avewright/proofwiki-math huggingface dataset:
Theorem: Closure of Half-Open Real Interval is Closed Real Interval

Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
Let $H_1 = \hointl a b$ and $H_2 = \hointr a b$ be half-open intervals of $\R$.
Then the closure of both $H_1$ and $H_2$ in $\R$ are the closed interval $\closedint a b$.

Proof:
For an arbitrary $H \subseteq \R$, let $H^-$ denote the closure of $H$ in $\R$.
By the definition of closure:
:$H^-$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $H$ as a subset.
Let $A := \openint a b$ be the open interval between $a$ and $b$.
By definition of $H_1$ and $H_2$:
:$A \subseteq H_1$
and:
:$A \subseteq H_2$
From Closure of Open Real Interval is Closed Real Interval:
:$\openint a b^- = \closedint a b$
Thus as $\closedint a b$ is the closure of $\openint a b$, it follows that:
:$\closedint a b$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $\openint a b$.
But $\closedint a b$ also contains as subsets both $H_1$ and $H_2$.
Thus we have:
:$A \subseteq H_1 \subseteq \closedint a b$
and:
:$A \subseteq H_2 \subseteq \closedint a b$
Thus it follows that $\closedint a b$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $H_1$ and $H_2$.
Hence the result.
{{qed}}
Category:Real Intervals
Category:Set Closures
e2fcgq4qmbpjesvpx91oj0d7xf7w5jz
Theorem: Closure of Irrational Interval is Closed Real Interval

Let $\struct {\R, \tau_d}$ be the real numbers under the usual (Euclidean) topology.
Let $\struct{\R \setminus \Q, \tau_d}$ be the irrational number space under the same topology.
Let $a, b \in \R$ such that $a < b$.
Let $\Bbb I \subseteq \R$ be an interval of $\R$
Then the closure of the set:
:$\Bbb I \cap \paren {\R \setminus \Q}$
is the closed real interval $\closedint a b$.

Proof:
Let $\Bbb I$ be an open real interval.
From Closure of Real Interval is Closed Real Interval:
:$\Bbb I^- = \closedint a b$
From Closure of Irrational Numbers is Real Numbers:
:$\paren {\R \setminus \Q}^- = \R$
From Closure of Intersection is Subset of Intersection of Closures:
:$\paren{\Bbb I \cap \paren{\R \setminus \Q}}^- \subseteq \Bbb I^- \cap \paren {\R \setminus \Q}^-$
From Intersection with Subset is Subset:
:$\Bbb I^- \cap \paren {\R \setminus \Q}^- = \closedint a b$
and so:
:$\paren {\Bbb I \cap \paren {\R \setminus \Q} }^- \subseteq \closedint a b$
From Irrationals are Everywhere Dense in Topological Space of Reals:
:$\closedint a b \subseteq \paren {\Bbb I \cap \paren {\R \setminus \Q} }^-$
and the result follows.
{{qed}}
Category:Real Intervals
Category:Irrational Number Space
Category:Set Closures
2sgds1491k5cm9vkp7ei81psyz1rljc
Theorem: Closure of Rational Interval is Closed Real Interval

Let $\struct {\R, \tau_d}$ be the real numbers under the usual (Euclidean) topology.
Let $\struct {\Q, \tau_d}$ be the rational number space under the same topology.
Let $a, b \in \R$ such that $a < b$.
Let $\Bbb I \subseteq \R$ be an interval of $\R$
Then the closure of the set :
:$\Bbb I \cap \Q$
is the closed real interval $\closedint a b$.

Proof:
Let $\Bbb I$ be an open real interval.
From Closure of Real Interval is Closed Real Interval:
:$\Bbb I^- = \closedint a b$
From Closure of Rational Numbers is Real Numbers:
:$\Q^- = \R$
From Closure of Intersection is Subset of Intersection of Closures:
:$\paren{\Bbb I \cap \Q}^- \subseteq \Bbb I^- \cap \Q^-$
From Intersection with Subset is Subset:
:$\Bbb I^- \cap \Q^- = \closedint a b$
and so:
:$\paren {\Bbb I \cap \Q}^- \subseteq \closedint a b$
From Rationals are Everywhere Dense in Topological Space of Reals:
:$\closedint a b \subseteq \paren {\Bbb I \cap \Q}^-$
and the result follows.
{{qed}}
Category:Real Intervals
Category:Irrational Number Space
m64xrdzlhmu195l3hgeur2klv6xxdnb
Theorem: Interior of Singleton in Real Number Line is Empty

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $a \in \R$ be a real number.
Then:
:$\set a^\circ = \O$
where $\set a^\circ$ denotes the interior of $\set a$ in $\R$.

Proof:
{{begin-eqn}}
{{eqn | l = \set a^\circ
      | r = \closedint a a^\circ
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | r = \openint a a
      | c = Interior of Closed Real Interval is Open Real Interval
}}
{{eqn | r = \set {x \in \R: a < x < a}
      | c = {{Defof|Open Real Interval}}
}}
{{eqn | r = \O
      | c = {{Defof|Empty Real Interval}}
}}
{{end-eqn}}
{{qed}}
Category:Examples of Set Interiors
Category:Singletons
Category:Real Number Line with Euclidean Topology
4d1pdk676zn43ptk0m9vuy79gruuyhp
Theorem: Kuratowski's Closure-Complement Problem/Proof of Maximum

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 no more than $14$ 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}}
Theorem: Closure of Complement of Closure is Regular Closed

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be a subset of $T$.
Let $A^-$ denote the closure of $A$ in $T$.
Let $A'$ denote the complement of $A$ in $S$: $A' = S \setminus A$.
Then $A^{-'-}$ is regular closed.

Proof:
Let $A^\circ$ denote the interior of $A$.
From Set is Closed iff Equals Topological Closure, $A^-$ is closed in $T$.
Since $A^-$ is a closed set, $A^{-'}$ is open.
Therefore:
:$\forall x \in A^{-'}: \exists \epsilon \in \R_{>0}$
{{questionable|The above statement makes no sense in this context|prime.mover}}
Additionally, because $A^{-'}$ is the complement of $A^-$ relative to $S$, we have:
:$\exists \epsilon \in \R_{>0}: \map {V_\epsilon} x \cap A^- = \O \iff x \in A^{-'}$
{{questionable|The above $\map {V_\epsilon} x$ cannot be applied in the context of a general topological space. If it were stated that $T$ were a metric space, then maybe (with some further work to establish links to results to back the statement up), but it isn't so it can't.|prime.mover}}
Therefore, for $y: y \in A^{-'-}$ and $ y \notin A^{- '}$ we know that:
:$\forall \epsilon \in \R_{>0}: \map {V_\epsilon} y \cap A^- \ne \O$
meaning that $y$ is a limit point of $A^-$.
Therefore:
:$y \notin A^{-'-\circ}$
Now:
:$x \in A^{-'-\circ} \iff x \in A^{-'}$
{{LinkWanted|the above statement}}
In conclusion, this means that:
:$A^{-'-\circ-} = A^{-'-},$ verifying the definition of $A^{-'-}$ being  regular closed. 
{{qed}}
Category:Regular Closed Sets
haoebnqqsbeuhjcl4nt80kr5sqfaujv
Theorem: Topology induced by Usual Metric on Positive Integers is Discrete

Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.
Then the metric topology for $d$ is a discrete topology.

Proof:
Let $\tau_d$ denote the metric topology for $d$.
Let $\epsilon \in \R_{>0}$ such that $\epsilon < 1$.
Let $a \in \Z_{>0}$.
Recall the definition of the open $\epsilon$-ball of $a$ in $\struct {\Z_{>0}, d}$:
:$\map {B_\epsilon} a := \set {x \in A: \map d {x, a} < \epsilon}$
But we have:
:$\forall x \in \Z_{>0}, x \ne a: \map d {x, a} \ge 1$
and so:
:$\forall x \in \Z_{>0}, x \ne a: x \notin \map {B_\epsilon} a$
It follows that:
:$\map {B_\epsilon} a := \set a$
Thus by definition of $\tau_d$:
:$\forall a \in \Z_{>0}: \set a \in \tau_d$
It follows from Basis for Discrete Topology that $\struct {\Z_{>0}, \tau_d}$ is a discrete topological space.
{{qed}}
Theorem: Topology induced by Scaled Euclidean Metric on Positive Integers is Discrete

Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the metric on $\Z_{>0}$ defined as:
:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
Then the metric topology for $\delta$ is a discrete topology.

Proof:
Let $\tau_\delta$ denote the metric topology for $\delta$.
In Scaled Euclidean Metric is Metric it is demonstrated that $\delta$ is indeed a metric on $\Z_{>0}$.
Let $a \in \Z_{>0}$.
Recall the definition of the open $\epsilon$-ball of $a$ in $\struct {\Z_{>0}, \delta}$:
:$\map {B_\epsilon} a := \set {x \in A: \map \delta {x, a} < \epsilon}$
Let $x \in \R_{>0}$.
Let $\epsilon \in \R_{>0}$ such that $\epsilon < \dfrac 1 {a \paren {a + 1} }$.
But we have:
{{begin-eqn}}
{{eqn | q = \forall x \in \Z_{>0}, x \ne a
      | l = \map \delta {x, a}
      | r = \frac {\size {x - a} } {x a}
      | c = 
}}
{{eqn | r = \size {\frac 1 x - \frac 1 a}
      | c = 
}}
{{eqn | o = \ge
      | r = \size {\frac 1 {a + 1} - \frac 1 a}
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
and so:
:$\forall x \in \Z_{>0}, x \ne a: x \notin \map {B_\epsilon} a$
It follows that:
:$\map {B_\epsilon} a := \set a$
Thus by definition of $\tau_d$:
:$\forall a \in \Z_{>0}: \set a \in \tau_\delta$
It follows from Basis for Discrete Topology that $\struct {\Z_{>0}, \tau_\delta}$ is a discrete topological space.
{{qed}}
Theorem: Scaled Euclidean Metric is Metric

Let $\R_{>0}$ be the set of (strictly) positive integers.
Let $\delta: \R_{>0} \times \R_{>0} \to \R$ be the metric on $\R_{>0}$ defined as:
:$\forall x, y \in \R_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
Then $\delta$ is a metric.

Proof:
=== Proof of {{Metric-space-axiom|1|nolink}} ===
{{begin-eqn}}
{{eqn | l = \map \delta {x, x}
      | r = \dfrac {\size {x - x} } {x^2}
      | c = Definition of $\delta$
}}
{{eqn | r = 0
      | c = as $\size {x - x} = 0$
}}
{{end-eqn}}
So {{Metric-space-axiom|1}} holds for $\delta$.
{{qed|lemma}}
Theorem: Topologies induced by Usual Metric and Scaled Euclidean Metric on Positive Integers are Homeomorphic

Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.
Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the metric on $\Z_{>0}$ defined as:
:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
Let $\tau_d$ denote the metric topology for $d$.
Let $\tau_\delta$ denote the metric topology for $\delta$.
Then $\struct {\Z_{>0}, \tau_d}$ and $\struct {\Z_{>0}, \tau_\delta}$ are homeomorphic.

Proof:
From Topology induced by Usual Metric on Positive Integers is Discrete:
:$\struct {\Z_{>0}, \tau_d}$ is a discrete space.
From Topology induced by Scaled Euclidean Metric on Positive Integers is Discrete:
:$\struct {\Z_{>0}, \tau_\delta}$ is a discrete space.
Let $I_{\Z_{>0} }$ be the identity mapping from $\Z_{>0}$ to itself.
From Mapping from Discrete Space is Continuous:
:$I_{\Z_{>0} }: \struct {\Z_{>0}, \tau_d} \to \struct {\Z_{>0}, \tau_\delta}$ is continuous
and:
:$I_{\Z_{>0} }: \struct {\Z_{>0}, \tau_\delta} \to \struct {\Z_{>0}, \tau_d}$ is continuous.
Hence the result by definition of homeomorphic.
{{qed}}
Theorem: Cauchy Sequence in Positive Integers under Usual Metric is eventually Constant

Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.
Let $\sequence {x_n}$ be a Cauchy sequence in $\struct {\Z_{>0}, d}$.
Then:
:$\exists m, n \in \Z_{>0}: \forall r > n: x_r = m$
That is, $\sequence {x_n}$ is eventually constant.

Proof:
Let $\sequence {x_n}$ be a Cauchy sequence in $\struct {\Z_{>0}, d}$.
By definition:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \map d {x_n, x_m} < \epsilon$
Let $\epsilon < 1$, say: $\epsilon = \dfrac 1 2$.
By the definition of $d$:
:$\forall m, n \in \N: x_m \ne x_n \implies \map d {x_m, x_n} \ge 1$
So the only possible way for:
:$\forall m, n \in \N: m, n \ge N: \map d {x_n, x_m} < \epsilon$
is for $x_m = x_n$.
Hence the result.
{{qed}}
Theorem: Positive Integers under Usual Metric is Complete Metric Space

Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.
Then $\struct {\Z_{>0}, d}$ is a complete metric space.

Proof:
Let $\sequence {x_n}$ be a Cauchy sequence in $\struct {\Z_{>0}, d}$.
From Cauchy Sequence in Positive Integers under Usual Metric is eventually Constant:
:$\sequence {x_n}$ is a convergent sequence to some $n \in \Z_{>0}$.
Hence the result by definition of complete metric space.
{{qed}}
Theorem: Cauchy Sequence in Positive Integers under Scaled Euclidean Metric

Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the scaled Euclidean metric on $\Z_{>0}$ defined as:
:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
The sequence $\sequence {x_n}$ in $\Z_{>0}$ defined as:
:$\forall n \in \N: x_n = n$
is a Cauchy sequence in $\struct {\Z_{>0}, \delta}$.

Proof:
For a general $x_m, x_n \in \sequence {x_n}$ as defined:
{{begin-eqn}}
{{eqn | l = \map \delta {x, y}
      | r = \frac {\size {x_m - x_n} } {x_m x_n}
      | c = Definition of $\delta$
}}
{{eqn | r = \size {\frac 1 {x_m} - \frac 1 {x_n} }
      | c = algebra
}}
{{eqn | n = 1
      | r = \size {\frac 1 m - \dfrac 1 n}
      | c = Definition of $\sequence {x_n}$
}}
{{end-eqn}}
Let $\epsilon \in \R_{>0}$.
Then by the Axiom of Archimedes:
:$\exists N \in \N: N > \dfrac 1 \epsilon$
from which it follows that:
:$\epsilon > \dfrac 1 N$
Thus:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
      | l = m, n
      | o = >
      | r = N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \delta {x_m, x_n}
      | r = \size {\frac 1 m - \frac 1 n}
      | c = from $(1)$ above
}}
{{eqn | o = <
      | r = \max \set {\frac 1 m, \frac 1 n}
      | c = 
}}
{{eqn | o = <
      | r = \frac 1 N
      | c = 
}}
{{eqn | o = <
      | r = \epsilon
      | c = 
}}
{{end-eqn}}
Therefore $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Z_{>0}, \delta}$.
{{qed}}
Theorem: Positive Integers under Scaled Euclidean Metric is not Complete Metric Space

Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the scaled Euclidean metric on $\Z_{>0}$ defined as:
:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
Then $\struct {\Z_{>0}, \delta}$ is not a complete metric space.

Proof:
Consider the sequence $\sequence {x_n}$ in $\Z_{>0}$ defined as:
:$\forall n \in \N: x_n = n$
From Cauchy Sequence in Positive Integers under Scaled Euclidean Metric:
:$\sequence {x_n}$ is a Cauchy sequence in $\struct {\Z_{>0}, \delta}$.
But $\sequence {x_n}$ is not convergent to any $m \in \Z_{>0}$.
Hence the result, by definition of complete metric space.
{{qed}}
Theorem: Subset of Euclidean Plane whose Product of Coordinates are Greater Than or Equal to 1 is Closed

Let $\struct {\R^2, \tau_d}$ be the real number plane with the usual (Euclidean) topology.
Let $A \subseteq R^2$ be the set of all points defined as:
:$A := \set {\tuple {x, y} \in \R^2: x y \ge 1}$
Then $A$ is a closed set in $\struct {\R^2, d}$.

Proof:
By definition, $\tau_d$ is the topology induced by the Euclidean metric $d$.
Consider the complement of $A$ in $\R^2$:
:$A' := \R^2 \setminus A$
Thus:
:$A := \set {\tuple {x, y} \in \R^2: x y < 1}$
Let $a = \tuple {x_a, y_a} \in A^2$.
Let $\epsilon = \size {1 - x_a y_a}$.
Then the open $\epsilon$-ball of $a$ in $\R^2$ lies entirely in $A'$.
As $a$ is arbitrary, it follows that any such $a$ has an open $\epsilon$-ball of $a$ in $\R^2$ which lies entirely in $A'$.
Thus, by definition, $A'$ is open in $\R^2$.
So, also by definition, $A$ is closed in $\R^2$.
{{qed}}
Theorem: Projection on Real Euclidean Plane is Open Mapping

Let $\struct {\R^2, d}$ be the real number plane with the usual (Euclidean) topology.
Let $\rho: \R^2 \to \R$ be the first projection on $\R^2$ defined as:
:$\forall \tuple {x, y} \in \R^2: \map \rho {x, y} = x$
Then $\rho$ is an open mapping.
The same applies with the second projection on $\R^2$.

Proof:
By definition, the real number plane with the usual (Euclidean) topology on $\R^2$ is the product space of $\struct {\R, d}$ with $\struct {\R, d}$, where $\struct {\R, d}$ is the real number line with the usual (Euclidean) topology
The result follows from Projection from Product Topology is Open.
{{qed}}
Theorem: Projection on Real Euclidean Plane is not Closed Mapping

Let $\struct {\R^2, d}$ be the real number plane with the usual (Euclidean) topology.
Let $\rho: \R^2 \to \R$ be the first projection on $\R^2$ defined as:
:$\forall \tuple {x, y} \in \R^2: \map \rho {x, y} = x$
Then $\rho$ is not a closed mapping.
The same applies with the second projection on $\R^2$.

Proof:
Consider the set $A \subseteq R^2$ of all points defined as:
:$A := \set {\tuple {x, y} \in \R^2: x y \ge 1}$
By Subset of Euclidean Plane whose Product of Coordinates are Greater Than or Equal to 1 is Closed:
:$A$ is a closed set in $\struct {\R^2, d}$.
By inspection, it can be seen that the image of $A$ under $\rho$ is:
:$\rho \sqbrk A = \openint \gets 0 \cup \openint 0 \to$
which by Union of Open Sets of Metric Space is Open is open.
Hence the result by definition of closed mapping.
{{qed}}
Theorem: Complement of Set of Rational Pairs in Real Euclidean Plane is Injectively Path-Connected

Let $\struct {\R^2, d}$ be the real number plane with the usual (Euclidean) topology.
Let $S \subseteq \R^2$ be the subset of $\R^2$ defined as:
:$\forall x, y \in \R^2: \tuple {x, y} \in S \iff x, y \in \Q$
Hence let $A := \R^2 \setminus S$:
:$\tuple {x, y} \in A$ {{iff}} either $x$ or $y$ or both is irrational.
Then $A$ is injectively path-connected.

Proof:
Let $\tuple {a, b} \in A$.
Consider any point $\tuple {x_1, y_1} \in A$ whose coordinates are both irrational.
By definition, either $a$ or $b$ is irrational.
{{WLOG}}, suppose $a$ is irrational.
Then the union of the straight lines $x = a, y = y_1$ is an injectively path-connected subset of $A$ connecting $\tuple {x_1, y_1}$ to $\tuple {a, b}$.
Hence any point in $A$ can be connected to $\tuple {x_1, y_1}$ by an injective path.
Hence the result, by definition of injectively path-connected.
{{qed}}
Theorem: Empty Set is Compact

Let $T = \struct {S, \tau}$ be a topological space.
Then the empty set $\O$ is a compact subspace of $T$.

Proof:
Recall the definition of compact subspace:
:$\struct {\O, \tau_\O}$ is '''compact in $T$''' {{iff}} every open cover $\CC \subseteq \tau_\O$ for $\O$ has a finite subcover.
The only open cover for $\O$ that is contained in $\O$ is $\set \O$ itself.
This has only one finite subcover, and that is $\set \O$.
This is a finite subcover.
Hence the result, by definition of compact subspace.
{{qed}}
Category:Empty Set
Category:Examples of Compact Topological Spaces
bng8btbu3y9u9r8uli695b0egjij9rp
Theorem: Alexandroff Extension is Topology

Let $T = \struct {S, \tau}$ be a non-empty topological space.
Let $p$ be a new element not in $S$.
Let $S^* := S \cup \set p$.
Let $T^* = \struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.
Then $\tau^*$ is a topology on $S^*$.

Proof:
Recall the definition of the Alexandroff extension on $S$:
$U$ is open in $T^*$ {{iff}}:
:$U$ is an open set of $T$
or
:$U$ is the complement in $T^*$ of a closed and compact subset of $T$.
Each of the open set axioms is examined in turn:
Theorem: Intersection of Closed Set with Compact Subspace is Compact

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be closed in $T$.
Let $K \subseteq S$ be compact in $T$.
Then $H \cap K$ is compact in $T$.

Proof:
Let $\tau_K$ be the subspace topology on $K$.
Let $T_K = \left({K, \tau_K}\right)$ be the topological subspace determined by $K$.
By Closed Set in Topological Subspace, $H \cap K$ is closed in $T_K$.
By Closed Subspace of Compact Space is Compact, $H \cap K$ is compact in $T_K$.
By Compact in Subspace is Compact in Topological Space, $H \cap K$ is compact in $T$.
{{qed}}
Theorem: Intersection of Closed Set with Compact Subspace is Compact

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be closed in $T$.
Let $K \subseteq S$ be compact in $T$.
Then $H \cap K$ is compact in $T$.

Proof:
Let $\family {U_\alpha}$ be an open cover of $H \cap K$:
:$\ds H \cap K \subseteq \bigcup_\alpha U_\alpha$
Then:
:$\ds K \subseteq \bigcup_\alpha U_\alpha \cup \paren {S \setminus H}$
Since $H$ is closed in $T$, $\paren {S \setminus H}$ is open in $T$.
Hence $\family {U_\alpha} \cup S \setminus H$ is an open cover of $K$.
We have that $K$ is compact in $T$.
It follows by definition that a finite subcover:
:$\set {U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n}, S \setminus H}$
of $K$ exists.
Thus:
:$H \cap K \subseteq \set {U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n} }$
and $H \cap K$ is compact in $T$.
{{qed}}
Theorem: Finite Union of Compact Sets is Compact

Let $T = \struct {S, \tau}$ be a topological space.
Let $n \in \N$ be a natural number.
Let $\sequence {U_i}_{1 \mathop \le i \mathop \le n}$ be a finite sequence of compact subsets of $T$.
Let $\UU_n := \ds \bigcup_{i \mathop = 1}^n U_i$ be the union of $\sequence {U_i}$.
Then $\UU_n$ is compact in $T$.

Proof:
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\UU_n := \ds \bigcup_{i \mathop = 1}^n U_i$ is compact in $T$.
$\map P 0$ is the case:
:$\UU_0 := \ds \bigcup_{i \mathop = 1}^0 U_i$
From Union of Empty Set:
:$\ds \bigcup_{i \mathop = 1}^0 U_i = \O$
From Empty Set is Compact it follows that:
:$\UU_0$ is compact in $T$.
$\map P 1$ is true, as this just says:
:$U_1$ is compact in $T$.
Theorem: Principle of Mathematical Induction

Let $\map P n$ be a propositional function depending on $n \in \Z$.
Let $n_0 \in \Z$ be given.
Suppose that:
:$(1): \quad \map P {n_0}$ is true
:$(2): \quad \forall k \in \Z: k \ge n_0 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \Z$ such that $n \ge n_0$.

Proof:
Let $S$ be the set defined as:
:$S := \set {n \in \N_{>0}: \map P n \text { is false} }$
{{AimForCont}} $S \ne \O$.
From the Well-Ordering Principle it follows that $S$ has a minimal element $m$.
From $(1)$ we have that $\map P 1$ holds.
Hence $1 \notin S$.
Therefore $m \ne 1$.
Therefore $m - 1 \in \N_{>0}$.
But $m$ is the minimal element of $S$.
So $m - 1 \notin S$.
Therefore $\map P {m - 1}$ is true.
Hence by $(2)$ it follows that $\map P m$.
But then $m \notin S$.
This contradicts our supposition that $m \in S$.
Hence there can be no such $m \in S$.
So $S = \O$ and the result follows.
{{qed}}