avewright/proofwiki-math huggingface dataset:
Theorem: Integer Reciprocal Space with Zero is not Extremally Disconnected

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
Then $A$ is not extremally disconnected.

Proof:
$\struct {A, \tau_d}$ is a metric space.
We have:
:Extremally Disconnected Metric Space is Discrete
We also have:
:Topological Space is Discrete iff All Points are Isolated
:Zero is Limit Point of Integer Reciprocal Space
From definition of limit point:
:$0$ is not an isolated point of $A$
Hence integer reciprocal space with zero is not the discrete space, and the result follows.
{{qed}}
Theorem: Interior of Intersection may not equal Intersection of Interiors

Let $T$ be a topological space.
Let $\mathbb H$ be a set of subsets of $T$.
That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.
Then it is not necessarily the case that:
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ = \bigcap_{H \mathop \in \mathbb H} H^\circ$
where $H^\circ$ denotes the interior of $H$.

Proof:
From Intersection of Interiors contains Interior of Intersection it is seen that it is always the case that:
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \subseteq \bigcap_{H \mathop \in \mathbb H} H^\circ$
From Interior of Finite Intersection equals Intersection of Interiors it is seen that if $\mathbb H$ is finite, then:
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ = \bigcap_{H \mathop \in \mathbb H} H^\circ$
It remains to be shown that in the case where $\mathbb H$ is infinite, then it is possible that:
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \ne \bigcap_{H \mathop \in \mathbb H} H^\circ$
;Proof by Counterexample
Let $A_n$ denote the open real interval:
:$A_n := \openint {1 - \dfrac 1 n} {1 + \dfrac 1 n}$
for all $n \in \Z_{>0}$.
From Open Sets in Real Number Line, each of $A_n$ is an open set in $\R$.
From Interior of Open Set:
:$\forall n \in \Z_{>0}: {A_n}^\circ = A_n$
Thus:
:$\ds \bigcap A_n = \bigcap {A_n}^\circ = \set 1$
However:
:$\ds \paren {\bigcap A_n}^\circ = \set 1^\circ = \O$
Hence the result.
{{qed}}
Theorem: Interior of Union of Adjacent Open Intervals

Let $a, b, c \in R$ where $a < b < c$.
Let $A$ be the union of the adjacent open intervals:
:$A := \openint a b \cup \openint b c$
Then:
:$A = A^\circ$
where $A^\circ$ is the interior of $A$.

Proof:
From Open Sets in Real Number Line, $A$ is open in $\R$.
The result follows from Interior of Open Set.
{{qed}}
Theorem: Interior of Closure of Interior of Union of Adjacent Open Intervals

Let $a, b, c \in R$ where $a < b < c$.
Let $A$ be the union of the adjacent open intervals:
:$A := \openint a b \cup \openint b c$
Then:
:$A^{\circ - \circ} = A^{- \circ} = \openint a c$
where:
:$A^\circ$ is the interior of $A$
:$A^-$ is the closure of $A$.

Proof:
From Interior of Union of Adjacent Open Intervals:
:$A^\circ = A$
From Closure of Union of Adjacent Open Intervals:
:$A^- = \closedint a c$
From Interior of Closed Real Interval is Open Real Interval:
:$\closedint a c^\circ = \openint a c$
whence the result.
{{qed}}
Theorem: Closure of Interior of Closure of Union of Adjacent Open Intervals

Let $a, b, c \in R$ where $a < b < c$.
Let $A$ be the union of the adjacent open intervals:
:$A := \openint a b \cup \openint b c$
Then:
:$A^{- \circ -} = A^{\circ -} = A^- = \closedint a c$
where:
:$A^\circ$ is the interior of $A$
:$A^-$ is the closure of $A$.

Proof:
{{begin-eqn}}
{{eqn | l = A^{\circ -}
      | r = A^-
      | c = Interior of Union of Adjacent Open Intervals: $A^\circ = A$
}}
{{eqn | r = \closedint a c
      | c = Closure of Union of Adjacent Open Intervals
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = A^{- \circ -}
      | r = \openint a c^-
      | c = Interior of Closure of Interior of Union of Adjacent Open Intervals
}}
{{eqn | r = \closedint a c
      | c = Closure of Open Ball in Metric Space
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Exterior of Exterior of Union of Adjacent Open Intervals

Let $A$ be the union of the adjacent open intervals:
:$A := \openint a b \cup \openint b c$
Then:
:$A^{ee} = \openint a c$
where $A^e$ is the exterior of $A$.

Proof:
By definition of exterior, $A^e$ is the complement relative to $\R$ of the closure of $A$ in $\R$.
Thus:
{{begin-eqn}}
{{eqn | l = A^{ee}
      | r = \paren {\relcomp \R {A^-} }^e
      | c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = \paren {\relcomp \R {\closedint a c} }^e
      | c = Closure of Interior of Closure of Union of Adjacent Open Intervals
}}
{{eqn | r = \paren {\openint \gets a \cup \openint c \to}^e
      | c = {{Defof|Relative Complement}}
}}
{{eqn | r = \relcomp \R {\paren {\openint \gets a \cup \openint c \to}^-}
      | c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = \relcomp \R {\openint \gets a^- \cup \openint c \to^-}
      | c = Closure of Finite Union equals Union of Closures
}}
{{eqn | r = \relcomp \R {\hointl \gets a \cup \hointr c \to}
      | c = Closure of Open Ball in Metric Space
}}
{{eqn | r = \openint a c
      | c = {{Defof|Relative Complement}}
}}
{{end-eqn}}
{{qed}}
Theorem: Interior may not equal Exterior of Exterior

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be a subset of the underlying set $S$ of $T$.
Let $A^e$ be the  exterior of $A$.
Let $A^\circ$ be the  interior of $A$.
Then it is not necessarily the case that:
:$A^{ee} = A^\circ$

Proof:
We have from Interior is Subset of Exterior of Exterior:
:$A^\circ \subseteq A^{ee}$
It remains to be shown that there exist $A \subseteq S$ such that:
:$A^\circ \ne A^{ee}$
Let $a, b, c \in R$ where $a < b < c$.
Let $A$ be the union of the adjacent open intervals:
:$A := \openint a b \cup \openint b c$
From Exterior of Exterior of Union of Adjacent Open Intervals:
:$A^{ee} = \openint a c$
From Interior of Union of Adjacent Open Intervals:
:$A^\circ := \openint a b \cup \openint b c$
Thus:
:$b \in A^{ee}$
but:
:$b \notin A^\circ$
and so:
:$A^{ee} \subsetneq A^\circ$
{{qed}}
Theorem: Interior may not equal Exterior of Exterior

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be a subset of the underlying set $S$ of $T$.
Let $A^e$ be the  exterior of $A$.
Let $A^\circ$ be the  interior of $A$.
Then it is not necessarily the case that:
:$A^{ee} = A^\circ$

Proof:
Proof by Counterexample:
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $T = \struct {S, \tau}$ be the Sorgenfrey line on $\struct {S, \preccurlyeq}$.
Let $A \subseteq S$ denote the subset of $S$ defined as:
:$A = \openint a {+\infty}$
By Exterior of Exterior in Sorgenfrey Line is not necessarily Interior:
:$A^{ee} = \hointr a {+\infty}$
while:
:$A^\circ = \openint a {+\infty}$
The result is apparent.
{{qed}}
Theorem: Open Real Interval is Regular Open

Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Let $\openint a b$ be an open interval of $\R$.
Then $\openint a b$ is regular open in $\struct {\R, \tau_d}$.

Proof:
From Open Sets in Real Number Line, $\openint a b$ is open in $\struct {\R, \tau_d}$.
From Closure of Open Real Interval is Closed Real Interval:
:$\openint a b^- = \closedint a b$
where $\openint a b^-$ denotes the closure of $\openint a b$.
From Interior of Closed Real Interval is Open Real Interval:
:$\closedint a b^\circ = \openint a b$
where $\closedint a b^\circ$ denotes the interior of $\closedint a b$.
Hence the result, by definition of regular open.
{{qed}}
Theorem: Closure of Open Real Interval is Closed Real Interval

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\openint a b$ be an open interval of $\R$.
Then the closure of $\openint a b$ is the closed interval $\closedint a b$.

Proof:
From Limit Points of Open Real Interval, the limit points of $\openint a b$ consist of:
:the points $\openint a b$ itself
and
:the points $a$ and $b$.
By definition, the closure of $\openint a b$ is the union of $\openint a b$ and its limit points.
Hence the result, by definition of the closed interval $\closedint a b$.
{{qed}}
Category:Open Sets
Category:Set Closures
Category:Real Number Line with Euclidean Topology
78lv3ewzobt6bmitc1gg8gsylcs6u2u
Theorem: Limit Points of Open Real Interval

Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
Let $\openint a b$ be an open interval of $\R$.
Then the limit points of $\openint a b$ are:
:all the points in $\openint a b$
as well as:
:the points $a$ and $b$.

Proof:
Let $x \in \openint a b$.
Then by definition $a < x < b$.
Let $N_x$ be an open neighborhood of $x$ in $\R$.
So by Open Superset is Open Neighborhood, $N_x$ is an open set of $\R$ containing $x$.
By Open Sets in Real Number Line, $N_x$ is a countable union of pairwise disjoint open intervals.
As $x \in N_x$, it follows that $\exists U \subseteq N_x$ such that $x \in U$ and:
:$\exists p, q \in \R: U = \openint p q$
Let $r = \max \set {a, p}$ and $s = \min \set {b, q}$.
Then as $x > a, x > p$ and $x < b, x < q$ it follows that $x > \max \set {a, p}$ and $x < \min \set {b, q}$.
That is:
:$x \in \openint r s$
From Real Numbers are Densely Ordered:
:$\exists y \in \R: r < y < x$
and so:
:$y \in \openint r s$
By construction, it follows that:
:$y \in \openint p q$
That is, $y$ is an element of $\R$ different from $x$ which is also in $N_x$.
As $N_x$ was arbitrary, it follows that every such open neighborhood of $x$ has the same property.
That is, $x$ is a limit point of $\openint a b$.
{{qed|lemma}}
Let $N_a$ be an open neighborhood of $a$ in $\R$.
So by Open Superset is Open Neighborhood, $N_a$ is an open set of $\R$ containing $a$.
By Open Sets in Real Number Line, $N_a$ is a countable union of pairwise disjoint open intervals.
As $a \in N_a$, it follows that $\exists U \subseteq N_a$ such that $a \in U$ and:
:$\exists p, q \in \R: U = \openint p q$
Let $s = \min \set {b, q}$.
As $a < b, a < q$ it follows that $a < \min \set {b, q}$.
That is:
:$a \in \openint p s$
From Real Numbers are Densely Ordered:
:$\exists y \in \R: a < y < s$
and so:
:$y \in \openint p s$
By construction, it follows that:
:$y \in \openint p q$
That is, $y$ is an element of $\R$ different from $a$ which is also in $N_a$.
As $N_a$ was arbitrary, it follows that every such open neighborhood of $a$ has the same property.
That is, $a$ is a limit point of $\openint a b$.
By a similar argument it is shown that $a$ is a limit point of $\openint a b$.
{{qed|lemma}}
Now let:
:$x \notin \openint a b \cup \set {a, b}$
Then either:
:$x < a$
or
:$x > b$
Let $x < a$.
Then by Real Numbers are Densely Ordered:
:$\exists q \in \R: x < q < a$
and so $\exists p \in \R$ such that:
:$x \in \openint p q$
such that:
:$\openint p q \cap \openint a b = \O$
That is, $\openint p q$ is an open set of $\R$ containing $x$ which contains no elements of $\openint a b$.
Hence by definition, $x$ is not a limit point of $\openint a b$.
By a similar argument it is shown that if $x > b$ then $x$ is not a limit point of $\openint a b$.
{{qed}}
Category:Examples of Limit Points
Category:Real Intervals
pi9tes5bljktjdpiickvy09erswk3tg
Theorem: Interior of Closed Real Interval is Open Real Interval

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\closedint a b$ be a closed interval of $\R$.
Then the interior of $\closedint a b$ is the open interval $\openint a b$.

Proof:
By definition, the interior of $\closedint a b$ is the largest open set contained in $\closedint a b$.
From Open Sets in Real Number Line it follows that $\openint a b$ is an open set of $\struct {\R, \tau_d}$.
By definition of open interval, $\openint a b$ is contained in $\closedint a b$.
Suppose $U$ is an open set of $\struct {\R, \tau_d}$ which is contained in $\closedint a b$.
The only way $U$ could be bigger than $\openint a b$ is if either $a \in U$ or $b \in U$ or both.
Suppose $a \in U$.
Then by Open Sets in Real Number Line it follows that $a \in \openint p q$ for some $p, q \in \R$ such that $\openint p q \subseteq U$.
From Real Numbers are Densely Ordered:
:$\exists r \in \R: p < r < a$
and so:
:$\exists r \in U: r < a$
which means:
:$r \notin \openint a b$
That is, $a$ is not in an open set of $\struct {\R, \tau_d}$ which is contained in $\closedint a b$.
Thus $a$ is not in the interior of $\closedint a b$.
By a similar argument it is shown that neither is $b$ in the interior of $\closedint a b$.
Hence the result.
{{qed}}
Theorem: Union of Regular Open Sets is not necessarily Regular Open

Let $T = \struct {S, \tau}$ be a topological space.
Let $U$ and $V$ be regular open sets of $T$.
Then $U \cup V$ is not also necessarily a regular open set of $T$.

Proof:
{{Recall|Regular Open Set|regular open set}}
{{:Definition:Regular Open Set}}
;Proof by Counterexample
By Open Real Interval is Regular Open, the open real intervals:
:$\openint 0 {\dfrac 1 2}, \openint {\dfrac 1 2} 1$
are both regular open sets of $\R$.
Consider $A$, the union of the adjacent open intervals:
:$A := \openint 0 {\dfrac 1 2} \cup \openint {\dfrac 1 2} 1$
From Interior of Closure of Interior of Union of Adjacent Open Intervals:
:$A^{- \circ} = \openint 0 1$
Thus $A$ is not a regular open set of $\R$.
{{qed}}
Theorem: Closure of Intersection may not equal Intersection of Closures

Let $T = \struct {S, \tau}$ be a topological space.
Let $H_1$ and $H_2$ be subsets of $S$.
Let ${H_1}^-$ and ${H_2}^-$ denote the closures of $H_1$ and $H_2$ respectively.
Then it is not necessarily the case that:
:$\paren {H_1 \cap H_2}^- = {H_1}^- \cap {H_2}^-$

Proof:
Let $\struct {\R, \tau}$ be the real number line under the usual (Euclidean) topology.
Let $\Q$ denote the set of rational numbers.
Let $\R \setminus \Q$ denote the set of irrational numbers.
From Closure of Intersection of Rationals and Irrationals is Empty Set:
:$\paren {\Q \cap \paren {\R \setminus \Q} }^- = \O$
From Intersection of Closures of Rationals and Irrationals is Reals:
:$\Q^- \cap \paren {\R \setminus \Q}^- = \R$
The result follows.
{{qed}}
Theorem: Closure of Intersection may not equal Intersection of Closures

Let $T = \struct {S, \tau}$ be a topological space.
Let $H_1$ and $H_2$ be subsets of $S$.
Let ${H_1}^-$ and ${H_2}^-$ denote the closures of $H_1$ and $H_2$ respectively.
Then it is not necessarily the case that:
:$\paren {H_1 \cap H_2}^- = {H_1}^- \cap {H_2}^-$

Proof:
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
Let $H_1 = \openint 0 {\dfrac 1 2}$ and $H_2 = \openint {\dfrac 1 2} 1$.
By inspection it can be seen that:
:$H_1 \cap H_2 = \O$
Thus from Closure of Empty Set is Empty Set:
:$\paren {H_1 \cap H_2}^- = \O$
From Closure of Open Real Interval is Closed Real Interval:
:$H_1 = \closedint 0 {\dfrac 1 2}, H_2 = \closedint {\dfrac 1 2} 1$
Thus:
:${H_1}^- \cap {H_2}^- = \set {\dfrac 1 2}$
So $\paren {H_1 \cap H_2}^- \ne {H_1}^- \cap {H_2}^-$
Hence the result.
{{qed}}
Theorem: Closure of Empty Set is Empty Set

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

Proof:
From Empty Set is Closed in Topological Space, $\O$ is closed in $T$.
The result follows from Closed Set equals its Closure.
{{qed}}
Category:Set Closures
Category:Empty Set
ev0qv3muvwdrhiss3hjy1kk3syrue4h
Theorem: Closed Real Interval is Regular Closed

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\closedint a b$ be a closed interval of $\R$.
Then $\closedint a b$ is a regular closed set in $\struct {\R, \tau_d}$.

Proof:
{{Recall|Regular Closed Set|regular closed set}}
{{:Definition:Regular Closed Set}}
From Closed Real Interval is Closed in Real Number Line, $\closedint a b$ is closed in $\struct {\R, \tau_d}$.
From Interior of Closed Real Interval is Open Real Interval:
:$\closedint a b^\circ = \openint a b$
where $\closedint a b^\circ$ denotes the interior of $\closedint a b$.
From Closure of Open Real Interval is Closed Real Interval:
:$\openint a b^- = \closedint a b$
where $\openint a b^-$ denotes the closure of $\openint a b$.
Hence the result, by definition of regular closed.
{{qed}}
Theorem: Closed Real Interval is Closed in Real Number Line

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\closedint a b$ be a closed interval of $\R$.
Then $\closedint a b$ is closed (in the topological sense) in $\struct {\R, \tau_d}$.

Proof:
From Open Sets in Real Number Line:
:$U := \openint \gets a \cup \openint b \to$
is an open set in $\struct {\R, \tau_d}$.
Consider the complement relative to $\R$:
:$V := \relcomp \R U = \closedint a b$
By definition, $\relcomp \R U$ is closed (in the topological sense) in $\struct {\R, \tau_d}$.
But by construction:
:$\relcomp \R U = \closedint a b$
Hence the result.
{{qed}}
Theorem: Intersection of Regular Closed Sets is not necessarily Regular Closed

Let $T = \struct {S, \tau}$ be a topological space.
Let $U$ and $V$ be regular closed sets of $T$.
Then $U \cap V$ is not also necessarily a regular closed set of $T$.

Proof:
;Proof by Counterexample
By Closed Real Interval is Regular Closed, the closed real intervals:
:$\closedint 0 {\dfrac 1 2}, \closedint {\dfrac 1 2} 1$
are both regular closed sets of $\R$.
Consider $A$, the intersection of the two half-unit closed intervals:
:$A := \closedint 0 {\dfrac 1 2} \cap \closedint {\dfrac 1 2} 1 = \set {\dfrac 1 2} = \closedint {\dfrac 1 2} {\dfrac 1 2}$
From Interior of Closed Real Interval is Open Real Interval:
:$A^\circ = \openint {\dfrac 1 2} {\dfrac 1 2} = \O$
From Closure of Empty Set is Empty Set:
:$A^{\circ -} = \O \ne A$
Thus $A$ is not a regular closed set of $\R$.
{{qed}}
Theorem: Interior of Union is not necessarily Union of Interiors

Let $T = \struct {S, \tau}$ be a topological space.
Let $H_1$ and $H_2$ be subsets of $S$.
Let ${H_1}^\circ$ and ${H_2}^\circ$ denote the interiors of $H_1$ and $H_2$ respectively.
Then it is not necessarily the case that:
:$\paren {H_1 \cup H_2}^\circ = {H_1}^\circ \cup {H_2}^\circ$

Proof:
From Union of Interiors is Subset of Interior of Union:
:$\paren {H_1 \cup H_2}^\circ \supseteq {H_1}^\circ \cup {H_2}^\circ$
It remains to be shown that it is not necessarily the case that:
:$\paren {H_1 \cup H_2}^\circ = {H_1}^\circ \cup {H_2}^\circ$
;Proof by Counterexample
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
Let $H_1 = \closedint 0 {\dfrac 1 2}$ and $H_2 = \closedint {\dfrac 1 2} 1$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {H_1 \cup H_2}^\circ
      | r = \paren {\closedint 0 {\dfrac 1 2} \cup \closedint {\dfrac 1 2} 1}^\circ
      | c = 
}}
{{eqn | r = \closedint 0 1^\circ
      | c = by definition
}}
{{eqn | r = \openint 0 1
      | c = Interior of Closed Real Interval is Open Real Interval
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = {H_1}^\circ \cup {H_2}^\circ
      | r = \closedint 0 {\dfrac 1 2}^\circ \cup \closedint {\dfrac 1 2} 1^\circ
      | c = 
}}
{{eqn | r = \openint 0 {\dfrac 1 2} \cup \openint {\dfrac 1 2} 1
      | c = Interior of Closed Real Interval is Open Real Interval
}}
{{eqn | o = \ne
      | r = \openint 0 1
      | c = as $\dfrac 1 2$ is in one but not the other
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Non-Homeomorphic Sets may be Homeomorphic to Subsets of Each Other

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H_1 \subseteq S_1$ and $H_2 \subseteq S_2$.
Then it is possible for:
{{begin-itemize}}
{{item|(1):|$T_1$ to be homeomorphic to $H_2$}}
{{item|(2):|$T_2$ to be homeomorphic to $H_1$}}
{{end-itemize}}
but:
{{begin-itemize}}
{{item|(3):|$T_1$ and $T_2$ to not be homeomorphic.}}
{{end-itemize}}

Proof:
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
Let $S_1 := \closedint 0 1$ be the closed unit interval.
Let $S_2 := \openint 0 1$ be the open unit interval.
Let $H_1 := \openint 0 1$ and $H_2 := \closedint {\dfrac 1 4} {\dfrac 3 4}$.
Then:
{{begin-itemize}}
{{item|(1):|$\struct {S_1, \tau_d}$ is homeomorphic to $\struct {H_2, \tau_d}$ by the mapping $f: S_1 \to H_2$ defined as:
:$\map f x {{=}} \dfrac x 2 + \dfrac 1 4$}}
{{item|(2):|$\struct {S_2, \tau_d}$ is trivially homeomorphic to $\struct {H_1, \tau_d}$}}
{{end-itemize}}
From Continuous Image of Compact Space is Compact, if $S_1$ and $S_2$ are homeomorphic then both must be compact.
From Closed Real Interval is Compact in Metric Space, $S_1$ is compact.
But from Open Real Interval is not Compact, $S_2$ is not compact.
Hence:
{{begin-itemize}}
{{item|(3):|$\struct {S_1, \tau_d}$ and $\struct {S_2, \tau_d}$ are not homeomorphic.}}
{{end-itemize}}
{{qed}}
Theorem: Superspace of Homeomorphic Subspaces may not have Homeomorphism to Itself containing Subspace Homeomorphism

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H_1 \subseteq S_1$ and $H_2 \subseteq S_2$.
Let $H_1$ and $H_2$ be a homeomorphic.
Then it may be the case that there does not exist a homeomorphism $g: T_1 \to T_2$ such that:
:$g \restriction_{H_1} = f$
where:
:$g \restriction_{H_1}$ is the restriction of $g$ to $H_1$
:$f: H_1 \to H_2$ is a homeomorphism.

Proof:
;Proof by Counterexample
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $H_1 := \set 0 \cup \closedint 1 2 \cup \set 3$, where $\closedint 1 2$ denotes the closed interval from $1$ to $2$.
Let $H_2 := \closedint 0 1 \cup \set 2 \cup \set 3$.
$H_1$ and $H_2$ are homeomorphic, as can be demonstrated by the mapping $\phi: H_1 \to \H_2$ defined as:
:$\forall x \in H_1: \map \phi x = \begin {cases} 2 & : x = 0 \\ 3 & : x = 3 \\ x - 1 & : x \in \closedint 1 2 \end {cases}$
which is trivially a homeomorphism.
So, let $g$ be a homeomorphism from $H_1$ and $H_2$.
Each of the singletons in $H_1$ has to map to one of the singletons in $H_2$.
As a result, $g$ is not monotone.
Let $f: \R \to \R$ be a homeomorphism such that $g$ is a restriction of $f$.
From Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone, $f$ is monotone.
{{explain|We need another result for the above over the whole range $\R$ -- for some reason this does not exist. We have Surjective Monotone Function is Continuous but not in the other direction.}}
Hence $f$ cannot be a bijection.
Hence $f$ cannot be a homeomorphism.
{{qed}}
Theorem: Expectation of Exponential Distribution

Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$
Then the expectation of $X$ is given by:
:$\expect X = \beta$

Proof:
The expectation of a continuous random variable $X$ with sample space $\Omega_X$ is given by:
:$\ds \expect X := \int_{x \mathop \in \Omega_X} x \map {f_X} x \rd x$
where $f_X$ is the probability density function of $X$.
For the exponential distribution:
:$\Omega_X = \hointr 0 \infty$
From Probability Density Function of Exponential Distribution:
:$\ds \map {f_X} x = \frac 1 \beta \map \exp {- \frac x \beta}$
So:
:$\ds \expect X = \int_0^\infty x \frac 1 \beta \map \exp {- \frac x \beta} \rd x$
where $\exp$ is the exponential function. 
Substituting $u = \dfrac x \beta$, we have:
:$\ds \expect X = \beta \int_0^\infty u \map \exp {-u} \rd u$
The integral evaluates to:
:$\ds \expect X = \bigintlimits {-\beta \paren {u + 1} \map \exp {-u} } 0 \infty$
So:
:$\ds \expect X = \beta - \beta \lim_{u \mathop \to \infty} \frac {u + 1} {\exp u}$
{{MissingLinks|The link below should be to the real counterpart of this result, but it's not up yet}}
By Limit at Infinity of Polynomial over Complex Exponential, it follows that this limit is zero, so that:
:$\expect X = \beta$
{{qed}}