avewright/proofwiki-math huggingface dataset:
Theorem: Riemann Removable Singularities Theorem

Let $U \subset \C$ be a domain.
Let $z_0 \in U$.
Let $f: U \setminus \set {z_0} \to \C$ be holomorphic.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f$ extends to a holomorphic function $f: U \to \C$}}
{{item|(2):|$f$ extends to a continuous function $f: U \to \C$}}
{{item|(3):|$f$ is bounded in a deleted neighborhood of $z_0$}}
{{item|(4):|$\map f z {{=}} \map o {\dfrac 1 {\cmod {z - z_0} } }$ as $z \to z_0$}}
{{end-itemize}}

Proof:
{{WLOG}}, we may assume that:
:$U = \Bbb D := \set {z \in \C: \cmod z < 1}$
and that $z_0 = 0$. 
Otherwise, restrict $f$ to a suitable disk centered at $z_0$, and precompose with a suitable affine map.
$(1) \implies (2)$ by Complex-Differentiable Function is Continuous.
$(2) \implies (3)$ by Continuous Function is Locally Bounded.
$(3) \implies (4)$ by definition. 
$(2) \implies (1)$ follows from the proof of Cauchy's Residue Theorem.
(For completeness, we sketch a self-contained argument below.)
{{Finish}}
Now assume that $(4)$ holds.
That is:
:$z \cdot \map f z \to 0$ as $z \to 0$.
We need to show that $(2)$ holds.
By assumption, the function:
:$\map g z := \begin{cases} z \cdot \map f z & : z \ne 0 \\ 0 & : \text{otherwise} \end{cases}$
is continuous in $\Bbb D$ and holomorphic in $\Bbb D \setminus \set 0$.
By applying the direction $(2) \implies (1)$ to this function, we see that $g$ is holomorphic in $\Bbb D$.
We have:
:$\ds \map {g'} 0 = \lim_{z \mathop \to 0} \frac {\map g z} z = \lim_{z \mathop \to 0} \map f z$
so $f$ extends continuously to $\Bbb D$, as claimed.
To prove $(2) \implies (1)$, we use Morera's Theorem and Cauchy's Residue Theorem.
By Morera's Theorem, we need to show that:
:$\ds \oint_C \map f z \rd z = 0$
for every closed contour $C$ in $\Bbb D$.
It follows from the Cauchy Integral Theorem that we only need to check that:
:$\ds \oint_C \map f z \rd z = 0$
for a simple closed loop surrounding $0$, and that this integral is independent of the loop $C$.
Letting $C = \map {C_\epsilon} 0$ be the circle of radius $\epsilon$ around $0$, we see by Estimation Lemma for Contour Integrals that:
:$\ds \size {\oint_{C_\epsilon} \map f z \rd z} \le 2 \pi \epsilon \max_{z \mathop \in C_\epsilon} \size {\map f z} \to 0$
as $\epsilon \to 0$ (because $f$ is continuous in $0$ by assumption).
This completes the proof.
{{qed}}
{{Namedfor|Georg Friedrich Bernhard Riemann|cat = Riemann}}
Category:Complex Analysis
Category:Continuous Functions
Category:Removable Singularities
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Theorem: Gaussian Integral

:$\ds \int_{-\infty}^\infty e^{-x^2} \rd x = \sqrt \pi$

Proof:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty e^{-x^2} \rd x
      | r = 2 \int_0^\infty e^{-x^2} \rd x
      | c = Definite Integral of Even Function
}}
{{eqn | r = \frac {2 \sqrt \pi} 2
      | c = Integral to Infinity of Exponential of -t^2
}}
{{eqn | r = \sqrt \pi
}}
{{end-eqn}}
{{qed}}
Category:Gaussian Integral
Category:Definite Integrals involving Exponential Function
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Theorem: Indiscrete Topology is not Metrizable

Let $S$ be a set with more than one element.
The indiscrete topology on $S$ is not metrizable.

Proof:
In order for $S$ to be a metrizable space, there needs to be a metric $d$ on $S$, so that $\struct {S, d}$ is a metric space.
As $S$ has more than one element, $\exists x, y \in S: x \ne y$.
Then:
:$\epsilon = \map d {x, y} > 0$
So the open $\epsilon$-ball $\map {B_{\epsilon / 2} } x$ is $d$-open.
Hence $\map {B_{\epsilon / 2} } x$ is in the topology which $d$ induces.
But $x \in \map {B_{\epsilon / 2} } x$ while $y \notin \map {B_{\epsilon / 2} } x$.
Thus $\map {B_{\epsilon / 2} } x \ne \O$ and $\map {B_{\epsilon / 2} } x \ne S$.
So the topology induced by $d$ is not the indiscrete topology.
{{qed}}
Theorem: Composite of Continuous Mappings is Continuous

Let $T_1, T_2, T_3$ be topological spaces.
Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be continuous mappings.
Then the composite mapping $g \circ f: T_1 \to T_3$ is continuous.

Proof:
Let $U \in T_3$ be open in $T_3$.
As $g$ is continuous, $g^{-1} \sqbrk U \in T_2$ is open in $T_2$.
As $f$ is continuous, $f^{-1} \sqbrk {g^{-1} \sqbrk U} \in T_1$ is open in $T_1$.
By Preimage of Subset under Composite Mapping, $f^{-1} \sqbrk {g^{-1} \sqbrk U} = \paren {g \circ f}^{-1} \sqbrk U$.
Hence the result.
{{qed}}
Theorem: Continuous Mapping is Continuous on Induced Topological Spaces

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\tau_{d_1}$ and $\tau_{d_2}$ be the topologies induced by the metrics $d_1$ and $d_2$.
Let $T_1 = \struct {A_1, \tau_{d_1} }$ and $T_2 = \struct {A_2, \tau_{d_2} }$ be the resulting topological spaces.
Let $f: A_1 \to A_2$ be a mapping.
Then $f$ is $\tuple {d_1, d_2}$-continuous {{iff}} $f$ is $\tuple {\tau_{d_1}, \tau_{d_2} }$-continuous.

Proof:
Follows directly from:
:the open set definition of continuity on a metric space
:the definition of continuity on a topological space.
{{qed}}
Theorem: Synthetic Basis and Analytic Basis are Compatible

Let $\struct {S, \tau}$ be a topological space.
Then $\BB$ is an analytic basis for $\tau$ {{iff}} $\tau$ is the topology on $S$ generated by the synthetic basis $\BB$.

Proof:
=== Necessary Condition ===
Suppose that $\BB$ is an analytic basis for $\tau$.
We proceed to check the axioms for $\BB$ to be a synthetic basis on $S$.
{{Open-set-axiom|3}} states that $S \in \tau$.
Therefore, by the definition of an analytic basis:
:$\ds \exists \SS \subseteq \BB: S = \bigcup \SS$
By Equivalent Conditions for Cover by Collection of Subsets, $\BB$ is a cover for $S$.
That is, axiom $(\text B 1)$ for a synthetic basis is satisfied by $\BB$.
Suppose that $A, B \in \BB$.
By the definition of an analytic basis, $\BB \subseteq \tau$.
Therefore $A, B \in \tau$.
{{Open-set-axiom|2}} states that $A \cap B \in \tau$.
Therefore, by the definition of an analytic basis:
:$\ds \exists \AA \subseteq \BB: A \cap B = \bigcup \AA$
That is, axiom $(\text B 2)$ for a synthetic basis is satisfied by $\BB$.
Therefore, $\BB$ is a synthetic basis on $S$.
Let $\tau'$ be the topology on $S$ generated by the synthetic basis $\BB$.
It follows from the definition of an analytic basis that $\tau \subseteq \tau'$.
Since the subset relation is transitive, we can apply {{Open-set-axiom|1}} for a topology to conclude that:
:$\ds \forall U \in \tau': \exists \AA \subseteq \BB \subseteq \tau: U = \bigcup \AA \in \tau$
That is, $\tau' \subseteq \tau$.
By definition of set equality:
:$\tau = \tau'$
{{qed|lemma}}
Theorem: Continuity Test using Basis

Let $\struct {X_1, \tau_1}$ and $\struct {X_2, \tau_2}$ be topological spaces.
Let $f: X_1 \to X_2$ be a mapping.
Let $\BB$ be an analytic basis for $\tau_2$.
Suppose that:
:$\forall B \in \BB: f^{-1} \sqbrk B \in \tau_1$
where $f^{-1} \sqbrk B$ denotes the preimage of $B$ under $f$.
Then $f$ is continuous.

Proof:
Let $U \in \tau_2$.
By the definition of an analytic basis, it follows that:
:$\ds \exists \AA \subseteq \BB: U = \bigcup \AA$
Hence:
{{begin-eqn}}
{{eqn | l = f^{-1} \sqbrk U
      | r = f^{-1} \sqbrk {\bigcup \AA}
}}
{{eqn | r = \bigcup_{B \mathop \in \AA} f^{-1} \sqbrk B
      | c = Preimage of Union under Mapping: General Result
}}
{{eqn | o = \in
      | r = \tau_1
      | c = {{hypothesis}}, and by {{Defof|Topology}}
}}
{{end-eqn}}
The result follows from the definition of continuity.
{{qed}}
Theorem: Sub-Basis for Real Number Line

Let the real number line $\R$ be considered as a topology under the usual (Euclidean) topology.
Then:
:$\BB := \set {\openint \gets a, \openint b \to: a, b \in \R}$ is a sub-basis for $\R$.

Proof:
Let $\openint c d$ be an open real interval.
Then by definition:
:$\openint c d = \openint \gets d \cap \openint c \to$
and so $\openint c d$ is the intersection of two elements of $\BB$.
From Open Sets in Real Number Line, any open set of $\R$ is the union of countably many open real intervals.
The result follows by definition of sub-basis.
{{qed}}
Theorem: Synthetic Basis formed from Synthetic Sub-Basis

Let $X$ be a set.
Let $\SS$ be a synthetic sub-basis on $X$.
Define:
:$\ds \BB = \set {\bigcap \FF: \FF \subseteq \SS, \text{$\FF$ is finite} }$
Then $\BB$ is a synthetic basis on $X$.
Thus, by this construction, an arbitrary collection of subsets of $X$ can form a synthetic basis and thus generate a topology on $X$.

Proof:
We consider $X$ as the universal set.
Thus, in accordance with Intersection of Empty Set, we take the convention that:
:$\ds \bigcap \O = X \in \BB$
By Set is Subset of Union: General Result, it follows that:
:$\ds X \subseteq \bigcup \BB$
That is, axiom $(\text B 1)$ for a synthetic basis is satisfied.
We have that $\BB \subseteq \powerset X$.
Let $B_1, B_2 \in \BB$.
Then there exist finite $\FF_1, \FF_2 \subseteq \SS$ such that:
:$\ds B_1 = \bigcap \FF_1$
:$\ds B_2 = \bigcap \FF_2$
It follows that:
:$\ds B_1 \cap B_2 = \bigcap \paren {\FF_1 \cup \FF_2}$
{{explain|proof that $\ds \bigcap_{i \mathop \in I} \bigcap \mathbb S_i$ is equal to $\ds \bigcap \bigcup_{i \mathop \in I} \mathbb S_i$}}
By Union is Smallest Superset, $\FF_1 \cup \FF_2 \subseteq \SS$.
We have that $\FF_1 \cup \FF_2$ is finite.
Hence $B_1 \cap B_2 \in \BB$, so it follows by definition that axiom $(\text B 2)$ for a synthetic basis is satisfied.
{{qed}}
Theorem: Epimorphism Preserves Distributivity

Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be algebraic structures.
Let $\phi: R_1 \to R_2$ be an epimorphism.
:If $\circ_1$ is left distributive over $+_1$, then $\circ_2$ is left distributive over $+_2$.
:If $\circ_1$ is right distributive over $+_1$, then $\circ_2$ is right distributive over $+_2$.
Consequently, if $\circ_1$ is distributive over $+_1$, then $\circ_2$ is distributive over $+_2$.
That is, epimorphism preserves distributivity.

Proof:
Throughout the following, we assume the morphism property holds for $\phi$ for both operations.
It remains to be shown that for non-empty $R_1$, $\paren {x +_1 y}, \paren {y +_1 z}, \paren {x \circ_1 y }, \paren {y \circ_1 z}, \paren {x \circ_1 z}, \paren { x \circ_1 \paren {y +_1 z} }, \paren { \paren {x +_1 y} \circ_1 z } \in \Dom \phi$.
{{ProofWanted| The domain of the epimorphism must be verified for a valid proof, since the algebraic structure are not assumed to be closed under its binary operation.}}
Theorem: Epimorphism Preserves Groups

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ be a group.
Then $\struct {T, *}$ is also a group.

Proof:
From Epimorphism Preserves Semigroups, if $\struct {S, \circ}$ is a semigroup then so is $\struct {T, *}$.
From Epimorphism Preserves Identity, if $\struct {S, \circ}$ has an identity $e_S$, then $\map \phi {e_S}$ is the identity for $*$.
From Epimorphism Preserves Inverses, if $x^{-1}$ is an inverse of $x$ for $\circ$, then $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
The result follows from the definition of group.
{{qed}}
Theorem: Totally Bounded Metric Space is Bounded

Let $M = \struct {A, d}$ be a totally bounded metric space.
Then $M$ is bounded.

Proof:
Let $M = \struct {A, d}$ be totally bounded.
Then there exist $n \in \N$ and points $x_0, \dots, x_n \in A$ such that:
:$\ds \inf_{0 \mathop \le i \mathop \le n} \map d {x_i, x} \le 1$
for all $x \in A$.
Let us set:
{{begin-eqn}}
{{eqn | l = a
      | o = :=
      | r = x_0
}}
{{eqn | l = D
      | o = :=
      | r = \max_{0 \mathop \le i \mathop \le n} \map d {x_0, x_i}
}}
{{eqn | l = K
      | o = :=
      | r = D + 1
}}
{{end-eqn}}
Now let $x \in A$ be arbitrary.
Then by assumption there exists $i$ such that:
:$\map d {x_i, x} \le 1$
Hence:
:$\map d {a, x} \le \map d {a, x_i} + \map d {x_i, x} \le 1 + D = K$
So $M$ is bounded, as claimed.
{{qed}}
Theorem: Existence and Uniqueness of Sigma-Algebra Generated by Collection of Subsets

Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map \sigma \GG$, the $\sigma$-algebra generated by $\GG$, exists and is unique.

Proof:
=== Existence ===
By Power Set is Sigma-Algebra, there is at least one $\sigma$-algebra containing $\GG$.
Next, let $\Bbb E$ be the collection of $\sigma$-algebras containing $\GG$:
:$\Bbb E := \set {\Sigma': \GG \subseteq \Sigma', \text{$\Sigma'$ is a $\sigma$-algebra} }$
By Intersection of Sigma-Algebras, $\Sigma := \bigcap \Bbb E$ is a $\sigma$-algebra.
Also, by Set Intersection Preserves Subsets:
:$\GG \subseteq \Sigma$
Now let $\Sigma'$ be a $\sigma$-algebra containing $\GG$.
By construction of $\Sigma$, and Intersection is Subset: General Result:
:$\Sigma \subseteq \Sigma'$
{{qed|lemma}}
Theorem: Basis for Topology on Cartesian Product

Let $T_1 = \struct{A_1, \tau_1}$ and $T_2 = \struct{A_2, \tau_2}$ be topological spaces.
Let $A_1 \times A_2$ be the Cartesian product of $A_1$ and $A_2$.
Let $\PP = \set {U_1 \times U_2 : U_1 \in \tau_1, U_2 \in \tau_2}$
Then $\PP$ is a synthetic basis on $A_1 \times A_2$.

Proof:
We need to show that conditions $(\text B 1)$ and $(\text B 2)$ in the definition for basis hold for $\PP$.
Theorem: Projection from Product Topology is Continuous

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Then both $\pr_1$ and $\pr_2$ are are continuous.

Proof:
From Natural Basis of Product Topology of Finite Product, a basis for $\tau$ is:
:$\BB = \set {U \times V: U \in \tau_1, V \in \tau_2}$
Let $U$ be open in $T_1$.
Then $\map {\pr_1^{-1} } U = U \times T_2$ is one of the open sets in the basis in the definition of product topology.
Thus $\pr_1$ is continuous.
The same argument can be applied to $\pr_2$.
Theorem: Projection from Product Topology is Continuous

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Then both $\pr_1$ and $\pr_2$ are are continuous.

Proof:
By definition of the product topology on $S$:
:$\tau$ is the initial topology on $S$ with respect to $\family {\pr_i}_{i \mathop \in I}$
By definition of the Initial Topoplogy: Definition 2:
:$\tau$ is the coarsest topology on $S$ such that each $\pr_i: S \to S_i$ is a $\struct{\tau, \tau_i}$-continuous.
Theorem: Continuous Mapping to Product Space

Let $T = T_1 \times T_2$ be the product space of two topological spaces $T_1$ and $T_2$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Let $T'$ be a topological space.
Let $f: T' \to T$ be a mapping.
Then $f$ is continuous {{iff}} $\pr_1 \circ f$ and $\pr_2 \circ f$ are continuous.

Proof:
=== Necessary Condition ===
Let $f$ be continuous.
Then by:
:Projection from Product Topology is Continuous
:Composite of Continuous Mappings is Continuous
so are $\pr_1 \circ f$ and $\pr_2 \circ f$.
{{qed|lemma}}
Theorem: Open and Closed Sets in Topological Space

Let $T = \struct {S, \tau}$ be a topological space.
Then $S$ and $\O$ are both both open and closed in $T$.

Proof:
From the definition of closed set, $U$ is open in $T$ {{iff}} $S \setminus U$ is closed in $T$.
From Underlying Set of Topological Space is Clopen, $S$ is both open and closed in $T$.
From Empty Set is Element of Topology, $\O$ is open in $T$.
From Empty Set is Closed in Topological Space, we have that $\O$ is closed in $T$.
{{qed}}
Theorem: Absolutely Continuous Real Function is Continuous

Let $I \subseteq \R$ be a real interval.
Let $f: I \to \R$ be an absolutely continuous real function.
Then $f$ is continuous.

Proof:
From Absolutely Continuous Real Function is Uniformly Continuous:
:$f$ is uniformly continuous.
Therefore, from Uniformly Continuous Real Function is Continuous:
:$f$ is continuous.
{{qed}}
Category:Absolutely Continuous Real Functions
Category:Continuous Real Functions
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Theorem: Closed Set in Topological Subspace

Let $T = \struct{S, \tau}$ be a topological space.
Let $T' = \struct{H, \tau_H}$ be a subspace of $T$ where $H \subseteq S$.
Then $V \subseteq T'$ is closed in $T'$ {{iff}} $V = T' \cap W$ for some $W$ closed in $T$.

Proof:
=== Necessary Condition ===
Suppose $V \subseteq T'$ is closed in $T'$.
Then $T' \setminus V$ is open in $T'$ by definition.
So, by definition of subspace topology, $T' \setminus V = T' \cap U$ for some $U$ open in $T$.
Then:
{{begin-eqn}}
{{eqn | l = V
      | r = T' \setminus \paren {T' \setminus V}
      | c = Relative Complement of Relative Complement
}}
{{eqn | r = T' \setminus \paren {T' \cap U}
      | c = from above
}}
{{eqn | r = T' \setminus U
      | c = Set Difference with Intersection is Difference
}}
{{eqn | r = T' \cap \paren {T \setminus U}
      | c = 
}}
{{end-eqn}}
Thus $V$ is closed in $T$.
{{qed|lemma}}
Theorem: Continuity Defined from Closed Sets

Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping.
Then $f$ is continuous {{iff}} for all $V$ closed in $T_2$, $f^{-1} \sqbrk V$ is closed in $T_1$.

Proof:
First we show the following.
Let $W \in T_2$.
We note that $f^{-1} \sqbrk {T_2} = T_1$.
Hence, from Preimage of Set Difference under Mapping, we have:
:$f^{-1} \sqbrk {T_2 \setminus W} = T_1 \setminus f^{-1} \sqbrk W$
Theorem: Real Numbers are Uncountably Infinite

The set of real numbers $\R$ is uncountably infinite.

Proof:
We prove the equivalent result that every sequence $\sequence {x_k}_{k \mathop \in \N}$ omits at least one $x \in \R$.
Let $\sequence {x_k}_{k \mathop \in \N}$ be a sequence of distinct real numbers.
Let a sequence of closed real intervals $\sequence {I_n}$ be defined as follows:
Let:
{{begin-eqn}}
{{eqn | l = a_k
      | r = \min \set {x_k, x_{k + 1} }
}}
{{eqn | l = b_k
      | r = \max \set {x_k, x_{k + 1} }
}}
{{eqn | l = I_k
      | r = \closedint {a_k} {b_k}
}}
{{end-eqn}}
Since the terms of $\sequence {x_k}_{k \mathop \in \N}$ are distinct, $a_k \ne b_k$.
Thus $I_k$ is not a singleton.
Let:
:$I_{n - 1} = \closedint {a_{n - 1} } {b_{n - 1} }$
It can be assumed that infinitely many of the $x_k$ lie inside $I_{n - 1}$.
Otherwise the proof is complete.
Let $y$ and $z$ be the first two such terms of $\sequence {x_k}_{k \mathop \in \N}$.
Let:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \min \set {y, z}
}}
{{eqn | l = b_n
      | r = \max \set {y, z}
}}
{{eqn | l = I_n
      | r = \closedint {a_n} {b_n}
}}
{{end-eqn}}
Thus we have sequences:
:$\sequence {a_k}_{k \mathop \in \N}$
:$\sequence {b_k}_{k \mathop \in \N}$
with:
:$ a_1 < a_2 < \cdots < b_2 < b_1$
So $\sequence {a_k}_{k \mathop \in \N}$ and $\sequence {b_k}_{k \mathop \in \N}$ are monotone, and bounded above and bounded below respectively.
Therefore by the Monotone Convergence Theorem (Real Analysis) both are convergent.
Let:
{{begin-eqn}}
{{eqn | l = A
      | r = \lim_{k \mathop \to \infty} a_k
}}
{{eqn | l = B
      | r = \lim_{k \mathop \to \infty} b_k
}}
{{end-eqn}}
Clearly we have $A \le B$.
So:
:$\closedint A B \ne \O$
Let $h \in \closedint A B$.
Then:
:$\forall k: h \ne a_k, b_k$
We claim that $h \ne x_k$ for all $k$. 
Suppose that $h = x_k$ for some $k$. 
Then there are only finitely many terms in the sequence before $h$ occurs.
{{Explain|"the sequence" -- which one in particular? There are a few under discussion here.}}
Therefore only finitely many of the $a_k$ precede $h$.
Let $a_d$ be the last term of the sequence $\sequence {a_k}_{k \mathop \in \N}$ preceding $h$.
We defined $a_{d + 1}$, $b_{d + 1}$ to be interior points of $I_d$, and also $h \in I_{d + 1}$ by the definition of $h$.
{{explain|Where are those interior points so defined?}}
Therefore $a_{d + 1}$ must precede $h$ in the sequence, for the sequence is increasing.
This is a contradiction.
{{explain|Specify precisely what assumption this contradiction falsifies.}}
{{qed}}
Theorem: Real Numbers are Uncountably Infinite

The set of real numbers $\R$ is uncountably infinite.

Proof:
By definition, a perfect set is a set $X$ such that every element $x \in X$ is the limit of a sequence of elements of $X$ distinct from $x$.
From Real Numbers form Perfect Set, $\R$ is perfect.
Therefore it is sufficient to show that a perfect subset of $X \subseteq \R^k$ is uncountable.
We prove the equivalent result that every sequence $\sequence {x_k}_{k \mathop \in \N} \subseteq X$ omits at least one element of $X$.
Let $y_1 \in X$.
Let $B_1:= \map {\BB_r} {y_1}$ be a closed ball whose center is at $y_1$.
Consider a closed ball:
:$B_{n - 1} \supseteq B_n := \map {\BB_{\map \delta n} } {y_n}$
such that:
:$\map \delta n \le \dfrac {\map \delta {n - 1} } 2$
:$y_n \in X$
:$x_n \notin B_n$
Note that $\map \delta 1 = r$.
We can satisfy the condition $x_n \notin B_n$ because $X$ is perfect, so every closed ball whose center is at an element of $X$ contains infinitely many elements of $X$.
Since $\map \delta n \le \dfrac r {2^{n - 1} }$, $y_n$ is Cauchy.
Therefore from Perfect Set is Closed, we may let:
:$\ds Y = \lim_{n \mathop \to \infty} y_n \in X$
For $n \in \N$ we have:
:$\set {y_m: m > n} \subseteq B_n$
so $Y \in B_n$.
But by construction:
:$\forall n \in \N: x_n \notin B_n$
Therefore:
:$\forall n \in \N: Y \ne x_n$
and the proof is complete.
{{qed}}
Theorem: Real Numbers are Uncountably Infinite

The set of real numbers $\R$ is uncountably infinite.

Proof:
It is sufficient to show that the real interval $I = \set {x \in \R: 0 < x \le 1}$ is uncountable.
Let $x \in I$.
From Existence of Base-N Representation, $x$ has a unique representation of the form:
:$x = \dfrac {\epsilon_1} 3 + \dfrac {\epsilon_2} {3^2} + \dfrac {\epsilon_3} {3^3} + \cdots$ 
where $\epsilon_k = 0, 1$ or $2$ and an infinite number of $\epsilon_k$ are different from $0$.
Let $S \subseteq I$ be countably infinite.
Let $S = \set {x_1, x_2, \ldots}$.
Let $\epsilon_{k 1}, \epsilon_{k 2}, \epsilon_{k 3}, \ldots$ be the ternary digits of $x_k$.
Let $\epsilon_k = 1 + 2 \epsilon_{k k} - \epsilon_{k k}^2$ so that:
:$\epsilon_k = 1$ if $\epsilon_{k k} = 0$ or $\epsilon_{k k} = 2$
:$\epsilon_k = 2$ if $\epsilon_{k k} = 1$
Then:
:$(1): \quad \forall k: \epsilon_k \ne 0$
:$(2): \quad \forall k: \epsilon_k \ne \epsilon_{k k}$
We have that $0 < x \le 1$ so $x \in I$.
But the real number $\displaystyle x = \sum \epsilon_k 3^{-k}$ is different from every $x_k \in I$.
Thus we have found an element of $I$ which is not an element of $S$.
Therefore $S$ is a proper subset of $I$.
It follows by definition that $I$ is uncountable.
{{qed}}
Theorem: Real Numbers are Uncountably Infinite

The set of real numbers $\R$ is uncountably infinite.

Proof:
Define a mapping $f: \powerset {\N_{>0} } \to \R$ thus:
:$\map f S = 0.d_1 d_2 \ldots$, interpreted as a ternary expansion where $\sequence {d_n}$ is the characteristic function of $S$.
That is:
:$\ds \map f S = \sum_{i \mathop \in S} 3^{-i}$
By the lemma, $f$ is an injection.
{{AimForCont}} that $\R$ is countable.
Then there is an injection $g: \R \to \N$.
By Composite of Injections is Injection, $g \circ f: \powerset \N \to \N$ is an injection.
But this contradicts No Injection from Power Set to Set.
Hence, by Proof by Contradiction, $\R$ is not countable.
{{qed}}