avewright/proofwiki-math huggingface dataset:
Theorem: Finite Complement Space is T1

Let $T = \struct {S, \tau}$ be a finite complement topology on any set $S$ which contains at least two points.
Then $T$ is a $T_1$ space.

Proof:
Let $x, y \in S$ such that $x \ne y$.
This is possible because $S$ contains at least two points.
By definition, both $\set x$ and $\set y$ are finite.
Hence by definition of finite complement topology:
:$\relcomp S {\set x}$ is an open set of $T$ containing $y$ but not $x$
:$\relcomp S {\set y}$ is an open set of $T$ containing $x$ but not $y$
Hence the result from definition of $T_1$ space.
{{qed}}
Theorem: Finite Complement Space is Irreducible

Let $T = \struct {S, \tau}$ be a finite complement space on an infinite set $S$.
Then $T$ is irreducible.

Proof:
Let $U_1, U_2 \in \tau$ be non-empty open sets of $T$.
By definition, $S$ is infinite.
By definition, $\relcomp S {U_1}$ and $\relcomp S {U_2}$ are finite.
From Complement of Finite Subset of Infinite Set is Infinite, it follows that $U_1$ and $U_2$ are both infinite.
From Infinite Subset of Finite Complement Space Intersects Open Sets:
:$U_1 \cap U_2\ne \O$
That is, $U_1$ and $U_2$ intersect each other.
As $U_1$ and $U_2$ are arbitrary, the result follows by definition of irreducible space.
{{qed}}
Theorem: Infinite Subset of Finite Complement Space Intersects Open Sets

Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Let $H \subseteq S$ be an infinite subset of $S$.
Then the intersection of $H$ with any non-empty open set of $T$ is infinite.

Proof:
Let $U \in \tau$ be any non-empty open set of $T$.
Then by definition of finite complement topology:
:$\relcomp S U$ is finite.
We have that:
:$H = H \cap \paren {U \cup \relcomp S U} = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$
{{AimForCont}} $H \cap U$ is finite.
From Intersection is Subset:
:$H \cap \relcomp S U \subseteq \relcomp S U$
From Subset of Finite Set is Finite:
:$H \cap \relcomp S U$ is also finite.
Thus we have:
:$H = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$
which is the union of two finite sets.
From Union of Finite Sets is Finite, $H$ is finite.
It is a contradiction that $H$ is infinite and finite at the same time.
Hence by Proof by Contradiction, $H \cap U$ must be infinite.
{{qed}}
Theorem: Irreducible Hausdorff Space is Singleton

Let $T = \struct {S, \tau}$ be a non-empty topological space which is irreducible and Hausdorff.
Then $S$ is a singleton.

Proof:
Suppose $S$ has exactly one element.
Then by definition $T = \struct {S, \tau}$ is the trivial topological space.
Hence, from Trivial Topological Space is Irreducible, $S$ is irreducible.
{{qed|lemma}}
Suppose $S$ has at least two distinct elements:
:$x, y \in S, x \ne y$
By definition of irreducible space:
:there are no two disjoint open sets $T$ such that $x$ is in one and $y$ is in the other:
:$\nexists U_1, U_2 \in \tau: U_1 \cap U_2 = \O, x \in U_1, y \in U_2$
This contradicts the fact that $T$ is Hausdorff.
Thus $S$ has only one element.
{{qed}}
Category:Irreducible Spaces
Category:Hausdorff Spaces
kmociah4inbzb9e60xvqzgfesqrjvvl
Theorem: Finite Complement Space is not T2

Let $T = \struct {S, \tau}$ be the finite complement topology on an infinite set $S$.
Then $T$ is not a $T_2$ (Hausdorff) space.

Proof:
We have:
:Finite Complement Space is Irreducible
:Irreducible Hausdorff Space is Singleton
{{qed}}
Theorem: Double Pointed Finite Complement Topology is Compact

Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is a compact space.

Proof:
The Finite Complement Topology is Compact.
Also, a Finite Topological Space is Compact.
So $D$, defined as the indiscrete space on a doubleton, is also compact.
The result follows directly from Tychonoff's Theorem.
{{qed}}
Theorem: Double Pointed Finite Complement Topology fulfils no Separation Axioms

Let $T = \struct {S, \tau}$ be a finite complement space on an infinite set $S$.
Let $T \times D$ be the double pointed topological space on $T$.
Then $T \times D$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space:
=== Double Pointed Finite Complement Space is not $T_0$ ===
{{:Double Pointed Finite Complement Space is not T0}}
=== Double Pointed Finite Complement Space is not $T_1$ ===
{{:Double Pointed Finite Complement Space is not T1}}
=== Double Pointed Finite Complement Space is not $T_2$ ===
{{:Double Pointed Finite Complement Space is not T2}}
=== Double Pointed Finite Complement Space is not $T_3$ ===
{{:Double Pointed Finite Complement Space is not T3}}
=== Double Pointed Finite Complement Space is not $T_4$ ===
{{:Double Pointed Finite Complement Space is not T4}}
=== Double Pointed Finite Complement Space is not $T_5$ ===
{{:Double Pointed Finite Complement Space is not T5}}

Proof:
From Separation Axioms on Double Pointed Topology, we have that:
:$T \times D$ is not a $T_0$ space.
:$T \times D$ is not a $T_1$ space.
:$T \times D$ is not a $T_2$ space.
Also from Separation Axioms on Double Pointed Topology, we have that:
:$T \times D$ is a $T_3$ space {{iff}} $T$ is a $T_3$ space
:$T \times D$ is a $T_4$ space {{iff}} $T$ is a $T_4$ space
:$T \times D$ is a $T_5$ space {{iff}} $T$ is a $T_5$ space
From Finite Complement Space is not $T_3$:
:$T$ is not a $T_3$ space
From Finite Complement Space is not $T_4$:
:$T$ is not a $T_4$ space
From Finite Complement Space is not $T_5$:
:$T$ is not a $T_5$ space
Hence the result.
{{qed}}
Theorem: Sum Over Divisors of von Mangoldt is Logarithm

Let $\Lambda$ be von Mangoldt's function.
Then for $n \ge 1$:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \ln n$

Proof:
Let $n \ge 1$, and by the Fundamental Theorem of Arithmetic write $n = p_1^{e_1} \cdots p_k^{e_k}$ with $p_1, \ldots, p_k$ distinct primes and $e_1, \ldots, e_k > 0$.
Now $d \divides n$ if any only if $d = p_1^{f_1} \cdots p_k^{f_k}$ with $0 \le f_i \le e_i$ for $i = 1, \ldots, k$.
By the definition of $\Lambda$, for such $d$ we have $\map \Lambda d \ne 0$ {{iff}} there is exactly one $i \in \set {1, \ldots, k}$ such that $f_i > 0$.
If this is the case, let $d = p_i^{f_i}$.
Then:
:$\map \Lambda d = \ln p_i$
Therefore:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \sum_{i \mathop = 1}^k e_i \ln p_i$
Also, we have:
{{begin-eqn}}
{{eqn | l = \ln n
      | r = \ln (p_1^{e_1} \cdots p_k^{e_k})
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^k e_i \ln p_i
      | c = Sum of Logarithms
}}
{{end-eqn}}
Thus we indeed have:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \ln n$
{{qed}}
Category:Analytic Number Theory
Category:Von Mangoldt Function
iuyuq7ucnmiatdzqlqkbfuhrvxvyf5c
Theorem: Multiplicative Function that Converges to Zero on Prime Powers

Let $f$ be a multiplicative function such that:
:$\ds \lim_{p^k \mathop \to \infty} \map f {p^k} = 0$
where $p^k$ runs though ''all'' prime powers.
Then:
:$\ds \lim_{n \mathop \to \infty} \map f n = 0$
where $n$ runs through the integers.

Proof:
By hypothesis, there exist only finitely many prime powers $p^k$ such that $\size {\map f {p^k} } > 1$.
Let $\ds A = \prod_{\size {\map f {p^k} } \mathop > 1} \size {\map f {p^k} }$.
Thus $A \ge 1$.
Let $0 < \dfrac \epsilon A$. 
There exist only finitely many prime powers $p^k$ such that $\size {\map f {p^k} } > \dfrac \epsilon A$.
Therefore there are only finitely many integers $n$ such that:
:$\size {\map f {p^k} } > \dfrac \epsilon A$
for every prime power $p^k$ that divides $n$.
Therefore if $n$ is sufficiently large there exists a prime power $p^k$ that divides $n$ such that:
:$\size {\map f {p^k} } < \dfrac \epsilon A$
Therefore $n$ can be written as:
:$\ds n = \prod_{i \mathop = 1}^ r p_i^{k_i} \prod_{i \mathop = r + 1}^{r + s} p_i^{k_i} \prod_{i \mathop = r + s + 1}^{r + s + t} p_i^{k_i}$
where $t \ge 1$ and:
{{begin-eqn}}
{{eqn | l = 1
      | o = \le
      | m = \size {\map f {p_i^{k_i} } }
      | mo= 
      | r = 
      | c = for $i = 1, \ldots, r$
}}
{{eqn | l = \frac \epsilon A
      | o = \le
      | m = \size {\map f {p_i^{k_i} } }
      | mo= <
      | r = 1
      | c = for $i = r + 1, \ldots, r + s$
}}
{{eqn | o = 
      | m = \size {\map f {p_i^{k_i} } }
      | mo= <
      | r = \frac \epsilon A
      | c = for $i = r + s + 1, \ldots, r + s + t$
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \size {\map f n}
      | r = \prod_{i \mathop = 1}^r \map f {p_i^{k_i} } \prod_{i \mathop = r + 1}^{r + s} \map f {p_i^{k_i} } \prod_{i \mathop = r + s + 1}^{r + s + t} \map f {p_i^{k_i} }
      | c = because $f$ is multiplicative
}}
{{eqn | o = <
      | r = A \paren {\frac \epsilon A}^t
      | c = 
}}
{{eqn | o = <
      | r = \epsilon
      | c = because $t \ge 1$
}}
{{end-eqn}}
This shows that $\map f n$ can be made arbitrarily small for sufficiently large $n$.
{{qed}}
Category:Analytic Number Theory
Category:Multiplicative Functions
scpfl4zmfe3fpx66ln65we7b0x1dtov
Theorem: Finite T1 Space is Discrete

Let $S$ be a finite set.
Let $T = \struct {S, \tau}$ be a $T_1$ space.
Then $\tau$ is the discrete topology on $S$.

Proof:
Let $T = \struct {S, \tau}$ be a $T_1$ space on a finite set $S$.
Let $U \subseteq S$ be any subset of $S$. 
Let $H = \relcomp S U$ be the complement of $U$ relative to $S$.
Then by Relative Complement of Relative Complement we have that $U = \relcomp S H$.
We can write $H$ as:
:$\ds H = \bigcup_{x \mathop \in H} \set x$
From Equivalence of Definitions of $T_1$ Space, $\forall x \in H: \set x$ is closed in $T$.
As $S$ is a finite set, it follows that $H$ is a finite union of closed sets of $T$.
By Topology Defined by Closed Sets, $H$ is therefore closed in $T$.
By definition of closed set, $U = \relcomp S H$ is open in $T$.
As $U$ is arbitrary, it follows that:
:$\forall U \subseteq S: S \in \tau$
by definition of a topology.
Hence the result, by definition of the discrete topology.
{{qed}}
Theorem: Finite T1 Space is Discrete

Let $S$ be a finite set.
Let $T = \struct {S, \tau}$ be a $T_1$ space.
Then $\tau$ is the discrete topology on $S$.

Proof:
We have from Finite Complement Topology is Minimal $T_1$ Topology that $\tau$ space is an expansion of a finite complement space.
But from the definition, any such finite complement space on a finite set is discrete.
But as the Discrete Topology is Finest Topology, $\tau$ must itself be the discrete topology on $S$.
{{qed}}
Theorem: Finite Complement Topology is Minimal T1 Topology

Let $T = \struct {S, \tau}$ be a finite complement space.
Let $\tau'$ be a topology on $S$ such that $T' = \struct {S, \tau'}$ is a $T_1$ space.
Then $\tau$ is comparable with $\tau'$ such that $\tau$ is coarser than $\tau'$.
That is, of all the topologies on $S$ fulfilling the $T_1$ separation axiom, the finite complement space is the smallest.
Thus the finite complement topology is known as the '''minimal $T_1$ topology''' on any given set.

Proof:
Let $T = \struct {S, \tau}$ be the finite complement space on $S$.
Let $U \in \tau$ be any open set of $T$.
Let $H = \relcomp S U$ be the complement of $U$ relative to $S$.
By definition of finite complement topology, $H \subseteq S$ is a finite subset of $S$.
Let $T' = \struct {S, \tau'}$ be any arbitrary $T_1$ space on $S$.
From Equivalence of Definitions of $T_1$ Space:
:$\forall x \in S: \set x$ is closed in $T'$.
We can write $H$ as:
:$\ds H = \bigcup_{x \mathop \in H} \set x$
As $H$ is finite, it follows that $H$ is a finite union of closed sets of $T'$.
By Topology Defined by Closed Sets, $H$ is therefore closed in $T'$.
By Relative Complement of Relative Complement we have that $U = \relcomp S H$.
So, by definition of closed set, we have that $U$ is open in $T'$.
So we have shown that for any arbitrary $U \in \tau$ it follows that $U \in \tau'$.
So $\tau \subseteq \tau'$ and so by definition $\tau$ is coarser than $\tau'$.
{{qed}}
Theorem: Finite Complement Space is Connected

Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is a connected space.

Proof:
We have that a Finite Complement Space is Irreducible.
The result follows from Irreducible Space is Connected.
{{qed}}
Theorem: Finite Complement Space is Locally Connected

Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is a locally connected space.

Proof:
We have that a Finite Complement Space is Irreducible.
The result follows from Irreducible Space is Locally Connected.
{{qed}}
Theorem: Mapping Induces Partition on Domain

Let $f: S \to T$ be a mapping.
Let $F$ be defined as:
:$F = \set {\map {f^{-1} } x: x \in T}$
where $\map {f^{-1} } x$ is the preimage of $x$.
Then $F$ is a partition of $S$.

Proof:
Let $\RR_f \subseteq S \times S$ be the relation induced by $f$:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Then from Relation Induced by Mapping is Equivalence Relation, $\RR_f$ is an equivalence relation.
The result follows from Relation Partitions Set iff Equivalence.
{{qed}}
Category:Mapping Theory
np3xh2gs37ojqomi640ch6hwvprcgcw
Theorem: Countable Finite Complement Space is not Path-Connected

Let $T = \struct {S, \tau}$ be a countable finite complement topology.
Then $T$ is not path-connected.

Proof:
{{AimForCont}} $f: \closedint 0 1 \to S$ is a path on $T$.
Then $f$ is by definition continuous.
By definition of the finite complement topology, for all $x \in S$, the set $\set x$ is closed.
Now consider the set:
:$F = \set {\map {f^{-1} } x: x \in S}$
From Continuity Defined from Closed Sets, each of the elements of $F$ is closed.
Also, from Mapping Induces Partition on Domain, the elements of $F$ are pairwise disjoint.
As $T$ is a countable finite complement topology, its underlying set $S$ is countably infinite by definition.
Hence $F$ is also countably infinite by nature of $f$ being a mapping.
Furthermore, we have $\ds \bigcup F = \closedint 0 1$.
Hence $F$ is a countably infinite set of pairwise disjoint closed sets whose union is $\closedint 0 1$.
From Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets, this is impossible.
From this contradiction, $f$ cannot be continuous, and so cannot be a path on $T$.
So $T$ cannot be path-connected.
{{qed}}
Theorem: Countable Finite Complement Space is not Locally Path-Connected

Let $T = \struct {S, \tau}$ be a countable finite complement topology.
Then $T$ is not locally path-connected.

Proof:
Let $\BB$ be a basis for $T$.
Let $B \in \BB$.
{{AimForCont}} $f: \closedint 0 1 \to B$ is a path on $T$.
Then $f$ is by definition continuous.
Now consider the set:
:$F = \set {\map {f^{-1} } x: x \in S}$
From Continuity Defined from Closed Sets, each of the elements of $F$ is closed.
Also, from Mapping Induces Partition on Domain, the elements of $F$ are pairwise disjoint.
From Basis of Countable Finite Complement Topology consists of Countably Infinite Sets, $B$ is a countably infinite set.
Hence $F$ is also countably infinite by nature of $f$ being a mapping.
Furthermore, we have $\ds \bigcup F = \closedint 0 1$.
Hence $F$ is a countably infinite set of pairwise disjoint closed sets whose union is $\closedint 0 1$.
From Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets, this is impossible.
From this contradiction, $f$ cannot be continuous, and so cannot be a path on $B$.
So $B$ cannot be path-connected.
So, by definition, $T$ is not locally path-connected.
{{qed}}
Theorem: Injective Path-Connectedness in Uncountable Finite Complement Space

Let $T = \struct {S, \tau}$ be a finite complement topology on a uncountable set $S$.
If the Continuum Hypothesis is accepted as true, then:
:$T$ is injectively path-connected
:$T$ is locally injectively path-connected.

Proof:
Let $a, b \in S$ such that $a \ne b$.
Let us assume the truth of the Continuum Hypothesis.
Then $a$ and $b$ are contained in a subset $X \subseteq S$ whose cardinality is the same as that of $\closedint 0 1$.
So by definition of cardinality we can set up a bijection $f: \closedint 0 1 \leftrightarrow X$ such that $\map f 0 = a$ and $\map f 1 = b$.
We have that Finite Complement Space is $T_1$.
Hence by definition of $T_1$ space, each element of $S$ is closed
Hence $f$ is continuous and therefore an injective path in $T$.
As $a$ and $b$ are arbitrary, $T$ is injectively path-connected.
Now let $\BB$ be a basis for $T$.
Let $B \in \BB$ and let $a, b \in B$.
By definition of a finite complement topology on a uncountable set, $B$ is uncountable.
So, by the same argument as above, we can set up an injective path $f$ between any points of $B$.
So $T$ is locally injectively path-connected.
{{qed}}
Theorem: Countable Complement Topology is Topology

Let $T = \struct {S, \tau}$ be a countable complement space.
Then $\tau$ is a topology on $T$.

Proof:
By definition, we have that $\O \in \tau$.
We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially countable.
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
{{begin-eqn}}
{{eqn | l = H
      | r = A \cap B
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \relcomp S H
      | r = \relcomp S {A \cap B}
      | c = 
}}
{{eqn | r = \relcomp S A \cup \relcomp S B
      | c = De Morgan's Laws: Complement of Intersection
}}
{{end-eqn}}
But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ are both countable.
Hence from Countable Union of Countable Sets is Countable their union is also countable and so $\relcomp S H$ is countable.
So $H = A \cap B \in \tau$ as its complement is countable.
Now let $\UU \subseteq \tau$.
Then:
:$\ds \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
by De Morgan's Laws: Complement of Union.
But as:
:$\forall U \in \UU: \relcomp S U \in \tau$
each of the $\relcomp S U$ is countable.
Hence so is their intersection.
So $\ds \relcomp S {\bigcup \UU}$ is countable which means:
:$\ds \bigcup \UU \in \tau$
So $\tau$ is a topology on $T$.
{{qed}}
Theorem: Countable Complement Topology is Expansion of Finite Complement Topology

Let $T = \struct {S, \tau}$ be the countable complement topology on an infinite set $S$.
Let $T' = \struct {S, \tau'}$ be the finite complement topology on the same infinite set $S$.
Then $\tau$ is an expansion of $\tau'$.

Proof:
Let $U \in \tau', U \ne \O$.
Then $\relcomp S U$ is finite by definition of finite complement topology.
Then by definition of countable complement topology, we have that $U \in \tau$.
So $\tau' \subseteq \tau$ and hence the result by definition of expansion.
{{qed}}
Theorem: Countable Complement Space is T1

Let $T = \struct {S, \tau}$ be a countable complement topology.
Then $T$ is a $T_1$ space.

Proof:
We have that the Countable Complement Topology is Expansion of Finite Complement Topology. 
We also have that a Finite Complement Space is $T_1$.
Then from Separation Properties Preserved under Expansion, we have that $T$ is a $T_1$ space.
{{qed}}
Theorem: Uncountable Subset of Countable Complement Space Intersects Open Sets

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $H \subseteq S$ be an uncountable subset of $S$.
Then the intersection of $H$ with an arbitrary non-empty open set of $T$ is uncountable.

Proof:
Let $U \in \tau$ be any non-empty open set of $T$.
Then $\relcomp S U$ is countable.
Suppose $H \cap U = \O$.
Then from Intersection with Complement is Empty iff Subset it follows that $H \subseteq \relcomp S U$ and so $H$ is countable.
So if $H$ is uncountable it is bound to have a non-empty intersection with every open set in $T$.
{{finish|It still needs to be shown that $U \cap H$ is uncountable {{iff}} $H$ is uncountable.}}
Theorem: Countable Complement Space is not T2

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a $T_2$ space.

Proof:
We have:
:Countable Complement Space is Irreducible
:Irreducible Hausdorff Space is Singleton
{{qed}}
Theorem: Countable Complement Space is Irreducible

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is an irreducible space.

Proof:
Let $U_1, U_2 \in \tau$ be non-empty open sets of $T$.
We have that both $U_1$ and $U_2$ are both uncountable if $S$ is.
From Uncountable Subset of Countable Complement Space Intersects Open Sets, they intersect each other.
Hence the result from definition of irreducible space.
{{qed}}
Theorem: Higher-Aleph Complement Topology is Topology

Let $T = \struct {S, \tau}$ be an $\aleph_n$ complement space.
Then $\tau$ is a topology on $T$.

Proof:
By definition, we have that $\O \in \tau$.
We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially finite.
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
{{begin-eqn}}
{{eqn | l = H
      | r = A \cap B
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \relcomp S H
      | r = \relcomp S {A \cap B}
      | c = 
}}
{{eqn | r = \relcomp S A \cup \relcomp S B
      | c = De Morgan's Laws: Complement of Intersection
}}
{{end-eqn}}
But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ both have a cardinality strictly smaller than that of $S$.
Hence their union also has a cardinality strictly smaller than that of $S$.
So $\relcomp S H$ has a cardinality strictly smaller than that of $S$.
So $H = A \cap B \in \tau$ as its complement has a cardinality strictly smaller than that of $S$.
Now let $\UU \subseteq \tau$.
Then from De Morgan's laws: Complement of Union:
:$\ds \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
But as:
:$\forall U \in \UU: \relcomp S U \in \tau$
each of the $\relcomp S U$ has a cardinality strictly smaller than that of $S$.
Hence so has their intersection.
So $\ds \relcomp S {\bigcup \UU}$ has a cardinality strictly smaller than that of $S$, which means:
:$\ds \bigcup \UU \in \tau$
So $\tau$ is a topology on $T$.
{{qed}}
Theorem: Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets

Let $\mathbb I = \closedint 0 1$ be the closed unit interval.
Then $\mathbb I$ cannot be expressed as the union of a countably infinite set of pairwise disjoint closed sets.

Proof:
{{AimForCont}} $\ds \mathbb I = \bigcup_{i \mathop = 1}^\infty C_i$ where $\set {C_i}$ is a set of pairwise disjoint closed sets.
Let:
:$\ds B = \bigcup \partial C_i = \mathbb I \setminus \bigcup C_i^\circ$
where $\partial C_i$ is the boundary of $C_i$ and $C_i^\circ$ is the interior of $C_i$.
Let $J \subseteq \mathbb I$ be a subinterval of $\mathbb I$.
Then $J$ is non-meager, and so some $C_k$ is dense in some open interval $L \subseteq J$.
Since $C_k$ is closed we have that $L \subseteq C_k^\circ$ and so $L \cap B = \O$.
So $J$ contains an open subset $L$ disjoint from $B$.
So $B$ is nowhere dense in $\mathbb I$.
So every open interval $U$ containing some $x \in \partial C_j$ must intersect $B \setminus \partial C_j$.
This is because $U$ is an open neighborhood of $x$ and therefore contains some point $u \in \mathbb I \setminus C_j$, say $u \in C_m$.
Then if $U \cap B \cap \partial C_m = \O$ it follows that $C_m^\circ \cap U$ is a non-empty open set in $U$. which is also relatively closed.
{{explain|Closed relative to what?}}
Now $B$ is itself non-meager, since it is a closed subset of $\mathbb I$.
Thus some $\partial C_k$ is dense in some non-empty open set $U \cap B$, for some open interval $U$ in $\mathbb I$.
Since $\partial C_k$ is closed, this means $\partial C_k \cap U = B \cap U$.
But this is impossible, since if $U \cap C_k \ne \O$ then $U \cap \paren {B \setminus \partial C_k} \ne \O$.
This contradiction shows that $\mathbb I$ cannot be expressed as $\ds \bigcup_{i \mathop = 1}^\infty C_i$.
{{qed}}
{{proofread}}