avewright/proofwiki-math huggingface dataset:
Theorem: Particular Point Space is Scattered

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is scattered.

Proof:
{{Recall|Scattered Space|scattered space}}
{{:Definition:Scattered Space/Definition 1}}
Let $H \subseteq S$ such that $p \notin H$.
From Subset of Particular Point Space is either Open or Closed, $H$ is closed in $T$.
We have that Closed Set in Particular Point Space has no Limit Points.
So if $p \notin H$ then $H$ has no limit points.
So by definition, every element of such a set $H$ is an isolated point.
So if $p \notin H$ then $H$ is by definition not dense-in-itself.
Now let $H \subseteq S$ such that $p \in H$.
We have that $\set p \in \tau_p$.
That is, $p$ is an open point of $T$.
From Point in Topological Space is Open iff Isolated, $p$ is an isolated point.
So if $p \in H$ then $H$ is by definition not dense-in-itself.
Hence the result, by definition of a scattered space.
{{qed}}
Theorem: Cook-Levin Theorem

The boolean satisfiability problem is NP-Complete.

Proof:
=== The Boolean Satisfiability Problem is NP ===
Given a boolean satisfiability problem with a set of variables $X$ and clauses $L$ and a possible solution to the problem, it is a trivial matter to evaluate all the clauses in $L$ to verify the solution in polynomial time.
{{Handwaving|The above needs to be justified with a page demonstrating this.}}
From NP Problem iff Solution Verifiable in Polynomial Time it is NP.
Theorem: Particular Point Space is Irreducible

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is irreducible.

Proof:
By definition, $T = \struct {S, \tau_p}$ is irreducible {{iff}} every two non-empty open sets of $T$ have non-empty intersection.
Let $U_1$ and $U_2$ be non-empty open sets of $T$.
By definition of particular point space, $p \in U_1$ and $p \in U_2$.
Thus:
:$p \in U_1 \cap U_2$
and so:
:$U_1 \cap U_2 \ne \O$
Hence the result.
{{qed}}
Theorem: Particular Point Space is not Ultraconnected

Let $T = \struct {S, \tau_p}$ be a particular point space with at least three points.
Then $T$ is not ultraconnected.

Proof:
Let $x, y \in S: x \ne p, y \ne p, x \ne y$.
Consider $\set x$ and $\set y$.
Neither are open as neither contain $p$.
So from Subset of Particular Point Space is either Open or Closed they are both closed.
We have that $\set x \cap \set y = \O$.
The result follows by definition of ultraconnected.
{{qed}}
Theorem: Dispersion Point in Particular Point Space

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $p$ is dispersion point of $T$.

Proof:
Let $H = S \setminus \set p$.
Let $T_H = \struct {H, \tau_H}$ be the topological subspace induced on $H$ by $\tau_p$.
From Particular Point Space less Particular Point is Discrete, the space $T_H$ is discrete.
We have Discrete Space is Locally Connected.
Thus from Totally Disconnected and Locally Connected Space is Discrete we have that $S \setminus \set p$ is totally disconnected.
Hence the result, from definition of dispersion point.
{{qed}}
Theorem: Infinite Particular Point Space is not Weakly Countably Compact

Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not weakly countably compact.

Proof:
{{Recall|Weakly Countably Compact Space|weakly countably compact space}}
{{:Definition:Weakly Countably Compact Space}}
So, let $T = \struct {S, \tau_p}$ be an infinite particular point space {{WRT}} the particular point $p$.
Let $H \subseteq S$ be an infinite subset of $S$ where $p \notin H$.
$H$ is not open in $T$ by definition.
So from Subset of Particular Point Space is either Open or Closed, $H$ is closed in $T$.
Then we have that a Closed Set in Particular Point Space has no Limit Points.
The result follows from definition of weakly countably compact.
{{qed}}
Theorem: Fundamental Theorem of Symmetric Polynomials

Let $K$ be a field.
Let $f$ a symmetric polynomial over $K$.
Then $f$ can be written uniquely as a polynomial in the elementary symmetric polynomials.

Proof:
Let $p \in R := K \sqbrk {x_1, \dotsc, x_n}$ be a symmetric polynomial.
Let $\sigma_1, \dotsc, \sigma_n$ denote the $n$ elementary symmetric polynomials.
Let $<$ be the degree reverse lexicographic monomial order.
Let us denote by $\map {LM} f$ the leading monomial of any $f \in R$, that is to say the unique $<$-largest monomial appearing in $f$.
Then it is straightforward to verify that $\map {LM} {\sigma_k} = x_1 \cdots x_k$ for $1 \le k \le n$:
For $k < n$, any other monomial appearing in $\sigma_k$ must contain a factor $x_j^{\alpha_1}$ with $j > k$ and $\alpha_1 > 0$.
This makes it smaller than $x_1 \cdots x_k$ with respect to $<$.
The case $k = n$ is trivial because $\sigma_n$ consists of only one monomial.
It follows the proof that $p$ may be written as a polynomial in $\sigma_1, \dotsc, \sigma_k$.
Firstly, since $p$ is symmetric, it must have the property:
:$\map {LM} p = x_1^{a_1} \dotsm x_n^{a_n}$
with $a_1 \ge a_2 \ge \ldots \ge a_n$.
To see this, assume that $a_i > a_j$ for some $1 \le i < j \le n$.
The monomial:
:$x_1^{a_1} \dotsm x_{i - 1}^{a_{i - 1} } \cdot x_i^{a_j} x_j^{a_i} \cdot x_{j + 1}^{a_{j + 1} } \cdots x_n^{a_n}$
must also appear in $p$, and if $a_i > a_j$ then it would be larger than $\map {LM} p$, a contradiction.
Now consider:
:$\sigma :=  c \sigma_1^{a_1 - a_2} \cdot \sigma_2^{a_2 - a_3} \cdot \dotsm \cdot \sigma_{n - 1}^{a_{n - 1} - a_n} \cdot \sigma_n^{a_n}$
where $c$ is the coefficient of $\map {LM} p$ in $p$.
Define $\bar p := p - \sigma$.
Because:
{{begin-eqn}}
{{eqn | l = \map {LM} \sigma
      | r = \prod_{i \mathop = 1}^n \map {LM} {\sigma_i^{a_i - a_{i + 1} } }
      | c = 
}}
{{eqn | r = x_1^{a_1 - a_2} \cdot \paren {x_1 x_2}^{a_2 - a_3} \cdot \dotsm \cdot \paren {x_1 \ldots x_{n - 1} }^{a_{n - 1} - a_n} \cdot \paren {x_1 \ldots x_n}^{a_n}
      | c = 
}}
{{eqn | r = x_1^{a_1} \dotsm x_n^{a_n}
      | c = 
}}
{{eqn | r = \map {LM} p
      | c = 
}}
{{end-eqn}}
the leading monomials in the difference $\bar p$ cancel.
Therefore:
:$\map {LM} {\bar p} < \map {LM} p$
Iterate this method, and after finitely many steps one has decomposed $p$ into a polynomial $F \in K \sqbrk {y_1, \dotsc, y_n}$ with $\map F {\sigma_1, \dotsc, \sigma_n} = p$.
The termination of this algorithm follows immediately from Dickson's Lemma.
{{qed}}
Theorem: Particular Point Space is Pseudocompact

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is a pseudocompact space.

Proof:
We have that:
:a Particular Point Space is Irreducible
:an Irreducible Space is Pseudocompact.
{{qed}}
Theorem: Particular Point Space is Path-Connected

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is path-connected.

Proof:
Let $q \in S$.
Let $\mathbb I$ be the closed unit interval in $\R$.
Let $f: \mathbb I \to S$ be the mapping defined as:
:<nowiki>$\forall x \in \mathbb I: \map f x = \begin{cases}
p & : x \in \hointr 0 1 \\
q & : x = 1
\end{cases}$</nowiki>
Let $U \in \tau_p$.
Then by definition of particular point space, $p \in U$
Either $q \in U$ or $q \notin U$.
If $q \in U$ then $f^{-1} \sqbrk U = \closedint 0 1$ which is open in $\mathbb I$ because $\closedint 0 1 = \mathbb I$.
If $q \notin U$ then $f^{-1} \sqbrk U = \hointr 0 1$ which is half open in $\R$ but open in $\mathbb I$.
So $f: \mathbb I \to S$ is a continuous mapping and so a path from $p$ to $q$.
As $q$ is any point in $S$, it follows from Path-Connected iff Path-Connected to Point that $T$ is path-connected.
{{qed}}
Theorem: Particular Point Space is not Injectively Path-Connected

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is not injectively path-connected.

Proof:
Let $q \in S$ be such that $q \ne p$.
Let $f: \closedint 0 1 \to T$ be an injection such that:
{{begin-eqn}}
{{eqn | l = \map f 0
      | r = q
}}
{{eqn | l = \map f 1
      | r = p
}}
{{end-eqn}}
Because $f$ is an injection, it must be that:
:$\map {f^{-1} } {\set p} = \set 1$
where $f^{-1}$ denotes the preimage under $f$.
Now $\set p$ is open in $\tau_p$ by definition of particular point topology.
By Closed Real Interval is not Open Set, $\set 1 = \closedint 1 1$ is not open.
Thus we have exhibited an open set whose preimage under $f$ is not open.
Hence $f$ is not continuous.
Since $f$ was an arbitrary injection, no injective path between $p$ and $q$ can exist.
Hence $T$ is not injectively path-connected.
{{qed}}
Theorem: Basis for Particular Point Space

Let $T = \struct {S, \tau_p}$ be a particular point space.
Consider the set $\BB$ defined as:
:$\BB = \set {\set {x, p}: x \in S} \cup \set p$
Then $B$ is a basis for $S$.

Proof:
Let $H \in \tau_p$ be open in $T$.
Then:
:$\forall y \in H: \exists \set {y, p} \in \BB$
which also holds when $y = p$ as $\set {y, p} = \set p \in \BB$.
Thus:
:$\ds H = \bigcup_{y \mathop \in H} \set {y, p}$
So $\BB$ is an analytic basis for $T$.
{{qed}}
It could equally well be shown that $\BB$ is also a synthetic basis for $T$.
Category:Particular Point Topologies
Category:Topological Bases
l1p3oezqv8ml3b4qu4t6j43wrrmz3nd
Theorem: Particular Point Space is Locally Path-Connected

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is locally path-connected.

Proof:
Consider the set $\BB$ defined as:
:$\BB = \set {\set {x, p}: x \in S}$
Then $\BB$ is a basis for $T$.
Now consider the open set $\set {p, q} \in \BB$.
Let $\mathbb I$ be the closed unit interval in $\R$.
Let $f: \mathbb I \to S$ be the mapping defined as:
:<nowiki>$\forall x \in \mathbb I: \map f x = \begin{cases}
p & : x \in \hointr 0 1 \\
q & : x = 1
\end{cases}$</nowiki>
Let $U \in \tau_p$.
Then either $q \in U$ or $q \notin U$.
If $q \in U$ then $f^{-1} \sqbrk U = \closedint 0 1$ which is open in $\mathbb I$ because $\closedint 0 1 = \mathbb I$.
If $q \notin U$ then $f^{-1} \sqbrk U = \hointr 0 1$ which is half open in $\R$ but open in $\mathbb I$.
So $f: \mathbb I \to S$ is a continuous mapping and so a path from $p$ to $q$.
As $\set {p, q}$ is any element of $\BB$ it follows that $T$ is locally path-connected.
{{qed}}
Theorem: Particular Point Space is Non-Meager

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is non-meager.

Proof:
Suppose $T$ were meager.
Then it would be a countable union of subsets which are nowhere dense in $T$.
Let $H \subseteq S$.
From Closure of Open Set of Particular Point Space, the closure of $H$ is $S$.
From the definition of interior, the interior of $S$ is $S$.
So the interior of the closure of $H$ is not empty.
So $T$ can not be the union of a countable set of subsets which are nowhere dense in $T$.
Hence $T$ is not meager and so by definition must be non-meager.
{{qed}}
Theorem: Particular Point Space is Non-Meager

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is non-meager.

Proof:
By definition of particular point space, any subset of $S$ which contains $p$ is open in $T$.
So $\left\{{p}\right\}$ itself is open in $T$.
That is, $p$ is an open point.
The result follows from Space with Open Point is Non-Meager.
{{qed}}
Theorem: Particular Point Space is Non-Meager

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is non-meager.

Proof:
By definition of particular point space, any subset of $S$ which contains $p$ is open in $T$.
{{AimForCont}} $T$ is meager.
By definition, $T$ is meager {{iff}} it is a countable union of subsets of $S$ which are nowhere dense in $T$.
At least one such nowhere dense subset $U$ of $S$ must contain $p$.
By definition, $U$ is nowhere dense in $T$ {{iff}}:
:$U^-$ contains no open set of $T$ which is non-empty
where $U^-$ denotes the closure of $U$.
By definition of particular point space, $U$ is open in $T$.
By Closure of Open Set of Particular Point Space, $U^- = S$.
But $S$ is itself open in $T$ and non-empty, and so $U$ is not nowhere dense.
From this contradiction it follows that $T$ is non-meager
{{qed}}
Theorem: Infinite Particular Point Space is not Countably Metacompact

Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not countably metacompact.

Proof:
{{Recall|Countably Metacompact Space|countably metacompact}}
{{:Definition:Countably Metacompact Space}}
;Countable case
Suppose $T$ is a countable particular point space.
Let $\CC$ be the open cover of $T$ defined as:
:$\CC = \set {\set {x, p}: x \in S}$
$\CC$ is countable and has no open refinement except $\CC$ itself.
But $\CC$ is not point finite because:
:$\forall U \in \CC: p \in U$
and $\CC$ is (countably) infinite.
{{qed|lemma}}
;Uncountable case
Now suppose $T$ is an uncountable particular point space.
Let $S' \subseteq S$ be a countable subset of $S$.
Let $\CC$ be the open cover of $T$ defined as:
:$\CC = \set {\set {x, p}: x \in S'} \cup \set {S \setminus S' \cup \set p}$
$\CC$ is also countable and has no open refinement except $\CC$ itself.
And similarly, $\CC$ is not point finite because:
:$\forall U \in \CC: p \in U$
and $\CC$ is (countably) infinite.
{{qed|lemma}}
Hence the result by definition of countably metacompact.
{{qed}}
Theorem: Exportation and Self-Conditional

:$p \land q \implies r \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$

Proof:
From the Rule of Exportation:
:$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$
Then by Self-Distributive Law for Conditional:
:$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$
{{qed}}
Category:Conditional
Category:Conjunction
mgps1nmi22at2befyb8y670spdkr545
Theorem: Łoś-Vaught Test

Let $T$ be a satisfiable $\LL$-theory with no finite models.
Let $T$ be $\kappa$-categorical for some infinite cardinal $\kappa \ge \card \LL$.
Then $T$ is complete.

Proof:
We prove the contrapositive.
The main idea is that if such a theory $T$ is incomplete, we can construct size $\kappa$ models which disagree on a sentence.
Suppose $T$ is not complete.
By the definition of complete, this means that there is some sentence $\phi$ such that both $T \not \models \phi$ and $T \not \models \neg \phi$.
This in turn means that both $T \cup \set {\neg \phi}$ and $T \cup \set \phi$ have models.
Since $T$ has no finite models, this means that $T \cup \set {\neg \phi}$ and $T \cup \set \phi$ both have infinite models.
We have that $\kappa$ is infinite and greater than the cardinality of the language.
We also have that these theories have infinite models.
From the Upward Löwenheim-Skolem Theorem one can prove that there are size $\kappa$ models $\MM_{\neg \phi}$ and $\MM_\phi$ of $T \cup \set {\neg \phi}$ and $T \cup \set \phi$ respectively.
In particular, $\MM_{\neg \phi}$ and $\MM_\phi$ are models of $T$ which disagree about the sentence $\phi$.
Such models cannot be isomorphic since isomorphisms preserve the truth of sentences.
Thus, $T$ is not $\kappa$-categorical.
{{qed}}
{{Namedfor|Jerzy Maria Michał Łoś|name2 = Robert Lawson Vaught|cat = Łoś|cat2 = Vaught}}
This result is also known as Vaught's Test.
Category:Model Theory for Predicate Logic
sdipbvnwnd8i7nj7dhus31v964o1r7l
Theorem: Sierpiński Space is Irreducible

Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is irreducible.

Proof:
A Sierpiński space is a particular point space by definition.
A Particular Point Space is Irreducible.
{{qed}}
Theorem: Sierpiński Space is Ultraconnected

Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is ultraconnected.

Proof:
The only closed sets of $T$ are $\O, \set 1$ and $\set {0, 1}$.
$\set 1$ and $\set {0, 1}$ are not disjoint.
Hence the result by definition of ultraconnected.
{{qed}}
Theorem: Sierpiński Space is Path-Connected

Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is path-connected.

Proof:
A Sierpiński space is a particular point space by definition.
A Particular Point Space is Path-Connected.
{{qed}}
Theorem: Sierpiński Space is not Injectively Path-Connected

Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is not injectively path-connected.

Proof:
A Sierpiński space is a particular point space by definition.
A Particular Point Space is not Injectively Path-Connected.
{{qed}}
Theorem: Sierpiński Space is T5

Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is a $T_5$ space.

Proof:
{{Recall|T5 Space|$T_5$ space}}
{{:Definition:T5 Space/Definition 1}}
The only closed sets in $T$ are $\O$, $\set 1$ and $\set {0, 1}$.
So there are no two separated sets $A, B \subseteq \set {0, 1}$.
So $T$ is a $T_5$ space vacuously.
{{qed}}
Theorem: Sierpiński Space is T4

Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is a $T_4$ space.

Proof:
We have that the Sierpiński Space is $T_5$.
Then we have that a $T_5$ Space is $T_4$.
{{qed}}
Theorem: Upward Löwenheim-Skolem Theorem

{{Disambiguate|Definition:Model|I suspect model of a first-order theory $\LL$, which is more specific than what is linked to now}}
Let $T$ be an $\LL$-theory with an infinite model.
Then for each infinite cardinal $\kappa \ge \card \LL$, there exists a model of $T$ with cardinality $\kappa$.

Proof:
{{Explain|Missing reference to the use of Axiom of Dependent Choice or Axiom of Choice/Zorn's Lemma}}
The idea is:
:to extend the language by adding $\kappa$ many new constants
and:
:to extend the theory by adding sentences asserting that these constants  are distinct.
It is shown that this new theory is finitely satisfiable using an infinite model of $T$.
Compactness then implies that the new theory has a model.
Some care needs to be taken to ensure that we  construct a model of exactly size $\kappa$.
Let $\LL^*$ be the language formed by adding new constants:
:$\set {c_\alpha: \alpha < \kappa}$ to $\LL$.
Let $T^*$ be the $\LL^*$-theory formed by adding the sentences:
:$\set {c_\alpha \ne c_\beta: \alpha, \beta < \kappa, \ \alpha \ne \beta}$
to $T$.
We show that $T^*$ is finitely satisfiable:
Let $\Delta$ be a finite subset of $T^*$.
Then $\Delta$ contains:
:finitely  many sentences from $T$
along with:
:finitely many sentences of the form $c_\alpha \ne c_\beta$ for the new constant symbols.
Since $T$ has an infinite model, it must have a model $\MM$ of cardinality at most:
:$\card \LL + \aleph_0$
This model already satisfies everything in $T$.
So, since we can find arbitrarily many distinct elements in it,  it can also be used as a model of $\Delta$ by interpreting the finitely many new constant symbols in $\Delta$ as distinct elements of $\MM$.
Since $T^*$ is finitely satisfiable, it follows by the Compactness Theorem that $T^*$ itself is satisfiable.
Since $T^*$ ensures the existence of $\kappa$ many distinct elements, this means it has models of size at least $\kappa$.
It can be proved separately or observed from the ultraproduct proof of the compactness theorem that $T^*$ then has a model $\MM^*$ of exactly size $\kappa$.
{{explain|That proof needs to be proved, and / or a link needs to be provided to that ultraproduct proof and its implications explained.}}
Since $T^*$ contains $T$, $\MM^*$ is a model of $T$ of size $\kappa$.
{{qed}}
{{Namedfor|Leopold Löwenheim|name2 = Thoralf Albert Skolem|cat = Löwenheim|cat2 = Skolem}}
Theorem: Compactness Theorem

Let $\LL$ be the language of predicate logic.
Let $T$ be a set of $\LL$-sentences.
Then $T$ is satisfiable {{iff}} $T$ is finitely satisfiable.

Proof:
{{Explain|Missing explicit reference to the use of Boolean Prime Ideal Theorem/Ultrafilter Lemma or Axiom of Choice/Zorn's Lemma}}
By definition, $T$ is finitely satisfiable means that every finite subset of $T$ is satisfiable.
Because the direction:
:$T$ satisfiable implies $T$ finitely satisfiable
is trivial, the proofs below all justify the converse:
:$T$ finitely satisfiable implies $T$ satisfiable.
Theorem: Compactness Theorem

Let $\LL$ be the language of predicate logic.
Let $T$ be a set of $\LL$-sentences.
Then $T$ is satisfiable {{iff}} $T$ is finitely satisfiable.

Proof:
Suppose $\mathbf H$ does '''not''' have a model.
By the Main Lemma of Propositional Tableaux, $\mathbf H$ has a tableau confutation $T$.
By Tableau Confutation contains Finite Tableau Confutation, $T$ may be assumed to be finite.
Hence the set $\mathbf H'$ of all WFFs in $\mathbf H$ used somewhere in $T$ is finite.
Now, let $T'$ be the labeled tree which is the same as $T$ but with root $\mathbf H'$ instead of $\mathbf H$.
Then $T'$ is a tableau confutation of $\mathbf H'$.
By the Tableau Confutation implies Unsatisfiable, $\mathbf H'$ has no models.
But this contradicts the assumption that all finite subsets of $\mathbf H$ have models.
Hence the result.
{{qed}}
Theorem: Compactness Theorem

Let $\LL$ be the language of predicate logic.
Let $T$ be a set of $\LL$-sentences.
Then $T$ is satisfiable {{iff}} $T$ is finitely satisfiable.

Proof:
If $\mathbf H$ is finite, the result is trivial.
So let $\mathbf H = \set {\mathbf A_n: n \in \N}$ be an enumeration of $\mathbf H$.
Define $\mathbf H_m = \set {\mathbf A_n: n \le m}$.
Let $T_1$ be the propositional tableau consisting of only a root node with hypothesis set $\mathbf H_1$.
For each $m \in \N$, construct $T_{m + 1}$ from $T_m$ by applying the Tableau Extension Lemma to $\mathbf H_{m + 1}$.
This provides us with a sequence $\sequence {T_m}_{m \mathop \in \N}$ of finished propositional tableaus.
The $T_m$ form an exhausting sequence of sets for the infinite propositional tableau $\ds T = \bigcup_{m \mathop = 1}^\infty T_m$.
If $T$ has a finished branch $\Gamma$, then by Finished Branch Lemma: Corollary:
:$\Phi \sqbrk \Gamma$ is satisfiable
and hence $\mathbf H$ is satisfiable (since $\mathbf H \subseteq \Phi \sqbrk \Gamma$).
Suppose $T$ has no finite finished branches.
Let $T'$ be the subtree of $T$ given by:
:$t \in T'$ {{iff}} $t$ is on a finished branch $\Gamma_m$ of every $T_m$ such that $t \in T_m$
Suppose the root node $r_T$ of $T$ were not in $T'$.
Then for some $T_m$, $r_T$ would not be on a finished branch of $T_m$.
But since $T_m$ is finished, every branch of $T_m$ is finished or contradictory.
Hence every branch of $T_m$ is contradictory.
But then $T_m$ is a tableau confutation of $\mathbf H_m$.
By Tableau Confutation implies Unsatisfiable, this contradicts the assumption that $\mathbf H_m$ is satisfiable.
Therefore, $r_T$ is in $T'$.
Suppose $T'$ were finite.
Let $t$ be a leaf of $T'$, which exists by Finite Tree has Leaf Nodes.
Suppose $t$ were a leaf of $T$.
Then $\Gamma_t$, the branch of $T$ identified by Leaf of Rooted Tree is on One Branch, is finished.
For, as any $T_m$ is a subtree of $T$, $\Gamma_t$ is the only branch of $T_m$ such that $t \in \Gamma_t$.
The conclusion follows from the definition of $T'$.
But then $\Gamma_t$ would be a finite finished branch of $T$, a contradiction.
Therefore, $T'$ cannot be finite.
Hence, $T'$ is a finitely branching tree.
By König's Tree Lemma, $T'$ has an infinite branch $\Gamma$.
By definition of $T'$, the branch $\Gamma_m := \Gamma \cap T_m$ of $T_m$ is finished for each $m \in \N$.
If $\Gamma$ were contradictory, then $\mathbf A \in \Gamma$ and $\neg\mathbf A \in \Gamma$ for some WFF $\mathbf A$.
But then $\mathbf A, \neg \mathbf A \in \Gamma_m$ for some $m \in \N$, contradicting that $\Gamma_m$ is finished.
Also, if $\mathbf A$ occurs on $\Gamma$, then it occurs on some $\Gamma_m$.
Since $\Gamma_m$ is finished, it follows that $\mathbf A$ is used on $\Gamma_m$, and hence on $\Gamma$.
In conclusion, $\Gamma$ is finished.
As established above, it follows that $\mathbf H$ is satisfiable for boolean interpretations.
{{qed}}