)}$. Then, observe that $\sum_{ξ<cf(κ)} κ_ξ = λ$ so the result follows. $\square$

It follows easily that the cofinality of continuum cannot be $ω$: notice that $(2^ω)^ω = 2^{2ω} = 2^ω$. Assume now that $cf(2^ω) = ω$. Let $\langle κ_i < 2^ω | i < ω \rangle$ be cofinal to $2^ω$. Then $\sum κ_i < (2^ω)^ω$ by König's theorem. However $\sum κ_i = 2^ω$, contradiction.

Therefore, the cardinality of continuum cannot be $ℝ_ω$ or $ℝ_{ω^2}$ or $ℝ_{ℝ_ℝ}$... because all these have cofinality $ω$. However, it can be bigger or smaller than any of these values.

# 5 Bullet-point outline of the proof

## 5.1 Part I: logical argument

1. Any sentence $φ$ is equivalent to $φ^M$ where $M$ is a countable transitive set. In particular, given any finite list of axioms of ZFC $φ_1, ..., φ_k, ∃M$ CntTrans($M$) ∧ $φ_1^M$ ∧ ... ∧ $φ_k^M$ is a ZFC theorem. This is established by the Reflection Principle and Mostowski collapse lemma.

2. Therefore, for any theorem $φ$ of ZFC, the proof can be copied to a proof $∃M$ CntTrans($M$) ∧ $φ^M$.

3. Extend ZFC to ZFC+ by adding the new constant symbol $M$, the axiom CntTrans($M$), and the axiom $φ^M$ for each axiom $φ$ of ZFC. If ZFC is consistent, so is ZFC+.

4. In ZFC+, $M$ is a symbol that represents a countable transitive set by fiat and $ℝ_0^M < ℝ_1^M < ℝ_2^M < ...$ are countable ordinals that are uncountable cardinals in $M$.

5. Assume there is a ZFC proof of CH for contradiction. The proof is a finite list of axioms $φ_1, ..., φ_k$ and their logical consequences $φ_{k+1}, ..., φ_n = CH$. The same proof will work in ZFC+ by relativizing each $φ_i$ to $M$ to show $CH^M$, that is, $(2^{ℝ_0})^M = ℝ_1^M$.

6. Extend $M$ to a bigger set $M[G]$ with a table $G$ that encodes $ℝ_2^M$-many new subsets of $ℝ_0$, so that (a) for any axiom $φ$ there is a ZFC+ proof that $φ^{M[G]}$; (b) there is a ZFC+ proof that $¬CH^{M[G]}$. This is accomplished in Part II, Forcing.

7. By (6a), the proof of $CH^M$ can be syntactically copied to a proof $CH^{M[G]}$ together with the ZFC+ proofs of $φ^{M[G]}$ for each axiom $φ$ needed. Those axioms include $φ_1, ..., φ_i$ as well as axioms required to complete the forcing construction in (6).
8. By 7 and (6b), ZFC+ is inconsistent, therefore by (4) ZFC is inconsistent. Assuming ZFC is consistent, the conclusion is that there is no proof of CH.

How to extend $M$ to $M[G]$ is Part II, Forcing. This part is done inside ZFC+.

## 5.2 Part II: forcing

Every statement below, unless otherwise indicated, is preceded with "it is derivable in ZFC+ that..." In particular, M is postulated by flat and its existence is impossible to derive in ZFC due to Gödel's incompleteness theorem.

1. Since M is countable, it contains countably many subsets of $\aleph_0^M = \aleph_0$, and $\aleph_2^M$ is countable. Therefore, there exist $\aleph_2^M$-many new subsets of $\aleph_0$ not in M. Represent this collection by an object $G$ such that $\bigcup G : \aleph_2^M \times \aleph_0 \to \{0, 1\}$ is a complete function whose "rows" are functions $\aleph_0 \to \{0, 1\}$ representing pairwise distinct subsets of $\aleph_0$ that are not in M. $G$ will be added to M to produce a new set $M[G]$ that is a model of ZFC+ with $\neg CH^{M[G]}$.

2. First, define the poset $\mathbb{P}$ consisting of all finite partial functions $p : \aleph_2^M \times \aleph_0 \to \{0, 1\}$ ordered by inclusion; each $p$ is called a forcing condition. By Pairing, each $p \in \mathbb{P}$ is in M and $\mathbb{P} \in M$. Each $p, q$ are compatible if $\exists r \ p \le r \wedge q \le r$ and can be patched together to a bigger partial function.

3. Any complete function $\bigcup G : \aleph_2^M \times \aleph_0 \to \{0, 1\}$ is the union of a maximal ideal of $\mathbb{P}$: a maximal downward closed subset of $\mathbb{P}$ in which members are pairwise compatible.

4. For $M[G]$ to be a model and satisfy Comprehension$^{M[G]}$, it has to include the preimages under $G$ of sets $r \subset x \times \mathbb{P}$ for $x, r \in M$. In addition, this operation of taking preimages under $G$ can be performed transfinitely. To do so, define the $\mathbb{P}$-name hierarchy $M^\mathbb{P} \subset M$ (a ZFC+ set that is a proper M-class), by $N_{\alpha+1} := \mathcal{P}(N_\alpha \times \mathbb{P}) \cap M$ for any successor ordinal $\alpha$, and $N_\alpha := \bigcup_{\beta<\alpha} N_\beta$ for limit ordinal $\alpha$.

5. $M^\mathbb{P} = \bigcup N_\alpha$. Define the $G$-evaluation of a $\mathbb{P}$-name $\tau$ to be $\tau^G := \{\sigma^G \mid (\sigma, p) \in \tau \wedge p \in G\}$. If $M[G]$ is a model satisfying Comprehension, it has to include each such $\tau^G$. Now by wishful thinking let $M[G] := \{\tau^G \mid \tau \in M^\mathbb{P}\}$, and aim to prove that $M[G]$ thus minimally defined is indeed a model.

6. Define the canonical name of $x \in M$ to be $\tilde{x} := \{(x, p) \mid p \in \mathbb{P}\}$ and $\Gamma := \{(\tilde{p}, p) \mid p \in \mathbb{P}\}$, show $\tilde{x}^G = x$ and $\Gamma^G = G$, concluding that $M \subsetneq M[G]$ and $G \in M[G]$.

7. For $M[G]$ to be a model, it has to satisfy all ZFC axioms. To prove Comprehension$^{M[G]}$, Comprehension$^M$ needs to be used. Specifically, $z := \{\sigma^G \in \tau^G \mid \phi(\sigma^G)\}$ has to be in $M[G]$ for every $\sigma, \tau \in M^\mathbb{P}$. It suffices to show that $z = v^G$ for some $v \in M^\mathbb{P}$.

8. The natural candidate is $v := \{(\sigma, p) \in \tau \mid \psi_\phi(\sigma, p)\}$ for some formula $\psi_\phi$, whereby $\psi_\phi(\sigma, p)$ is true when $p \in G$ forces $\phi(\sigma^G)$ to be true.

9. It immediately follows that $G$ has to be generic: no new infinite "AND" information can be encoded in $G$. In particular, for every $p$ and every sentence $\phi$, if every $q \ge p$ has an extension $r \ge q$ that makes $\phi$ true in every ideal $G \ni r$, then $p$ is forced to make $\phi$ true: $\phi$ has to be decided in $M[G]$, and no $q \ge p$ can possibly decide $\neg\phi$ is true. Iterating this requirement by structural induction, $p$ forces $\phi$ to be true whenever every $q \ge p$ has $r \ge q$ forcing $\phi$ to be true.

10. Genericity of $G$ is defined by the concept of density. A subset $D \subset \mathbb{P}$ is **dense** if for all $q \in \mathbb{P}$ there is a $r \ge q$ such that $r \in D$. $G$ is *generic* over M if $G \cap D \ne \emptyset$ for any $D \in M$ that is a dense subset of $\mathbb{P}$. It follows that if $G$ is generic, for any $p \in G$ and any sentence $\phi$, if the set $D$ of $q \in \mathbb{P}$ that force $\phi$ to be true is dense above $p$, then $p$ forces $\phi$ to be true.

11. Generic ideals $G$ exist as long as the ground model is countable: enumerate all dense $D_1, D_2, \dots$ in M; at each step $i+1$ pick $p_{i+1} \in D_{i+1}$ such that $p_{i+1} \ge p_i$; let $G = \{p \in \mathbb{P} \mid p \le p_i \text{ for some } i\}$. $G$ is not in M if $\mathbb{P}$ is separative: for any $f \in M$ of the same domain and range as $\bigcup G$, $\{p \in \mathbb{P} \mid p$ disagrees with $f$ in some position\} is dense in M.

12. A condition $p$ **forces** $\phi(\tau_1, \dots, \tau_k)$ if $p \in G$ implies $\phi^{M[G]}(\tau_1^G, \dots, \tau_k^G)$. The fundamental theorem of forcing (FTF) states that for any $\phi$, the collection of $(p, \tau_1, \dots, \tau_k)$ where $p \Vdash \phi$ and the $\tau_i$s have max name rank $\alpha$ is a set $F_\alpha^\phi \in M$ (Definability), and $\phi^{M[G]}$ iff there is some $p \in G$ that forces $\phi$
(Truth). Moreover, $p \Vdash \phi$ and $q \ge p$ implies $q \Vdash \phi$. FTF is a metatheorem: for any $\phi$ there is a separate ZFC$+$ proof of FTF for $\phi$.

13. Using the FTF, the metatheorem that $M[G]$ is a model is proven. For any axiom $\phi$ of ZFC, there is a ZFC$+$ proof of $\phi^{M[G]}$. Therefore $CH^{M[G]}$ goes through.

14. It remains to show that forcing does not collapse cardinals. This is achieved by showing that for each function $f$ in $M[G]$ each element $x$ in $f$'s domain is forced to map to at most countably many distinct $y$ in $f$'s range by elements $q_y$ in $\mathbb{P}$. A combinatorial argument finishes the proof.

# 6 Coming back to omitted steps

## 6.1 Reflection Principle

Any sentence can be reflected in a countable transitive set $M$. This is a metatheorem, or better, an algorithm that takes any $\phi$ as input and returns a ZFC proof of $\exists M \operatorname{CntTrans}(M) \wedge \phi \leftrightarrow \phi^M$. We sketch the algorithm here following the excellent exposition in Weaver (2014).

The first step is a lemma schema that takes as input a finite list of formulas $\phi_1, \dots, \phi_n$ and produces a ZFC proof of existence of a countable set $M$ that contains all the elements needed so that the formulas have “similar” behavior with their variables restricted to $M$ as when their variables are unconstrained. More precisely, for any $\phi(x_1, \dots, x_k)$ in the list with all free variables displayed and for any $a_1, \dots, a_k \in M$, if there exists $x_j$ such that $\phi(a_1, \dots, a_{j-1}, x_j, a_{j+1}, \dots, a_k)$, then there exists $x_j \in M$ such that $\phi(a_1, \dots, a_{j-1}, x_j, a_{j+1}, \dots, a_k)$.

**Lemma Schema.** Let $\phi_1, \dots, \phi_n$ be any formulas with free variables among $x_1, \dots, x_k$. There is a ZFC proof that there exists a countable set $M$ such that for any formula $\phi_i$ in the list and for any variable $x_j$ and any constants $a_1, \dots, a_k \in M$, whenever $\exists x_j \phi_i(a_1, \dots, x_j, \dots, a_k)$ holds, $\exists x_j \in M \phi_i(a_1, \dots, x_j, \dots, a_k)$ also holds.<fnref n="64" /> That is to say, whenever there exists $x_j$ that makes formula $\phi_i$ true with other variables assigned to specific elements of $M$, keeping the same assignments for other variables there exists $x_j \in M$ making $\phi_i$ true.

**Proof Sketch.** We create a countable sequence of sets $M_0 = \omega \subset M_1 \subset M_2 \subset \dots$ whereby $M := \cup_l M_l$. At each step $l+1$, for every formula $\phi_i$ and variable $x_j$, and for all possible assignments of the other variables to elements from $M_l$, if there exists $x_j$ making the formula true, choose one such element and add it to $M_{l+1}$.<fnref n="65" />

In other words, for each $i, j$ and for each $a_1, \dots, a_k \in M_l$ if $\exists x_j \phi_i(a_1, \dots, x_j, \dots, a_k)$ then choose one element $a$ such that $\phi_i(a_1, \dots, x_j = a, \dots, a_k)$ and add $a$ to $M_{l+1}$.

The set $M_{l+1}$ is countable by induction: assuming $M_l$ is countable, there are countably many tuples $a_1, \dots, a_k$ times finitely many formulas times finitely many variables, and each such combination generates at most one additional element to be added to $M_{l+1}$. Therefore $M$ is countable.

Now assume that for some $a_1, \dots, a_j - 1, a_j + 1, \dots, a_k \in M$, $\exists x_j \phi_i(a_1, \dots, x_j, \dots, a_k)$. In other words for some $a$ not necessarily in $M$, $\phi_i(a_1, \dots, a_j - 1, a, a_j + 1, \dots, a_k)$. Well, each one of $a_1, \dots, a_k$ is in some $M_l$, so choose $l^*$ such that $a_1, \dots, a_k \in M_{l^*}$. But then some $a^*$ should have been added to $M_{l^*+1}$ such that $\phi_i(a_1, \dots, a^*, \dots, a_k)$. Therefore some such $a^*$ is in $M$. Therefore $\exists x_j \in M \phi_i(a_1, \dots, x_j, \dots, a_k)$.

Given any input formulas $\phi_1, \dots, \phi_n$, the above proof can be written down mechanically in full ZFC formality. $\square$

**Theorem: The Reflection Principle Algorithm.** Given any input sentence $\phi$ there is a ZFC proof that there is a countable transitive set $M$ such that $\phi \leftrightarrow \phi^M$.

**Proof.** Given $\phi$, the strategy to generate $M$ will be to break down $\phi$ into the list $\phi_1, \dots, \phi_n$ of all its component formulas, apply the above lemma to the list, and then show by structural induction on $\phi$ that all subformulas as well as $\phi$ are reflected in the resulting set $M$. For this to work technically, the list

<footnote n="64">⁶⁴Here, $a_j$ is a dummy constant but it is not called out as such in order to reduce notational clutter.</footnote>

<footnote n="65">⁶⁵This requires AC, or at least that $\{x_j | \phi(...x_j... )\}$ can be well-ordered. With AC, we can choose an element from the set of $x_j$ of lowest rank making $\phi_i$ true, and details are omitted.</footnote>
of formulas needs to be constructed carefully. First, the axiom of extensionality will be needed (for all $x$ and $y$, $x = y \leftrightarrow (\forall z z \in x \leftrightarrow z \in y)$). So let's assume our starting list is $\phi_1 = \phi, \phi_2 = \text{Extensionality}$. Now we break down to components to expand the list: if $\phi_i = \neg\psi$ is in the list, add $\psi$ to the list; if $\phi_i = \psi \wedge \chi$ or $\phi_i = \psi \rightarrow \chi$ is in the list, add both $\psi$ and $\chi$ to the list: if $\phi_i = \forall x \psi$, add $\psi$ to the list. Continue until no more formulas can be added. Now we need another trick: if the resulting list so far is $\phi_1, \dots, \phi_n$, apply the above lemma to the negations: $\neg\phi_1, \dots, \neg\phi_n$.

Now let $M$ be the countable set resulting from the lemma. Using structural induction starting from atomic $\phi_i$ (i.e., of the form $x = y$ or $x \in y$) and moving up in formula complexity, we will show for each $\phi_i$ that for all $a_i, \dots, a_k \in M$, $\phi_i(a_1, \dots, a_k) \leftrightarrow \phi_i^M(a_1, \dots, a_k)$.

Let $\phi_i$ be atomic: well, $(x_1 = x_2)^M$ is $x_1 = x_2$ and $(x_1 \in x_2)^M$ is $x_1 \in x_2$ so this case is trivial.

Let $\phi_i = \neg\psi$. By inductive assumption we have $\psi \leftrightarrow \psi^M$ for all $a_j \in M$, and therefore $\neg\psi \leftrightarrow \neg\psi^M$ therefore $\neg\psi \leftrightarrow (\neg\psi)^M$.

Similarly the cases of $\psi \rightarrow \chi$ or $\psi \wedge \chi$ are trivial.

The only case that is interesting is when $\psi(x, x_1, \dots, x_k) \leftrightarrow \psi^M(x, x_1, \dots, x_k)$ has been proven for a subformula that has the unbound variable $x$ (with other unbound variables $x_j$ displayed), and we need to show that for all $a_1, \dots, a_k \in M$:

$$ \forall x \psi(x, a_1, \dots, a_k) \leftrightarrow \forall x \in M \psi^M(x, a_1, \dots, a_k) $$

Fix the $a_j$s in $M$. If $\forall x \psi(x, a_1, \dots, a_k)$, then certainly $\forall x \in M \psi(x, a_1, \dots, a_k)$. By inductive assumption, $\psi(\dots) \leftrightarrow \psi^M(\dots)$ therefore $\forall x \in M \psi^M(x, a_1, \dots, a_k)$. Conversely, if $\forall x \psi(x, a_1, \dots, a_k)$ is false, then $\exists x \neg\psi(x, a_1, \dots, a_k)$ and by the lemma some $x \in M$ should have been included (here is why we needed to apply the lemma with all negations of the $\phi_i$s) and therefore $\exists x \in M \neg\psi(x, a_1, \dots, a_k)$, and by inductive assumption $\exists x \in M \neg\psi^M(x, a_1, \dots, a_k)$ or equivalently $\forall x \in M \psi^M(x, a_1, \dots, a_k)$ is false.

Since the sentence $\phi$ is in the list, $\phi \leftrightarrow \phi^M$ is established. Moreover, $M$ is countable and extensional: $\forall x, y \in M x = y \leftrightarrow (\forall z \in M z \in x \leftrightarrow z \in y)$. Now the Mostowski collapse theorem, discussed below, can be applied to $M$ to obtain countable transitive set $M'$ where $\phi \leftrightarrow \phi^{M'}$.

Given any input $\phi$, the above proof can be written down mechanically in full ZFC formality. $\square$

It remains to establish that an extensional set $M$ can be converted to a transitive set $N$ so that for every formula $\phi, \phi^M \leftrightarrow \phi^N$. The map $f$ from $M$ to $N$ will be an $\in$-isomorphism: a bijection such that $x \in y$ iff $f(x) \in f(y)$ for all $x, y \in M$.

**Theorem.** Let $f: M \rightarrow N$ be an $\in$-isomorphism. Let $\phi(x_1, \dots, x_k)$ be any formula with free variables displayed. Then for any $a_1, \dots, a_k \in M$, the formula

$$ \phi^M(a_1, \dots, a_k) \leftrightarrow \phi^N(f(a_1), \dots, f(a_k)) $$

is provable in ZFC.

**Proof sketch.** This is an easy induction on the structure of $\phi$. The base cases $x = y$ and $x \in y$ as well as the cases of $\rightarrow$ and $\wedge$ are trivial. Therefore, assume we have a ZFC proof that $\phi^M(a_1, \dots, a_k) \leftrightarrow \phi^N(f(a_1), \dots, f(a_k))$ for all $a_1, \dots, a_k \in M$.

We need to show $\forall x_j \phi^M(a_1, \dots, x_j, \dots, a_k) \leftrightarrow \forall x_j \phi^N(f(a_1), \dots, x_j, \dots, f(a_k))$. Because $f$ is a bijection, if $\forall x_j \phi^M(\dots)$ fails, the $a_j$ that makes it fail is mapped 1-1 to an $f(a_j)$ that makes $\phi^N(\dots f(a_j)\dots)$ fail by inductive assumption, and therefore $\forall x_j \phi^N(\dots)$ fails. Similarly if $\forall x_j \phi^M(\dots)$ holds, then for any $a_j \phi^M(\dots a_j \dots)$ holds, by inductive assumption $\phi^N(\dots f(a_j)\dots)$ holds and therefore by $f$ being surjective $\forall x_j \phi^N(\dots)$ holds. $\square$

Finally, given a set $M$ where extensionality holds, we can find an $\in$-isomorphism $f$ to a transitive set $N$. The following theorem is provable in a much more general form, where the relationship does not need to be $\in$ as long as it is well founded. However, we won't need the general form.

**Theorem: Mostowski Collapse Lemma.** Let $M$ be a set where Extensionality$^M$ holds. $M$ is $\in$-isomorphic to a transitive set $N$.
**Proof.** Define $f: M \to V$ by $f(x) = \{f(y) | y \in x\}$. This definition is by transfinite recursion and it makes sense because of Foundation.<fnref n="66" />

The set $N := \{y | \exists x \in M y = f(x)\}$ exists by Replacement. We need to show that $f$ is an $\in$-isomorphism and that $N$ is transitive.

To show $f$ is an $\in$-isomorphism, let $y \in x \in M$. Then $f(x) = \{f(y) | y \in x\}$ so by definition it follows that $f(y) \in f(x)$.

To show that $N$ is transitive, let $w \in z \in N$, to show $w \in N$. Well, $z = f(x)$ for some $x \in M$, therefore $z = \{f(y) | y \in x\}$ therefore $w = f(y)$ for some $y \in x$ therefore $w$ is in the image of $f$, i.e., $w \in N$. □

Therefore, given any sentence $\phi$, a ZFC proof of $\exists M CntTrans(M) \wedge \phi \leftrightarrow \phi^M$ can be written down mechanically. This is a ridiculously powerful result. Given any finite number of axioms of ZFC, they hold in a countable transitive set $M$. Equivalently, given any ZFC proof of some $\phi$, we can convert it to a ZFC proof of $\exists M CntTrans(M) \wedge \phi^M$.

**Example of Mostowski Collapse.** The reader may wonder what, if anything, is the effect of the Mostowski Collapse procedure in a set. Let's give a simple example of $M$ and calculate the Mostowski collapse:

$$M := \{1, 2, 2022, \omega, \omega \cdot 2022, \{2022, \omega \cdot 2022\}\}$$

By definition $f(1) = \{f(y) | y \in x\}$. Now, $\emptyset \in 1$, however because $\emptyset \notin M$ it is not in the domain of $f$ therefore $f(1)$ becomes $\emptyset$ or if you prefer, $0$. Similarly, $f(2) = 1$ and $f(2022) = 2$.

The reader can easily verify that $f(\omega) = \{0, 1, 2\} = 3$ and similarly $f(\omega \cdot 2022) = 4$. Finally, $f(\{2022, \omega \cdot 2022\}) = \{2, 4\}$.

Therefore the resulting $N$ is $\{0, 1, 2, 3, 4, \{2, 4\}\}$, a transitive set $\in$-isomorphic to $M$ and quite a simplification from $M$.

**Existence of $M$ versus existence of a proof of existence of $M$.** Given $\phi$, the Reflection Principle algorithm outputs a ZFC proof of $\exists M CntTrans(M) \wedge \phi \leftrightarrow \phi^M$. It always terminates and can be implemented in software. The algorithm goes through all subformulas of $\phi$ and writes down the respective ZFC subproofs as sketched above. The resulting ZFC proof states the existence of an countable set $M$ with the requisite properties. This latter proof, by contrast, is profoundly non-constructive. It involves infinitely many iterative applications of the axiom of choice. There is no ontological conclusion that can be drawn about the existence of any such $M$. $M$ is neither shown to truly exist (whatever that would mean), neither is anything of this sort needed for the proof of the independence of CH. All we need is that we can syntactically and mechanically convert any ZFC theorem $\phi$ to a ZFC theorem $\exists M CntTrans(M) \wedge \phi^M$.

## 6.2 Forcing consistency of CH

Consistency of CH is the second part of independence of CH. Godel (1939) originally showed it by building an inner model of ZFC — a formula $L(x)$ that describes a proper class of sets such that all axioms $\phi$ hold as $\phi^L$ and such that $CH^L$ holds. $L(x)$ is the class of constructible sets, a topic beyond scope here. The reader is referred to Smullyan and Fitting (2010).

The forcing machinery can be deployed to provide a much simpler proof of this part. The argument below follows Weaver (2014).

Starting from the same ground model $M$ in ZFC+, the goal now is to extend through forcing to a model $M[G]$ that contains a bijection between $(2^{\aleph_0})^{M[G]}$ and $\aleph_1^{M[G]}$ so that $(|2^{\aleph_0}| = \aleph_1)^{M[G]}$. Can the reader think of what object $G$ we may want to add to $M$?

The first thing that comes to mind is a bijection between the countable sets $(2^{\aleph_0})^M$ and $\aleph_1^M$. Let's define a separative poset $\mathbb{P}$ that contains partial constructions $p$ of this bijection such that $p \in M$ for all

<footnote n="66">⁶⁶More generally, instead of '∈' a relation $R$ that is well founded, i.e., where any set of elements has an $R$-minimal element, suffices because well-foundedness enables transfinite recursion on $R$.</footnote>
$p \in \mathbb{P}$. Analogously to proving consistency of $\neg CH$, we may be inclined to let $p$ be finite partial bijections $(2^{\aleph_0})^M \to \aleph_1^M$. So let's follow this through and see what properties $\bigcup G$ and $M[G]$ have:

* For any subset $s \subset \aleph_0$ in $M$, the set $D_s := \{p \in \mathbb{P} \mid s \in \text{Dom}(p)\}$ is a dense set in $M$, therefore $s \in \text{Dom}(\bigcup G)$.

* For any element $\alpha \in \aleph_1^M$, the set $D'_\alpha := \{p \in \mathbb{P} \mid \alpha \in \text{Ran}(p)\}$ is a dense set in $M$, therefore $\alpha \in \text{Ran}(\bigcup G)$.

* In conclusion, $\bigcup G$ is a complete bijection and $((2^{\aleph_0})^M = \aleph_1^M)^{M[G]}$.

Are we done? Not quite: the set $(2^{\aleph_0})^{M[G]}$ could be bigger than $(2^{\aleph_0})^M$ if forcing with $\mathbb{P}$ introduced new subsets of $\aleph_0$. Moreover, $\aleph_1^M$ could be countable in $M[G]$ if forcing with $\mathbb{P}$ introduces a new surjection $\aleph_0 \to \aleph_1^M$. And indeed, just like in the case of forcing for consistency of $\neg CH$, forcing with finite partial functions $(2^{\aleph_0})^M \to \aleph_1^M$ introduces new functions with domain $\aleph_0$ and in particular new subsets of $\aleph_0$.

Here is some intuition why. Take any injective function $f \in M$, $f : \omega \to (2^{\aleph_0})^M$. There are many such functions that are always $\in M$. For example, $f(n) = n$ is such a function because $n \in (2^{\aleph_0})^M$ when $n$ is viewed as an ordinal. Also, $f_k(n) := kn$ is such a function, as is really any injective function $f : \omega \to 2^{\aleph_0}$ definable by a formula $\phi(x, y)$ that is true iff $x \in \omega$ and $y \in f(x)$.

Now take $\bigcup G \circ f$ to produce a function $s_f := \omega \to \aleph_1^M$. Can the reader see how new subsets of $\omega$ are formed?

Let's cut the range into two infinite parts within $M$, and call one of them 0 and the other 1. For example (as in Weaver (2014)) we could say that $\alpha < \omega$ corresponds to 0 and $\omega \le \alpha < \aleph_1^M$ corresponds to 1. Any partition in two infinite pieces works as long as they are in $M$.

This automatically converts $s_f$ to a function $\omega \to \{0, 1\}$, or equivalently a subset of $\omega$. Could this subset possibly be in $M$ already? Not if $G$ is generic: given any $p \in \mathbb{P}$, and given any $M$-set $t \subset \omega$, it is necessarily possible to extend $p$ to $q \ge p$ that makes $s_f$ and $t$ differ in some position. That is where we need both pieces of the partition to be infinite so that $p$ exhausts neither of them.

The problem of generating new $\omega$-subsets arises because any $p \in \mathbb{P}$ only forces a finite part of $s_f$. How can we change $\mathbb{P}$ in a way to ensure that $s_f \in M$ whenever $f \in M$? The solution of how to do this emerges by observing that given $\bigcup G : (2^{\aleph_0})^M \to \aleph_1^M$, any $f : \omega \to (2^{\aleph_0})^M$ in $M$ induces a countable partial bijection $p : (2^{\aleph_0})^M|_{\text{Ran}(f)} \to \aleph_1^M$, whose domain is $\text{Ran}(f)$ and such that $p \subset \bigcup G$.

So, if we ensure that $\mathbb{P}$ contains all countable partial bijections $p : (2^{\aleph_0})^M \to \aleph_1^M$ such that $p \in M$, the hope is that $G$ won't introduce any new ones, and therefore it won't introduce any new subsets $s$ of $\omega$. Here is the rationale: consider a chain of elements of $\mathbb{P}$, $p_1 < p_2 < p_3 < \dots$. Given $f : \omega \to (2^{\aleph_0})^M$ in $M$, and given a partition of $\aleph_1^M$ in two parts mapping to 0 and 1, each $p_i$ partially defines a subset $s_f$ of $\omega$ by $p_i \circ f : \omega \to \{0, 1\}$. In $M[G]$, the complete subset $s_f$ defined by $\bigcup G|_{\text{Ran}(f)} \circ f$ is present. However, notice that $\bigcup G|_{\text{Ran}(f)} = \bigcup_i p_i \circ f$, where $p := \bigcup_i p_i$ is just a countable partial bijection $(2^{\aleph_0})^M \to \aleph_1^M$. Because each $p_i \in M$ we have $p \in M$ and consequently $p \in \mathbb{P}$. Therefore $s_f$ is already a member of $M$!

We can formalize this with a lemma:

**Definition.** A poset $\mathbb{P}$ is $\omega$-closed if for any infinite chain $p_1 < p_2 < p_3 < \dots$ of elements $p_i \in \mathbb{P}$, $\bigcup_i p_i \in \mathbb{P}$.

**Lemma.** Let $\mathbb{P}$ be $(\omega$-closed)$^M$ and let $X \in M$. Any function $f : \omega \to X$ in $M[G]$ is already in $M$.

**Proof.** Let $S \in M$ be the set of all functions $\omega \to X$ in $M$. Let $\tau$ be a $\mathbb{P}$-name for $f$ and let $p$ force "$\tau$ is a function $\tilde{\omega} \to \tilde{X}$". If we show that the set of $r$ that force $\tau \in \tilde{S}$ is dense above $p$, i.e., that at $p$ it is necessarily possible that $f$ is in $S$, then $(f \in S)^{M[G]}$ and therefore $f \in S$.

Because $p$ forces "$\tau$ is a function $\tilde{\omega} \to \tilde{X}$", it forces "$\forall x \in \tilde{\omega} \exists y \in \tilde{X}$ such that $(x, y)$ is a pair of values in $\tau$". Therefore for any $n \in \omega$ and $\sigma = \tilde{n}$, the set of $q$ that force "$\exists y \in \tilde{X}$ such that $(\sigma, y)$ is a pair of values in $\tau$" is dense above $p$. Therefore the set of $r$ that force "$\langle \tilde{n}, \tilde{c}_n \rangle$ is a pair of values in $\tau$" for some $c_n \in X$ is dense above $q$.

Now starting with any $q \ge p$ let $p_0 \ge q$ force "$\tau$ maps $\tilde{0}$ to $\tilde{c}_0$" for some $\tilde{c}_0$, let $p_1 \ge p_0$ force "$\tau$ maps $\tilde{1}$ to $\tilde{c}_1$" for some $\tilde{c}_1$, and so on defining a sequence $p_0 \le p_1 \le p_2 \le \dots$. Since $(\mathbb{P}$ is $\omega$-closed)$^M$,
$r := \bigcup p_i \in \mathbb{P}$ and $r$ forces “$(\tilde{n}, \tilde{c}_n)$ is a pair of values in $\tau$” for every $n \in \omega$ therefore $r \Vdash \tau \in \tilde{S}$. Because $q$ was arbitrary, the set of $r$ that force $\tau \in \tilde{S}$ is indeed dense above $p$ and therefore $f \in S$. $\square$

**Theorem.** $CH^{M[G]}$ is a ZFC+ theorem.

**Proof.** By the lemma above, in $M[G]$ there are no new functions $f : \omega \to X$ for any $X \in M$. Therefore there are no new subsets of $\aleph_0$ and moreover there is no surjection $\aleph_0 \to \aleph_1^M$, which means that $(2_{\aleph_0}^M)^{M[G]} = (2_{\aleph_0}^M)^M$ and $\aleph_1^{M[G]} = \aleph_1^M$.

In conclusion, $\bigcup G$ is a bijection from $(2_{\aleph_0}^M)^{M[G]}$ to $\aleph_1^{M[G]}$. $\square$

# 7 Alternative forcing proofs of consistency of $\neg CHw$

There are many alternative proofs of consistency of $\neg CH$. However, all of them are based on forcing. They share the underlying structure of the poset $\mathbb{P}$ that encodes a collection of new subsets of $\aleph_2$ and some nonstandard version of logic that implements the nonstandard logic of forcing—the “necessarily possible”, or $\forall q \ge p \exists r \ge q$ such that “...”.

## 7.1 Original syntactic approach by Cohen

The very original construction by Cohen described in two parts in his 1963 and 1964 PNAS papers was perhaps the cleanest after all. In the first part (1963) he defined forcing. Here is part of the definition, with notation changed to match this manuscript:

"**Definition 6.** By induction, we define the concept of “$p$ forces $\phi$” as follows: I. If $r > 0$, $p$ forces $\forall_\alpha x \phi(x)$ if for all $q \ge p$ $q$ does not force $\neg\phi(\mathcal{F}_\beta)$ for $\beta < \alpha$. $p$ forces $\exists_\alpha x \phi(x)$ if for some $\beta < \alpha$, $p$ forces $\phi(\mathcal{F}_\beta)$. [...] The most important part of Definition 6 is I, the other parts are merely obvious derivatives of it."

Here, $\forall_\alpha$ and $\exists_\alpha$ are quantifiers limited to rank < $\alpha$. Describing Cohen’s construction of sets $\mathcal{F}$ would be too big a digression.

In Cohen’s original construction, the poset $\mathbb{P}$ is defined in the ground model and then truth is assigned in $M[G]$ as follows. The collection of all sentences of ZFC, allowing for parameters ranging across $M[G]$, is countable. Enumerate all sentences:

$$ \psi_1, \psi_2, \psi_3, \dots, \psi_n, \dots $$

Now it is shown that if $p \Vdash \neg\phi$ then there is $q \ge p$ such that $q \Vdash \neg\phi$. Then, starting with $p_0 = \{\}$, at each step $i$ extend to $p_{i+1} \ge p_i$ by considering $\psi_i$. If $p_i \Vdash \phi$, then let $p_{i+1} = p_i$. Otherwise there is $q \ge p_i$ such that $q \Vdash \neg\psi_i$. Let $p_{i+1} := q$. At the end of the construction we have a *complete sequence* $p_0 \le p_1 \le p_2 \le \dots$ that decides the truth of every sentence $\psi_i$ in the resulting model.

There are many tricky technical details for all this to work. The ordering of the sentences as well as ranks of each element of the model are quite involved. Notice also that the ground model has to be countable from the perspective of the “outside” universe $V$: all possible sentences are enumerated in $V$.

Despite the technical complexities, the main idea is perhaps the simplest: Cohen wanted to add $\aleph_2$-many subsets of $\aleph_0$ and obtain a model $M[G]$. To achieve that, he picked an order that works for all sentences $\psi_i$ of $M[G]$ and constructed a $G$ that decides them one-by-one by brute force. The later installations of forcing achieve the same task in a more elegant and less syntactical way by using dense sets, but obscure the honest clarity of the goal.

## 7.2 Standard versus nonstandard models

It is worthwhile to comment on the role of countable models versus nonstandard models. A standard model $M$ of ZFC is one where $\in_M := \in_V \cap (M \times M)$ is true set membership and sentences follow the standard binary logic. To produce the standard model $M[G]$ where CH breaks, a generic $G$ is needed, which in turn necessitates that $M$ is countable. However, the consistency of $\neg CH$ does not strictly require that a standard model is produced. The forcing relation itself provides a notion of $\in$ in the ground
model $M^\mathbb{P}$ with nonstandard rules—$\sigma$ “in” $\tau$ with respect to $p$ if $\sigma^G \in \tau^G$ in $M[G]$ when $p \in G$—and nonstandard logic that defines different truth values of every sentence localized to every $p \in \mathbb{P}$. It can be shown that all axioms are true in this nonstandard assignment of truth—they are true at $p = \{\}$—and therefore all theorems go through as before. Then, it can be shown that $\neg\text{CH}$ holds at $p = \{\}$ in this nonstandard logic.<fnref n="67" /> In this way, the ground model can be any $M$ or even $V$ itself. With that idea in mind, there are proofs of consistency of $\neg\text{CH}$ that are based on nonstandard models such as Boolean valued models and S4 modal logical models. In these models, existence of a generic $G$ is not needed. Instead, the properties that such a generic $G$ would have, had it existed, are just the guiding principle in assigning truth values to sentences for every $p$ in the forcing notion $\mathbb{P}$. The forcing notion itself plays pretty much the same role in all these approaches.

## 7.3 Boolean valued models

A Boolean algebra is an algebra on the subsets of a set $S$, with operations $a \wedge b := a \cap b$, $a \vee b := a \cup b$, $\neg a := S \setminus a$, where $1 := S$ and $0 := \{\}$. The partial order $\leq$ is defined by $\subseteq$. Boolean algebras generalize truth values: the algebra of $\{0,1\}$ is the simplest Boolean algebra, generated by $S = \{\emptyset\}$.

The poset $\mathbb{P}$ can give rise to a Boolean algebra, and then a Boolean valued model can be defined. The reader is referred to Bell (1977) or Jech (2002) for the full details, as well as to the highly readable exposition by Timothy Chow (2007). Risking confusion of the reader, below is a very brief introduction to Boolean algebras and a few words on how this is done.

An ideal of a Boolean algebra $B$ is a set $I$ such that (1) $0 \in I$; $1 \notin I$; (2) for every $a, b \in I$, $a \vee b \in I$; (2) if $a \in I$ and $b \leq a$ then $b \in I$. A maximal ideal of a Boolean algebra, also called a prime ideal, is one where for every $a \in B$ either $a \in I$ or $\neg a \in I$.

Often Boolean algebras are defined in terms of the algebraic operations $\vee$ (sometimes denoted by $+$), $\wedge$ (sometimes denoted by $\cdot$) and $\setminus$ (sometimes denoted by $-$), and then **Stone's representation theorem** applies: every Boolean algebra is isomorphic to an algebra of sets.

The natural partial order on a Boolean algebra allows definition of the supremum and infimum of a set of elements: $\sup\{a_1, \dots, a_k\} = a_1 \vee \dots \vee a_k$, which in an algebra of sets is just $\bigcup_1^k a_i$, and similarly $\inf\{a_1, \dots, a_k\} = a_1 \wedge \dots \wedge a_k = \bigcap_1^k a_i$. These can generalize to infinite joins and meets: $\bigwedge_{a \in X} a := \inf X$ and $\bigvee_{a \in X} a := \sup X$. However, these may not be elements of $B$. If $\bigwedge, \bigvee$ exist for all $X \subset B$ then $B$ is called **complete**. If they exist for any $X$ of cardinality $< \kappa$ where $\kappa$ is a regular cardinal, then $B$ is **$\kappa$-complete**.

*Every Boolean algebra can be densely embedded in a complete Boolean algebra, its completion.* This is accomplished in a similar way that $\mathbb{R}$ completes $\mathbb{Q}$ by the method of Dedekind cuts. The elements $\bigwedge_X, \bigvee_X$ for any $X \subset B$ are defined similarly to the way $\pi$ is defined as the limit of all $a \in \mathbb{Q}$ $a < \pi$.

A separative poset $\mathbb{P}$ can be embedded in a complete Boolean algebra $B$: $\mathbb{P} \subseteq B$, $\leq_P$ agrees with $\leq_B$, and $\mathbb{P}$ is dense in $B$. Moreover, $B$ is unique up to isomorphism. If $\mathbb{P}$ is a forcing notion defined on a ground model $M$, the resulting Boolean algebra can be used to define a Boolean-valued model for ZFC.

Each set $x \in M$ is converted to a Boolean-valued set in a structure $M^B$ that is analogous to the M-class of $\mathbb{P}$-names. In particular, instead of $x$ containing members $y \in x$, it becomes a collection of values $x(y) \in B$: each membership of a set $y$ in $x$ is associated with a value in $B$. Then, sentences from the ground ones $x = y$ and $x \in y$ up to composites such as $\forall x \phi$ are given Boolean values in $B$ instead of true/false.

Once these definitions are made, $M^B$ is proven to be a model of ZFC in the sense that each axiom evaluates to a Boolean value of 1 and consequently each theorem evaluates to 1 as well.

The role of the generic ideal $G$ is played by a generic ultrafilter. Filters are the duel of ideals; in this exposition here we reversed $\leq$ of the poset $\mathbb{P}$, and proved the existence of a generic ideal. Once $\leq$ reverses back, the object needed is a generic ultrafilter. This object defines a homomorphism of the Boolean algebra to the algebra $\{0,1\}$, i.e., decides binary truth and falsehood of every sentence. Formally, the

<footnote n="67">⁶⁷In fact, even a sentence $\phi$ that is not false at $\{\}$ and is true in some $p \in \mathbb{P}$ is automatically shown to be consistent; for $\neg\text{CH}$ the stronger statement of truth at $p = \{\}$ applies.</footnote>
quotient $M^B/G$ decides truth. If $G$ is generic, $M^B/G$ is shown to be isomorphic to a standard transitive model of ZFC and to contain $G$. And again, if $M$ is countable then a generic ultrafilter $G$ is proven to exist.

An exceptionally interesting proof version by Dana Scott (1967) uses a Boolean valued model of ordinary real algebra axiomatized in just three levels: the reals, functions over reals, and *functionals*—functions over functions of the reals. To quote Scott (1967) “To those readers familiar with Cohen’s original proof [...] our approach may seem at first very different. Actually it is not. [...] Cohen’s definition of forcing could be viewed as a way of assigning Boolean values to formulas.”

My personal opinion is that the Boolean algebra machinery is an overkill for the forcing methodology. As a matter of taste, I find this approach overpowered and consequently less intuitive. Little is achieved by embedding $\mathbb{P}$ to a complete Boolean algebra because $\mathbb{P}$ already has the needed information. Most of the machinery of the Boolean algebra is unused by forcing. Notably, $\neg CH$ evaluates to 1 as opposed to some other nonzero value. What is used is just the logic of “necessarily possibly true”, which naturally leads to the concept of genericity. This key idea is obscured in an exposition of forcing with Boolean valued models.

## 7.4 Modal logical models

Modal logic is the study of the deductive behavior of expressions that include the modal “necessarily” (□) and its dual “possibly” (◇ := ¬□¬). The semantics of modal logic involves Saul Kripke’s frames of *possible worlds* formulated as a directed graph (*W*, *E*) of vertices *w* that represent worlds and edges (*w*, *w*′) that represent reachability of world *w*′ from world *w*. Each world *w* has truth assignments for the atomic formulas. The common logical connectives work as usual, and □*φ* is true at world *w* if for all (*w*, *w*′) ∈ *E*, *φ* is true in *w*′. This is the meaning of *φ* to be necessarily true at *w*. Importantly, reachability is not transitive or reflexive in principle: all reachable worlds *w*′ from *w* are exposed with explicit edges (*w*, *w*′) ∈ *E*. The edge structure can be encoded in logical rules governing □ and its dual ◊. In particular, if (*w*, *w*) is always present then □*φ* → *φ* is an axiom: if *φ* is necessarily true at *w* and *w* is reachable from *w*, then *φ* is true at *w*.

The basic rules **K** (for Kripke) apply to all modal structures: (1) *Necessitation deduction rule*: if *φ* is a theorem, then it holds in all worlds and therefore □*φ* is also a theorem; (2) *Distribution axiom*: □(*φ* → *ψ*) → (□*φ* → □*ψ*). Both of these rules hold in any graph structure (*W*, *E*) as the reader may wish to verify.

In addition, the rule “M”, □*φ* → *φ*, corresponds to *E* being reflexive, and the rule □*φ* → □□*φ* corresponds to *E* being transitive. The system resulting from adding these two rules to **K** is called S4.

In forcing, a poset $\mathbb{P}$ is formed with reachability structure that is reflexive and transitive, neatly fitting in the S4 framework. The whole forcing construction can take place within an S4 modal model. Every formula $\phi$ is embedded in the model by attaching □◇ in front of it and all its subformulas. For example, $\forall x \forall y \exists z x \in z \wedge y \in z$ becomes:

$$ \Box \diamond \forall x (\Box \diamond \forall y (\Box \diamond \exists z (\Box \diamond (\Box \diamond (x \in z) \wedge \Box \diamond (y \in z)))))) $$

It is shown that every axiom $\phi$ of ZFC is a valid sentence of the S4 modal model if thus embedded, and therefore every theorem of ZFC is a theorem of the modal model under the above embedding.<fnref n="68" /> To show that, similar machinery as the one we described above is employed. Some of the proofs are easier because of the syntactic power of modal logic, and some are in my opinion a bit more cumbersome. To prove Comprehension and Replacement, the forcing relation is again defined with first order formulas just like in this manuscript. Then, it is shown that CH is not valid by creating the same poset $\mathbb{P}$ of finite partial functions representing $\aleph_2$-many subsets of $\aleph_0$, and building an S4 modal model over it.

The beauty of this approach is that it formalizes precisely the amount of nonstandard logic needed—no less and no more. The approach described in this manuscript arguably uses less: it deploys “$\forall q \ge p \exists r \ge q \dots$” to express “necessarily possibly”. The Boolean approach uses too much power: it builds

<footnote n="68">⁶⁸Notice the similarity with proving that every axiom $\phi$ is necessarily possibly forced at $p = \{\}$ ∈ $\mathbb{P}$ with parameters from $M^\mathbb{P}$ using induction on $\phi$'s structure with "necessarily possibly forces" logic.</footnote>
a complete Boolean model just to use the “necessarily possibly” logic within it. The modal approach
distills just what is needed.

8 Discussion

What does the proof of independence of CH tell us about the truth value of CH? Here is a list of
possibilities: (1) there is a truth value for CH and we know it; (2) there is a truth value for CH and we
don't know it; (3) there is no truth value for CH; (4) the question is meaningless. The positions among
philosophers and set theorists today span all these possibilities. There is extensive literature on the topic,
which I will not attempt to summarize other than give the reader a few references for further reading
(Shelah 2000, Hamkins 2001, Shelah 2001, Woodin 2001, Woodin 2001b, Bellotti 2005, Woodin 2011,
Hamkins 2015, Rittberg 2015, Barton and Friedman 2021).

The device employed in the proof holds a clue. A countable ground model *M* that contains its
versions of $2^N$ and $\aleph_\alpha$ is extended with countably many distinct new subsets of *N* mapped surjectively
to *M*'s countable $\aleph_\alpha$ impostor. Unwrapping this, there is a syntactic layer and a semantic layer. The
semantic layer concerns *M*'s nature as a model and the membership of certain sets in *M*. However, the
existence of *M* is not needed. It is safe to assume *M* exists syntactically because every proof is finite and
employs a finite list of axioms; for any such finite list *Φ*, there is a ZFC proof that there is a model $M_\Phi$.
So if an inconsistency proof $\phi_1, \dots, \phi_k := \emptyset \neq \emptyset$ arises from assuming *M*, then pick the *Φ* used in the
proof, write down the ZFC proof of existence of $M_\Phi$, and copy the proof to $\phi_1^{M_\Phi}, \dots, \phi_k^{M_\Phi}$ resulting in an
inconsistency proof without assuming *M*. The syntactic layer is an algorithm that shows it is no worse to
assume *M* than to assume ZFC. *M* is never demonstrated and the ZFC proof of existence of any $M_\Phi$ is
highly non-constructive. The question of existence of *M* is not meaningful while the syntactic argument
that it is safe to assume *M* is a terminating algorithm.

From this point on, we are already on shaky ontological ground. Since the ground model is
syntactically concrete and semantically dubious, anything we build on top of it is definitely semantically
dubious, even if it is syntactically concrete. One may object: what is semantically dubious and who is the
arbiter of that?

Here is another clue. Consider the question of whether ZFC is consistent, call it *Con(ZFC)*.
We know from Gödel’s second incompleteness theorem that *Con(ZFC)* is independent of ZFC. Is the
statement *Con(ZFC)* semantically dubious? Not at all: it is the arithmetic statement that there is
no finite string of letters *φ*₁, ..., *φ*ₖ := 0 = 1 with the requisite syntactic properties. We can write a
computer program that will go through all syntactically valid strings and if ¬*Con(ZFC)* it will terminate
reporting the fact, while the longer the program takes, the more evidence we get in favor of *Con(ZFC)*.
The semantic statement *is* the syntactic statement in the case of *Con(ZFC)* even though *Con(ZFC)* is
independent of ZFC. The situation is different with 2<sup>aleph₀</sup> = ℵ₁. We already know that the vast majority of
members of 2<sup>aleph₀</sup> can never be listed or named by a formula. That is why any new axiom or argument
to further settle the question of the size of 2<sup>aleph₀</sup> can only hope to do so by further specifying 2<sup>aleph₀</sup>. *Our
concept of the real line is not specific, regardless how familiar we feel we are with R*. The independence of
2<sup>aleph₀</sup> = ℵ₁ tells us that the statement is syntactically independent of ZFC, and does not bother telling us
anything about what we were hoping to answer—whether 2<sup>aleph₀</sup> = ℵ₁.

Two views that I find especially interesting for philosophical debate are the multiverse view (Hamkins
2012), and countabilism (Barton and Friedman 2021). The multiverse position is that the concept of
a set has infinitely many different formulations depending on the axioms chosen, and the solutions to
CH vary in each formulation. According to Hamkins, the multiverse view “holds that there are diverse
distinct concepts of set, each instantiated in a corresponding set-theoretic universe, which exhibit diverse
set-theoretic truths.” This view can be taken syntactically, where each universe is a syntactic construct.
However, Hamkins goes further and sees the multiverse as literally existing. “Each such universe exists
independently in the same Platonic sense that proponents of the universe view regard their universe to
exist.” According to this view, the solution to CH is simply the totality of theorems of how it behaves
across the multiverse. Intriguingly, Hamkins proposes the “Countability Principle” stating that every
universe is countable from the perspective of another, “better” universe. We saw this with *M* being
countable from the perspective of *V*. If we view the universe *V* as truly existing, an immediate question
is whether “our” universe—what we perceive as V—is countable from the perspective of some outside, better universe that is not accessible to us.

Countabilism (Barton and Friedman 2021) goes in the opposite direction and explores the idea of rejecting the Power axiom and asserting that every set is countable and that the continuum is a proper class. To quote Scott as also quoted in this paper, “Perhaps we would be pushed in the end to say that all sets are countable (and that the continuum is not even a set) when at last all cardinals are absolutely destroyed.” (Scott 1977). The fact that forcing can collapse any cardinal to be countable in a given model is an intuitive argument in favor of this view. As the authors observe, (a) Power implies that there are uncountable sets; (b) given a model $M$ and an $M$-cardinal $κ$, there is a forcing notion $P$ of partial functions $ω \to κ$ forcing $κ$ to be countable in the extension; and (c) there is another forcing notion $P'$ that adds $κ$ new subsets of $ω$ and pushes the continuum to be $\geq \kappa$. The mainstream resolution of this paradoxical situation is that all these impostor models $M$ are missing the “real” Power operation. The authors explore instead the removal of Power and addition of alternative axioms that enhance the ability of the remaining ZFC to be a foundation of mathematics while keeping all sets countable and elevating the continuum to a proper class.

Personally I find countabilism attractive. If we accept that every number exists by virtue of being the successor of a previous number that exists all the way down to 0 and 1, it makes sense to say that $N$ exists as simply the collection of all of them. Beyond that, regardless of how syntactically convenient it is to postulate Power and AC for mathematical proofs, these axioms are not self evident. Power in particular is super convenient in allowing definition of higher structures such as arbitrary functions and relations, functionals and so on, but it does let all hell break loose. To get to “the set of all subsets of $N$” one has to first get to a single subset of $N$. If that subset is generic, it contains Shakespeare’s complete works and everything else infinitely many times, so it will be quite a while before one finishes with one subset to get to the next. In perceiving Cantor’s infinity of infinities, we may be fooled philosophically by what is nothing more than syntactic sugar. The collection of all possible mathematical statements is countable and therefore our ability to reason about the uncountable is a leap of faith.
9 Acknowledgements

I thank David McAllester who introduced me to the result when I was a sophomore intern in his lab at
MIT back in 1993, and Bob Givan, his PhD student at the time who was supervising my "research": my
task was to read and understand the proof as a step towards machine-verifying it. I did read much of the
proof line-by-line but didn't at the time "get" it. In this manuscript, I attempt to provide what could
have been helpful for a student like me at the time. I especially thank Vassilis Papakonstantinou and
Timothy Chow for providing feedback and edits. I thank Ilias Kastanas and Pekka Kauranen for edits,
and Nik Weaver, Scott Aaronson, Richard Durbin and Bob Givan for positive feedback on an earlier
version of the manuscript.
# 10 Formula Abbreviations

$s[i] = x$: $(i, x) \in s$.

$\mathcal{P}(x)$: the power set of $x$.

Aleph($\kappa, \alpha$): $\alpha$ is an ordinal and $\kappa$ is precisely the $\alpha$th infinite cardinal.

Bij($f$): $f$ is a bijection.

Card($x$) : $x$ is a cardinal.

Cnt($x$): $x$ is countable.

CntTrans($x$): $x$ is countable and transitive.

Comp($\phi, x$): Comprehension axiom given formula $\phi$ and domain set $x$.

Dom($f, A$): The domain of function $f$ is $A$.

Dom($f$): The domain of function $f$.

Empty($x$): $x$ is empty.

Inf($x$): $x$ is infinite.

Inj($f$): $f$ is an injection.

LimOrd($\alpha$): $\alpha$ is a limit ordinal.

Ord($\alpha$): $\alpha$ is an ordinal.

Pow($x, y$): $y$ is the power set of $x$.

Ran($f, B$): The range of function $f$ is $B$.

Ran($f$): The range of function $f$.

Succ($x, y$): $y = x \cup \{x\}$

SuccOrd($\alpha$): $\alpha$ is a successor ordinal.

Surj($f$): $f$ is a surjection.

Trans($x$): $x$ is transitive.

Uncountable($x$): $x$ is uncountable.
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