avewright/proofwiki-math huggingface dataset:
Theorem: Cardinality of Set Union/General Case

Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of sets.
Then:
{{begin-eqn}}     
{{eqn | l = \card {\bigcup_{i \mathop = 1}^n S_i}
      | r = \sum_{i \mathop = 1}^n \card {S_i}
      | c =
}}
{{eqn | o =
      | ro= -
      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \card {S_i \cap S_j}
      | c =
}}
{{eqn | o =
      | ro= +
      | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \card {S_i \cap S_j \cap S_k}
      | c =
}}
{{eqn | o =
      | r = \cdots
      | c =
}}
{{eqn | o =
      | ro= +
      | r = \paren {-1}^{n - 1} \card {\bigcap_{i \mathop = 1}^n S_i}
      | c =
}}
{{end-eqn}}

Proof:
By Cardinality is Additive Function, we can apply the Inclusion-Exclusion Principle:
If $f: \SS \to \R$ is an additive function, then:
{{Explain|What is $\SS$?}}
{{begin-eqn}}     
{{eqn | l = \map f {\bigcup_{i \mathop = 1}^n S_i}
      | r = \sum_{i \mathop = 1}^n \map f {S_i}
      | c =
}}
{{eqn | o =
      | ro= -
      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {S_i \cap S_j}
      | c =
}}
{{eqn | o =
      | ro= +
      | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {S_i \cap S_j \cap S_k}
      | c =
}}
{{eqn | o =
      | r = \cdots
      | c =
}}
{{eqn | o =
      | ro= +
      | r = \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n S_i}
      | c =
}}
{{end-eqn}}
{{qed}}
Theorem: Structure of Inverse Completion of Commutative Semigroup

Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then:
:$T = S \circ' C^{-1}$
where:
:$C^{-1}$ is the inverse of $C$ in $T$
:$S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$.

Proof:
Let $a \in C$.
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = S
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = x \circ \paren {a \circ' a^{-1} }
      | c = {{Defof|Invertible Element}}
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \paren {x \circ a} \circ' a^{-1}
      | c = {{Defof|Associative Operation}}
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = S \circ' C^{-1}
      | c = {{Defof|Subset Product}}
}}
{{eqn | ll= \leadsto
      | l = S
      | o = \subseteq
      | r = S \circ' C^{-1}
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | r = C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y^{-1}
      | o = \in
      | r = C^{-1}
      | c = {{Defof|Inverse of Subset of Monoid}}
}}
{{eqn | ll= \leadsto
      | l = y^{-1}
      | r = a \circ' a^{-1} \circ' y^{-1}
      | c = {{Defof|Invertible Element}}
}}
{{eqn | ll= \leadsto
      | l = y^{-1}
      | r = a \circ' \paren {y \circ' a}^{-1}
      | c = Inverse of Product
}}
{{eqn | ll= \leadsto
      | l = y^{-1}
      | r = a \circ' \paren {y \circ a}^{-1}
      | c = {{Defof|Extension of Operation}}
}}
{{eqn | ll= \leadsto
      | l = y^{-1}
      | o = \in
      | r = S \circ' C^{-1}
      | c = {{Defof|Subset Product}}
}}
{{eqn | ll= \leadsto
      | l = C^{-1}
      | o = \subseteq
      | r = S \circ' C^{-1}
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
Thus, as:
:$C^{-1} \subseteq S \circ' C^{-1}$
and:
:$S \subseteq S \circ' C^{-1}$
by Union is Smallest Superset it follows that:
:$S \cup C^{-1} \subseteq S \circ' C^{-1}$
From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup:
:$S \circ' C^{-1}$ is a commutative semigroup.
So $S \circ' C^{-1}$ is a semigroup which contains $S \cup C^{-1}$.
By definition of generator of semigroup, it follows that:
:$\gen {S \cup C^{-1} } \subseteq S \circ' C^{-1}$
Let $z = x \circ' y^{-1} \in S \circ' C^{-1}$.
Then by definition:
:$x \in S$
and:
:$y^{-1} \in C^{-1}$
and so by definition of generator of semigroup:
:$x \circ' y^{-1} \in \gen {S \cup C^{-1} }$
Thus:
$S \circ' C^{-1} \subseteq \gen {S \cup C^{-1} }$
By definition of set equality:
:$S \circ' C^{-1} = \gen {S \cup C^{-1} }$
and so by definition of the inverse completion:
:$T = S \circ' C^{-1}$
{{qed}}
Theorem: Inverse in Monoid is Unique

Let $\struct {S, \circ}$ be a monoid.
Then an element $x \in S$ can have at most one inverse for $\circ$.

Proof:
Let $e$ be the identity element of $\struct {S, \circ}$.
Suppose $x \in S$ has two inverses: $y$ and $z$.
Then:
{{begin-eqn}}
{{eqn | l = y
      | r = y \circ e
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = y \circ \paren {x \circ z}
      | c = {{Defof|Inverse Element}}
}}
{{eqn | r = \paren {y \circ x} \circ z
      | c = {{MonoidAxiom|1}}
}}
{{eqn | r = e \circ z
      | c = {{Defof|Inverse Element}}
}}
{{eqn | r = z
      | c = {{Defof|Identity Element}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = y
      | r = e \circ y
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \paren {z \circ x} \circ y
      | c = {{Defof|Inverse Element}}
}}
{{eqn | r = z \circ \paren {x \circ y}
      | c = {{MonoidAxiom|1}}
}}
{{eqn | r = z \circ e
      | c = {{Defof|Inverse Element}}
}}
{{eqn | r = z
      | c = {{Defof|Identity Element}}
}}
{{end-eqn}}
So whichever way round you do it, $y = z$ and the inverse of $x$ is unique.
{{qed}}
Theorem: Inverse of Inverse/Monoid

Let $\struct {S, \circ}$ be a monoid.
Let $x \in S$ be invertible, and let its inverse be $x^{-1}$.
Then $x^{-1}$ is also invertible, and:
:$\paren {x^{-1} }^{-1} = x$

Proof:
By Inverse in Monoid is Unique, any inverse of $x$ is unique, and can be denoted $x^{-1}$.
From Inverse of Inverse in General Algebraic Structure:
:$x^{-1}$ is invertible and its inverse is $x$.
That is:
:$\paren {x^{-1} }^{-1} = x$
{{qed}}
Theorem: Inverse of Product/Monoid

Let $\struct {S, \circ}$ be a monoid whose identity is $e$.
Let $a, b \in S$ be invertible for $\circ$, with inverses $a^{-1}, b^{-1}$.
Then $a \circ b$ is invertible for $\circ$, and:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }
      | r = \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}
      | c = {{MonoidAxiom|1}}
}}
{{eqn | r = \paren {a \circ \paren {b \circ b^{-1} } } \circ a^{-1}
      | c = {{MonoidAxiom|1}}
}}
{{eqn | r = \paren {a \circ e} \circ a^{-1}
      | c = {{Defof|Inverse Element}}
}}
{{eqn | r = a \circ a^{-1}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = e
      | c = {{Defof|Inverse Element}}
}}
{{end-eqn}}
Similarly for $\paren {b^{-1} \circ a^{-1} } \circ \paren {a \circ b}$.
{{qed}}
Theorem: Right Identity Element is Idempotent

Let $\struct {S, \circ}$ be an algebraic structure.
Let $e_R \in S$ be a right identity with respect to $\circ$.
Then $e_R$ is idempotent under $\circ$.

Proof:
By the definition of a right identity:
:$\forall x \in S: x \circ e_R = x$
Thus in particular:
:$e_R \circ e_R = e_R$
Therefore $e_R$ is idempotent under $\circ$.
{{qed}}
Theorem: Left Identity Element is Idempotent

Let $\struct {S, \circ}$ be an algebraic structure.
Let $e_L \in S$ be a left identity with respect to $\circ$.
Then $e_L$ is idempotent under $\circ$.

Proof:
By the definition of a left identity:
:$\forall x \in S: e_L \circ x = x$
Thus in particular:
:$e_L \circ e_L = e_L$
Therefore $e_L$ is idempotent under $\circ$.
{{qed}}
Theorem: More than one Right Identity then no Left Identity

Let $\struct {S, \circ}$ be an algebraic structure.
If $\struct {S, \circ}$ has more than one right identity, then it has no left identity.

Proof:
Let $\struct {S, \circ}$ be an algebraic structure with more than one right identity.
Take any two of these, and call them $e_{R_1}$ and $e_{R_2}$, where $e_{R_1} \ne e_{R_2}$.
Suppose $\struct {S, \circ}$ has a left identity.
Call this left identity $e_L$.
Then, by the behaviour of $e_L$, $e_{R_1}$ and $e_{R_2}$:
:$e_{R_1} = e_L \circ e_{R_1} = e_L$
:$e_{R_2} = e_L \circ e_{R_2} = e_L$
So $e_{R_1} = e_L = e_{R_2}$, which contradicts the supposition that $e_{R_1}$ and $e_{R_2}$ are different.
Therefore, in an algebraic structure with more than one right identity, there can be no left identity.
{{qed}}
Theorem: Inverse Completion is Commutative Monoid

Let $\left({S, \circ}\right)$ be a commutative semigroup.
Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$.
Let $\left({T, \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$.
Then $\left({T, \circ'}\right)$ is a commutative monoid.

Proof:
From Inverse Completion is Commutative Semigroup:
:$\left({T, \circ'}\right)$ is a commutative semigroup.
By definition of inverse completion:
:$\forall x \in C: \exists x^{-1} \in C: x \circ x^{-1} \in T$
Thus by definition of inverse element:
:$e := x \circ x^{-1}$ is an identity element of $T$.
By Identity is Unique, $e$ is the only such element of $T$.
Thus $\left({T, \circ'}\right)$ is a semigroup with an identity element.
So, by definition, $\left({T, \circ'}\right)$ is a commutative monoid.
{{qed}}
Category:Inverse Completions
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Theorem: Inverse Completion is Commutative Semigroup

Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then $T = S \circ' C^{-1}$, and is a commutative semigroup.

Proof:
From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup:
:$S \circ' C^{-1}$ is a commutative semigroup.
From Structure of Inverse Completion of Commutative Semigroup:
:$T = S \circ' C^{-1}$
{{qed}}
Theorem: Complement of G-Delta Set is F-Sigma Set

Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be an $G_\delta$ set of $T$.
Then its complement $S \setminus X$ is an $F_\sigma$ set of $T$.

Proof:
Let $X$ be a $G_\delta$ set of $T$.
Let $X = \bigcap \UU$ where $\UU$ is a countable set of open sets in $T$.
Then from De Morgan's Laws: Difference with Intersection we have:
:$\ds S \setminus X = S \setminus \bigcap \UU = \bigcup_{U \mathop \in \UU} \paren {S \setminus U}$
By definition of closed set, each of the $S \setminus U$ are closed sets.
So $\ds \bigcup_{U \mathop \in \UU} \paren {S \setminus U}$ is a countable union of closed sets in $T$.
Hence $S \setminus X$ is, by definition, an $F_\sigma$ set of $T$.
{{qed}}
Theorem: Bounded Piecewise Continuous Function has Improper Integrals

Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$. 
Let $f$ be piecewise continuous and bounded on $\closedint a b$.
Then $f$ is a piecewise continuous function with improper integrals.

Proof:
Let $f$ be piecewise continuous and bounded on $\closedint a b$.
It is sufficient to prove that the improper integral $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$.
Let $i \in \set {1, 2, \ldots, n}$.
Let $c$ be a point in $\openint {x_{i − 1} } {x_i}$.
By definition, the improper integral $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists {{iff}}:
:$\ds \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$
and:
:$\ds \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$
both exist.
By Bounded Piecewise Continuous Function is Darboux Integrable we know that $f$ is Darboux integrable on $\closedint a b$.
Therefore, $f$ is Darboux integrable on every closed subinterval of $\closedint a b$. 
Accordingly, the following Darboux integrals:
:$\ds \int_{x_{i - 1} }^c \map f x \rd x$
:$\ds \int_{\gamma_1}^c \map f x \rd x$
:$\ds \int_c^{\gamma_2} \map f x \rd x$
:$\ds \int_c^{x_i} \map f x \rd x$
all exist where $\gamma_1 \in \openint {x_{i − 1} } c$ and $\gamma_2 \in \openint c {x_i}$.
Note that $f$ is bounded on $\closedint a b$ as $f$ is piecewise continuous and bounded.
Therefore, a bound $B$ exists for $f$ on $\closedint a b$.
We have:
{{begin-eqn}}
{{eqn | l = \size {\int_{\gamma_1}^c \map f x \rd x - \int_{x_{i - 1} }^c \map f x \rd x}
      | r = \size {\int_{x_{i - 1} }^{\gamma_1} \map f x \rd x}
}}
{{eqn | o = \le
      | r = B \size {\gamma_1 - x_{i - 1} }
      | c = {{Corollary|Darboux's Theorem|1}}
}}
{{end-eqn}}
which approaches $0$ as $\gamma_1$ approaches $x_{i − 1}$.
This shows that $\ds \lim_{\gamma \mathop \to x_{i - 1}+} \int_\gamma^c \map f x \rd x$ equals $\ds \int_{x_{i - 1} }^c \map f x \rd x$.
From this we gather that $\ds \lim_{\gamma \mathop \to  {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ exists.
Also:
{{begin-eqn}}
{{eqn | l = \size {\int_c^{x_i} \map f x \rd x - \int_c^{\gamma_2} \map f x \rd x}
      | r = \size {\int_{\gamma_2}^{x_i} \map f x \rd x}
}}
{{eqn | o = \le
      | r = B \size {x_i - \gamma_2}
      | c = {{Corollary|Darboux's Theorem|1}}
}}
{{end-eqn}}
which approaches $0$ as $\gamma_2$ approaches $x_i$.
This shows that $\ds \lim_{\gamma \mathop \to x_i-} \int_c^\gamma \map f x \rd x$ equals $\ds \int_c^{x_i} \map f x \rd x$.
From this we gather that $\ds \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ exists.
Since $i$ is arbitrary, we have shown that $\ds \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ and $\ds \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ exist for every $i \in \set {1, \ldots, n}$.
Therefore, $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$.
That is, $f$ is a piecewise continuous function with improper integrals.
{{qed}}
Theorem: F-Sigma Set is not necessarily Closed Set

Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be an $F_\sigma$ set of $T$.
Then it is not necessarily the case that $X$ is a closed set of $T$.

Proof:
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $X \subseteq S$ be an $F_\sigma$ set of $T$.
From $F_\sigma$ and $G_\delta$ Subsets of Uncountable Finite Complement Space:
:$X$ is neither open nor closed in $T$.
Hence the result.
{{qed}}
Theorem: Not every Closed Set is G-Delta Set

Let $T = \struct {S, \tau}$ be a topological space.
Let $V$ be a closed set of $T$.
Then it is not necessarily the case that $V$ is a $G_\delta$ set of $T$.

Proof:
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $V$ be a closed set of $T$.
From Closed Set of Uncountable Finite Complement Topology is not $G_\delta$:
:$V$ is not a $G_\delta$ set of $T$.
Hence the result.
{{qed}}
Theorem: Open Set of Uncountable Finite Complement Topology is not F-Sigma

Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $U \in \tau$ be an open set of $T$.
Then $U$ is not an $F_\sigma$ set.

Proof:
Let $U$ be an open set of $T$.
As $S$ is uncountable, then so is $U$.
By the definition of a finite complement topology, all closed sets of $T$ are finite.
From Countable Union of Countable Sets is Countable, $U$ can not be expressed as the union of a countable number of closed sets.
So by definition $U$ is not an $F_\sigma$ set.
{{qed}}
Theorem: Closed Set of Uncountable Finite Complement Topology is not G-Delta

Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $V \in \tau$ be a closed set of $T$.
Then $V$ is not a $G_\delta$ set.

Proof:
Let $V$ be a closed set of $T$.
{{AimForCont}} $V$ is $G_\delta$ set.
Then by Complement of $G_\delta$ Set is $F_\sigma$ Set:
:$S \setminus V$ is an $F_\sigma$ set.
By definition of closed set, $S \setminus V$ is an open set of $T$.
But by Open Set of Uncountable Finite Complement Topology is not $F_\sigma$:
:$S \setminus V$ is not an $F_\sigma$ set.
It follows by Proof by Contradiction that $V$ is not a $G_\delta$ set.
{{qed}}
Theorem: G-Delta Set is not necessarily Open Set

Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be a $G_\delta$ set of $T$.
Then it is not necessarily the case that $X$ is a open set of $T$.

Proof:
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $X \subseteq S$ be a $G_\delta$ set of $T$.
From $F_\sigma$ and $G_\delta$ Subsets of Uncountable Finite Complement Space:
:$X$ is neither open nor closed in $T$.
Hence the result.
{{qed}}
Theorem: Not every Open Set is F-Sigma Set

Let $T = \struct {S, \tau}$ be a topological space.
Let $V$ be an open set of $T$.
Then it is not necessarily the case that $V$ is an $F_\sigma$ set of $T$.

Proof:
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $U$ be an open set of $T$.
From Closed Set of Uncountable Finite Complement Topology is not $F_\sigma$:
:$U$ is not an $F_\sigma$ set of $T$.
Hence the result.
{{qed}}
Theorem: Limit Point of Sequence may only be Adherent Point of Range

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
Let $\sequence {x_n}$ be a sequence in $A$.
Let $\alpha$ be a limit of $\sequence {x_n}$.
Then $\alpha$ may be only an adherent point of $A$ and not a limit point of $A$.

Proof:
Let $T = \struct {S, \tau}$ be the discrete space on $S$.
Let $x \in S$.
Then by definition of discrete space:
:$U = \set x$ is an open set of $T$.
Consider the sequence $\sequence {x_n}$ defined as:
:$\forall n \in \N: x_n = x$
That is:
:$\sequence {x_n} = \tuple {x, x, x, \ldots}$
From Limit Point of Sequence in Discrete Space not always Limit Point of Open Set:
:$x$ is not a limit point of $U$.
But from Limit Point of Sequence is Adherent Point of Range:
:$x$ is an adherent point of $\set x$.
Hence the result.
{{qed}}
Theorem: Limit Point of Sequence in Discrete Space not always Limit Point of Open Set

Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $U \in \tau$ be an open set of $T$.
Let $\sequence {x_n}$ be a sequence in $U$.
Let $x$ be the limit of $\sequence {x_n}$.
Then $x$ is not always a limit point of $U$.

Proof:
Let $x \in S$.
By definition of discrete space:
:$U = \set x$ is an open set of $T$.
Consider the sequence $\sequence {x_n}$ defined as:
:$\forall n \in \N: x_n = x$
That is:
:$\sequence {x_n} = \tuple {x, x, x, \ldots}$
Thus $x$ is the limit point of $\sequence {x_n}$.
But:
:$U \setminus \set x = \O$
and so $x$ is not a limit point of $U$.
Hence the result.
{{qed}}
Theorem: Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range

Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {x_n}$ be a sequence of distinct terms of $S$.
Let $\alpha$ be an accumulation point of $\sequence {x_n}$.
Then $\alpha$ is also an $\omega$-accumulation point of $\set {x_n: n \in \N}$.

Proof:
Let $U$ be an open set of $T$ containing $\alpha$.
By definition of accumulation point of $\sequence {x_n}$, $U$ contains infinitely many terms of $\sequence {x_n}$.
As $\sequence {x_n}$ is a sequence of distinct terms:
$U$ contains infinitely many elements of $\set {x_n: n \in \N}$.
Thus by definition $\alpha$ is an $\omega$-accumulation point of $\set {x_n: n \in \N}$.
{{qed}}
Theorem: Limit Point of Sequence is Adherent Point of Range

Let $T = \struct{S, \tau}$ be a topological space.
Let $\sequence{x_n}$ be a sequence in $S$.
Let $\alpha$ be a limit of $\sequence {x_n}$.
Then $\alpha$ is an adherent point of $\set{x_n: n \in \N}$.

Proof:
By definition of Limit of sequence:
:$\forall U \in \tau : \exists N \in \N : \forall n \ge N : x_n \in U$
Hence:
:$\forall U \in \tau : U \cap \set{x_n: n \in \N} \ne \O$.
By definition $\alpha$ is an adherent point of $\set{x_n: n \in \N}$.
{{qed}}
Category:Adherent Points of Sets
Category:Limits
20738664di2sm6nkc6li0hk05ao9dm1
Theorem: Limit Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range

Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {x_n}$ be a sequence of distinct terms of $S$.
Let $\alpha$ be a limit point of $\sequence {x_n}$.
Then $\alpha$ is also an $\omega$-accumulation point of $\set {x_n: n \in \N}$.

Proof:
Let $\alpha$ be an limit point of $\sequence {x_n}$.
From Limit of Sequence is Accumulation Point, $\alpha$ is an accumulation point of $\sequence {x_n}$.
From Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range, $\alpha$ is an $\omega$-accumulation point of $\set {x_n: n \in \N}$.
{{qed}}
Theorem: Equivalence of Definitions of Isolated Point

{{TFAE|def = Isolated Point of Subset|view = isolated point}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.

Proof:
=== Definition 1 implies Definition 2 ===
Let $x$ be an isolated point of $H$ by definition 1.
Then by definition:
:$\exists U \in \tau: U \cap H = \set x$
Thus we have an open set in $T$ such that $x \in U$ contains no other point of $H$ than $x$.
Thus, by definition, $x$ is not a limit point of $H$.
Thus $x$ is an isolated point of $H$ by definition 2.
{{qed|lemma}}
Theorem: Equivalence of Definitions of Closed Set

{{TFAE|def = Closed Set (Topology)|view = closed set|context = Topology (Mathematical Branch)|contextview = topology}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.

Proof:
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Theorem: Kuratowski's Closure-Complement Problem

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be a subset of $T$.
By successive applications of the operations of complement relative to $S$ and the closure, there can be as many as $14$ distinct subsets of $S$ (including $A$ itself).

Proof:
That there can be as many as $14$ will be demonstrated by example.
Theorem: Kuratowski's Closure-Complement Problem

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be a subset of $T$.
By successive applications of the operations of complement relative to $S$ and the closure, there can be as many as $14$ distinct subsets of $S$ (including $A$ itself).

Proof:
Consider an arbitrary subset $A$ of a topological space $T = \struct {S, \tau}$.
To simplify the presentation:
:let $a$ be used to denote the operation of taking the complement of $A$ relative to $S$: $\map a A = S \setminus A$
:let $b$ be used to denote the operation of taking the closure of $A$ in $T$: $\map b A = A^-$
:let $I$ be used to denote the identity operation on $A$, that is: $\map I A = A$.
:let the parentheses and the reference to $A$ be removed, so as to present, for example:
::$\map a {\map b {\map a A} }$
:as:
::$a b a$
From Relative Complement of Relative Complement:
:$\map a {\map a A} = A$
or, using the compact notation defined above:
:$(1): \quad a a = I$
and from Closure of Topological Closure equals Closure:
:$\map b {\map b A} = \map b A = A^-$
or, using the compact notation defined above:
:$(2): \quad b b = b$
Let $s$ be a finite sequence of elements of $\set {a, b}$.
By successive applications of $(1)$ and $(2)$, it is possible to eliminate all multiple consecutive instances of $a$ and $b$ in $s$, and so reduce $s$ to one of the following forms:
:$\text{a)}: \quad a b a b \ldots a$
:$\text{b)}: \quad b a b a \ldots a$
:$\text{c)}: \quad a b a b \ldots b$
:$\text{d)}: \quad b a b a \ldots b$
From Closure of Complement of Closure is Regular Closed:
:$b a b$ is regular closed.
By Interior equals Complement of Closure of Complement, the interior of $A$ is:
:$a b a$
Recall the definition of regular closed:
:a set $A$ is regular closed {{iff}} it equals the closure of its interior.
And so as $b a b$ is regular closed:
:$b a b = b a b a \paren {b a b}$
So, adding an extra $b$ to either end of $a b a b a b a$ or $b a b a b a$ will generate a string containing $b a b a b a b$ which can be reduced immediately to $b a b$.
It follows that the possible different subsets of $S$ that can be obtained from $A$ by applying $a$ and $b$ can be generated by none other than:
:$I$
:$a$
:$a b$
:$a b a$
:$a b a b$
:$a b a b a$
:$a b a b a b$
:$a b a b a b a$
:$b$
:$b a$
:$b a b$
:$b a b a$
:$b a b a b$
:$b a b a b a$
... a total of $14$.
Hence the result.
{{qed}}