avewright/proofwiki-math huggingface dataset:
Theorem: Open Real Interval is not Closed Set

Let $\R$ be the real number line with the usual (Euclidean) metric.
Let $I = \openint a b$ be an open real interval.
Then $I$ is not a closed set of $\R$.

Proof:
Consider the relative complement of $I$ in $\R$:
:$J = \relcomp \R I = \R \setminus I = \hointl \gets a \cup \hointr b \to$
Let $\epsilon \in \R_{>0}$.
Consider the open $\epsilon$-ball $\map {B_\epsilon} a$.
Whatever the value of $\epsilon$ is, $a + \epsilon$ is not in $\map {B_\epsilon} a$.
So, by definition, $J$ is not an open set of $\R$.
By Relative Complement of Relative Complement, $\relcomp \R J = I$.
By definition, it follows that $I$ is not a closed set of $\R$.
{{qed}}
Category:Real Intervals
Category:Real Number Line with Euclidean Metric
o7ynhtgi5e7o9q23ih5o29t9ub074b1
Theorem: Open Real Interval is not Compact

Let $\R$ be the real number line considered as an Euclidean space.
Let $I = \openint a b$ be an open real interval.
Then $I$ is not compact.

Proof:
It suffices to demonstrate this for a particular open interval: we use $\openint 0 1$.
Consider the set of open intervals $\openint {\dfrac 1 n} 1$ for all $n \in \Z_{>1}$.
Each of these is an open set in $\openint 0 1$.
Also:
:$\openint 0 1 = \ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$
Thus $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$ is an open cover for $\openint 0 1$.
Note that:
:$\forall x \in \openint 0 1: \exists n \in \Z: n > \dfrac 1 x$
and so:
:$x \in \openint {\dfrac 1 n} 1$
Consider a finite subcover of $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$.
It will be in the form:
:$\set {\openint {\dfrac 1 {n_1} } 1, \openint {\dfrac 1 {n_2} } 1, \ldots, \openint {\dfrac 1 {n_r} } 1}$
But this is just $\openint {\dfrac 1 N} 1$, where $N = \max \set {n_1, n_2, \ldots, n_r}$.
So no finite subcover of $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$ can be a cover for $\openint 0 1$.
{{qed}}
Theorem: Set of Integers is not Compact

Let $\Z$ be the set of integers.
Then $\Z$ is not compact.

Proof:
Let $\R$ be the real number line considered as an Euclidean space.
From Set of Integers is not Bounded, $\Z$ is not bounded in $\R$.
The result follows by definition of compact.
{{qed}}
Theorem: Heine-Borel Theorem/Euclidean Space

Let $n \in \N_{> 0}$.
Let $C$ be a subspace of the Euclidean space $\R^n$.
Then $C$ is closed and bounded {{iff}} it is compact.

Proof:
=== Necessary Condition ===
{{:Heine-Borel Theorem/Euclidean Space/Necessary Condition}}{{qed|lemma}}
Theorem: Heine-Borel Theorem/Euclidean Space

Let $n \in \N_{> 0}$.
Let $C$ be a subspace of the Euclidean space $\R^n$.
Then $C$ is closed and bounded {{iff}} it is compact.

Proof:
Let $C \subseteq \R^n$ be closed and bounded.
Since $C$ is bounded, $C \subseteq \closedint a b^n = B$ for some $a, b \in \R$.
By the Heine-Borel Theorem: Real Line and by Topological Product of Compact Spaces, it follows that $B$ is compact.
From Euclidean Topology is Product Topology, it follows that $B$ is compact in the usual Euclidean topology.
By {{Corollary|Closed Set in Topological Subspace}}, we have that $C$ is closed in $B$.
By Closed Subspace of Compact Space is Compact, $C$ is compact.
{{qed}}
Theorem: Heine-Borel Theorem/Euclidean Space

Let $n \in \N_{> 0}$.
Let $C$ be a subspace of the Euclidean space $\R^n$.
Then $C$ is closed and bounded {{iff}} it is compact.

Proof:
The proof holds for $n = 1$, as follows.
Suppose $C$ is a closed, bounded subspace of $\R$.
Then $C \subseteq \closedint a b$ for some $a, b \in \R$.
Moreover, $C$ is closed in $\closedint a b$ by {{Corollary|Closed Set in Topological Subspace}}.
Hence $C$ is compact, by Closed Subspace of Compact Space is Compact.
Now suppose $C \subseteq \R^n$ is closed and bounded.
Since $C$ is bounded, $C \subseteq \closedint a b \times \closedint a b \times \cdots \times \closedint a b = B$ for some $a, b \in \R$.
Now $B$ is compact by Topological Product of Compact Spaces.
Also, $C$ is closed in $B$ by {{Corollary|Closed Set in Topological Subspace}}.
Hence $C$ is compact, by Closed Subspace of Compact Space is Compact.
{{qed}}
Theorem: Continuous Function on Compact Space is Bounded

Let $\struct {X, \tau}$ be a compact topological space.
Let $\struct {Y, \norm {\, \cdot \, } }$ be a normed vector space.
Let $f: X \to Y$ be a continuous mapping.
Then $f$ is bounded.

Proof:
{{AimForCont}} $f$ is not bounded.
Let $g : X \to \R$ be defined by:
:$\map g x = \norm {\map f x}$
From Norm is Continuous and Composite of Continuous Mappings is Continuous, it follows that $g$ is continuous.
For all $n \in \N$, set $A_n := g^{-1} \sqbrk {\map {B_n} 0}$, where $\map {B_n} 0$ denotes the open ball in $\R$ with radius $n$ and center $0$.
From Open Ball is Open Set in Normed Vector Space and the definition of continuity, it follows that all $A_n$ are open in $X$.
For all $x \in X$, we have:
:$x \in A_n$ for all $n \ge \norm {\map f x}$
It follows that:
:$\ds X \subseteq \bigcup_{n \mathop \in \N} A_n$
By definition of open cover, $\set {A_n}_{n \mathop \in \N}$ is an open cover of $X$.
As $X$ is compact, there exist $m \in \N$ and $n_1, \dots, n_m \in \N$ such that $\set {A_{n_k} }_{k \mathop \in \set {1, \ldots, m} }$ is a finite subcover of $X$.
It follows that:
:$\ds X \subseteq \bigcup_{k \mathop \in \set {1, \ldots, m} } A_{n_k}$
As $f$ is not bounded, there exists $x_0 \in X$ such that:
:$\norm {\map f {x_0} } > \max \set {n_1, \ldots, n_m}$
It follows that:
:$x_0 \notin \ds \bigcup_{k \mathop \in \set {1, \ldots, m} } A_{n_k}$
which is a contradiction.
{{qed}}
Theorem: Continuous Function on Compact Space is Bounded

Let $\struct {X, \tau}$ be a compact topological space.
Let $\struct {Y, \norm {\, \cdot \, } }$ be a normed vector space.
Let $f: X \to Y$ be a continuous mapping.
Then $f$ is bounded.

Proof:
From Continuous Image of Compact Space is Compact, $f \sqbrk X$ is a compact subset of $Y$. 
From Compact Subset of Normed Vector Space is Closed and Bounded, $f \sqbrk X$ is bounded.
Hence there exists a real number $M > 0$ such that:
:$\norm {\map f x} \le M$ for all $x \in X$. 
So $f$ is bounded.
{{qed}}
Theorem: Continuous Function on Compact Space is Uniformly Continuous

Let $\R^n$ be the $n$-dimensional Euclidean space.
Let $S \subseteq \R^n$ be a compact subspace of $\R^n$.
Let $f: S \to \R$ be a continuous function.
Then $f$ is uniformly continuous on $S$.

Proof:
{{ProofWanted|Use Heine-Cantor Theorem}}
Theorem: T2 Space is Noetherian iff Finite

Let $\struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Then $\struct {S, \tau}$ is Noetherian {{iff}} $S$ is finite.

Proof:
=== Necessary Condition ===
Let $\struct {S, \tau}$ be Noetherian.
{{Recall|Noetherian Topological Space|Noetherian topological space|index = 5}}
{{:Definition:Noetherian Topological Space/Definition 5}}
Let $H \subseteq S$ be a subspace of $\struct {S, \tau}$.
From Compact Subspace of Hausdorff Space is Closed, $H$ is closed.
Thus for all $H \subseteq S$, we find that $H$ is closed in $S$.
It follows by definition of discrete topology that $\tau$ is discrete.
But since $S$ is compact and Discrete Space is Compact iff Finite, $S$ is also finite.
{{qed|lemma}}
Theorem: Indiscrete Space is Noetherian

Let $\struct {S, \tau}$ be an indiscrete topological space.
Then $\struct {S, \tau}$ is Noetherian.

Proof:
Let $\struct {H, \tau_H}$ be a subspace of $T$.
From Subset of Indiscrete Space is Compact, $\struct {H, \tau_H}$ is compact.
The result follows by definition of Noetherian topological space.
{{qed}}
Theorem: Complete and Totally Bounded Metric Space is Sequentially Compact

Let $M = \struct {A, d}$ be a metric space.
Let $M$ be complete and totally bounded.
Then $M$ is sequentially compact.

Proof:
Let $\sequence {x_m}_{m \mathop \in \N}$ be an infinite sequence in $A$.
By the definition of a totally bounded metric space, we can use the {{Axiom-link|Countable Choice}} to obtain a sequence $\sequence {F_n}_{n \mathop \in \N}$ such that:
:For all $n \in \N$, $F_n$ is a finite $2^{-n}$-net for $M$.
For all $n \in \N$ and $y \in F_n$, define:
:$\map {S_n} y = \set {m \in \N: \map d {x_m, y} < 2^{-n} }$
It follows from the definition of a net that:
:$(1): \quad \ds \N = \bigcup_{y \mathop \in F_n} \map {S_n} y$
For all $n \in \N$, define:
:$G_n = \leftset {y \in F_n: \map {S_n} y}$ is infinite$\rightset{}$
Since $F_n$ is finite by definition, it follows by $(1)$ that $G_n$ is non-empty.
For all $y \in G_n$, define:
:$\ds \map {T_n} y = \leftset {z \in G_{n + 1}: \map {S_n} y \cap \map {S_{n + 1} } z}$ is infinite$\rightset{}$
By $(1)$, it follows from the distributivity of intersection over union that:
:$\ds \map {S_n} y = \bigcup_{z \mathop \in F_{n + 1} } \paren {\map {S_n} y \cap \map {S_{n + 1} } z}$
Hence, by the definition of $G_n$, it follows that $\map {T_n} y$ is non-empty.
From Countable Union of Countable Sets is Countable, it follows that the disjoint union $\ds \bigsqcup_{n \mathop \in \N} G_n$ is countable.
Using the {{Axiom-link|Countable Choice}}, there exists a sequence $\sequence {\phi_n: G_n \to G_{n + 1} }_{n \mathop \in \N}$ of mappings such that:
:$\forall n \in \N: \forall y \in G_n: \map {\phi_n} y \in \map {T_n} y$
{{explain|It is rather unclear what the connection with PoRD is here}}
Now, we use the Principle of Recursive Definition to construct a strictly increasing sequence $\sequence {m_k}_{k \mathop \in \N}$ in $\N$.
Let $y_0 \in G_0$.
Let $m_0 \in \map {S_0} {y_0}$.
For all $k \in \N$, let:
:$y_{k + 1} = \map {\paren {\phi_k \circ \cdots \circ \phi_1 \circ \phi_0} } {y_0}$
where $\circ$ denotes composition of mappings.
Let $m_{k + 1} > m_k$ be the smallest natural number such that:
:$m_{k + 1} \in \map {S_k} {y_k} \cap \map {S_{k + 1} } {y_{k + 1} }$
Such an $m_{k + 1}$ exists by:
:the well-ordering principle
and because:
:$\map {S_k} {y_k} \cap \map {S_{k + 1} } {y_{k + 1} }$ is infinite by the definitions of $\map {T_k} {y_k}$ and $\phi_k$.
Note that:
:$m_k, m_{k + 1} \in \map {S_k} {y_k}$
Let $\hat x_k = x_{m_k}$.
Let $i < j$, where $i, j \in \N$.
Then, by Sum of Infinite Geometric Progression:
:$\ds \map d {\hat x_i, \hat x_j} \le \sum_{k \mathop = i}^{j - 1} \paren {\map d {\hat x_k, y_k} + \map d {\hat x_{k + 1}, y_k} } < \sum_{k \mathop = i}^\infty 2^{1 - k} = 2^{2 - i}$
Hence, by Sequence of Powers of Number less than One, the sequence $\sequence {\hat x_k}_{k \mathop \in \N}$ is Cauchy.
By the assumption that $M$ is complete, the sequence $\sequence {\hat x_k}$ converges in $M$.
Since $\sequence {\hat x_k}$ is a convergent subsequence of $\sequence {x_m}$, it follows that $M$ is sequentially compact by definition.
{{qed}}
Theorem: Complete and Totally Bounded Metric Space is Sequentially Compact

Let $M = \struct {A, d}$ be a metric space.
Let $M$ be complete and totally bounded.
Then $M$ is sequentially compact.

Proof:
Let $M$ be both complete and totally bounded.
Let $\sequence {a_k}$ be any infinite sequence in $A$.
Let $\epsilon \in \R_{>0}$.
Let $x_1, \ldots, x_n \in X$ be a finite set of points such that:
:$\ds A = \bigcup_{i \mathop = 1}^n \map {B_\epsilon} {x_i}$
where $\map {B_\epsilon} {x_i}$ represents the open $\epsilon$-ball of $x_i$.
This is known to exist as $M$ is totally bounded.
Then for every $k \in \N$, there is some $j_k \in \set {0, \dots, n}$ such that $\map d {a_k, x_{j_k} } \le \epsilon$.
For some $j \in \set {0, \dots, n}$, we must have $j_k = j$ for infinitely many $k$, and it follows by setting $x := x_{j_k}$.
Setting $x := x_{j_k}$, we see that:
:$(1): \quad$ There is some $x \in X$ such that $\map d {a_k, x} \le \epsilon$ for infinitely many $k$.
Now let $\sequence {a_k}$ be any infinite sequence in $A$.
By $(1)$, there is some $x_1 \in X$ such that $\map d {a_k, x_1} \le \dfrac 1 2$ for infinitely many $k$. 
{{AxiomReview|use only axiom of countable choice, not axiom of dependent choice<br>Keep this in case someone out there can figure out a way to fix this without making it like Proof 1}}
Now we can apply $(1)$ to the subsequence of $\sequence {a_k}$ which consisting of those elements for which $\map d {a_k, x_1} \le \dfrac 1 2$.
Thus we can find $x_2 \in A$ such that infinitely many $k$ satisfy both $\map d {a_k, x_2} \le \dfrac 1 4$ and $\map d {a_k, x_1} \le \dfrac 1 2$.
Now we proceed inductively, to obtain a sequence $\sequence {x_m}$ with the property that there exist infinitely many $k$ such that, for $1 \le j \le m$:
:$(2) \quad \map d {a_k, x_j} \le 2^{-j}$
Now define a subsequence $\sequence {a_{k_m} }$ inductively by letting $k_0$ be arbitrary, and choosing $k_{m + 1}$ minimal such that $k_{m + 1} > k_m$ and such that $(2)$ holds for $k = k_m$ and all $1 \le j \le m$. 
Let $\epsilon > 0$, and choose $n$ sufficiently large that $\paren {\dfrac 1 2}^{n - 1} < \epsilon$.
Then:
:$\map d {a_{k_r}, a_{k_s} } \le \map d {a_{k_r}, x_n} + \map d {a_{k_s}, x_n} \le 2 \cdot 2^{-n} < \epsilon$
whenever $r, s \ge n$.
So this subsequence is a Cauchy sequence and hence, because $M$ is complete by assumption, it is convergent.
Thus we see that $\sequence {a_k}$ has a convergent subsequence.
Hence, by definition, $M$ is sequentially compact.
{{qed}}
Theorem: Complete and Totally Bounded Metric Space is Sequentially Compact

Let $M = \struct {A, d}$ be a metric space.
Let $M$ be complete and totally bounded.
Then $M$ is sequentially compact.

Proof:
We use the following {{Lemma|Complete and Totally Bounded Metric Space is Sequentially Compact}}, which depends on the {{Axiom-link|Countable Choice}}.
=== Lemma ===
{{:Complete and Totally Bounded Metric Space is Sequentially Compact/Lemma}}{{qed|lemma}}
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $A$.
In the following, we will construct recursively nested subsequences:
:$\sequence {x^{\paren 0}_n}_{n \mathop \in \N}\supseteq \sequence {x^{\paren 1}_n}_{n \mathop \in \N}\supseteq \sequence {x ^{\paren 2}_n}_{n \mathop \in \N} \supseteq\cdots$
such that:
:$\forall N\in\Z_{>0} : \forall k,l \in \N: \map d {x^{\paren N}_k, x^{\paren N}_l} \le \dfrac 1 N$
First, let:
:$\sequence {x^{\paren 0}_n}_{n \mathop \in \N} = \sequence {x_n}_{n \mathop \in \N}$
Then, for each $N \in \Z_{>0}$, applying $f$ from lemma, let:
:$\sequence {x^{\paren N}_n}_{n \mathop \in \N} := \map f {N, \sequence {x^{\paren {N - 1} }_n}_{n \mathop \in \N}}$
Now, let:
:$y_n := x^{\paren n}_n$
for all $n \in \N$.
By construction:
:$\sequence {y_n}_{n \mathop \in \N}$ is a subsequence of $\sequence {x_n}_{n \mathop \in \N}$
and:
:$\forall N \in \Z_{>0}: \sequence {y_n}_{n \mathop \ge N}$ is a subsequence of $\sequence {x^{\paren N}_n}_{n \mathop \in \N}$
From the latter follows:
:$\forall N\in\Z_{>0} : \forall m, n \ge N : \map d {y_m, y_n} \le \dfrac 1 N $
that implies that $\sequence {y_n}_{n \mathop \ge N}$ is a Cauchy sequence.
As $\struct {A, d}$ is complete, $\sequence {y_n}_{n \mathop \in \N}$ is convergent.
{{qed}}
Theorem: Closed Bounded Subset of Real Numbers is Compact

Let $\R$ be the real number line considered as an Euclidean space.
Let $S \subseteq \R$ be a closed and bounded subspace of $\R$.
Then $S$ is compact in $\R$.

Proof:
Let $S$ be closed and bounded.
As $S$ is bounded, there exist some $a, b \in \R$ such that:
:$S \subseteq \openint a b$
where $\openint a b$ is the open interval between $a$ and $b$.
It follows that $S \subseteq \closedint a b$.
Consider the set:
:$U = \relcomp \R S \cap \openint {a - 1} {b + 1}$
By inspection it can be seen that:
:$U = \openint {a - 1} a \cup \openint b {b + 1}$
By Union of Open Sets of Metric Space is Open, it follows that $U$ is open in $\R$.
Let $\CC$ be an open cover for $S$.
Then we can construct the open cover $\CC' = \CC \cup \set U$ for $\closedint a b$.
Let $\FF' \subseteq \CC'$ be a finite subcover of $\CC'$ for $\closedint a b$.
Then $\FF = \FF' \setminus \set U$ is the desired finite subcover for $S$.
Hence it is sufficient to prove that any open cover for $\closedint a b$, has a finite subcover.
So, suppose $S = \closedint a b$ and create the set $T \subseteq \closedint a b$ as follows:
Let $a \le x \le b$.
Then $x \in T$ {{iff}} $\closedint a x$ has a finite subcover of $\CC$.
We have that $a \in T$ and that $b$ is an upper bound for $T$.
Let $l = \map \sup T$ be the supremum of $T$.
Let $L \in \CC$ such that $l \in L$.
Since $L$ is open:
:$\exists \epsilon \in \R_{>0}: \openint {l - \epsilon} {l + \epsilon} \subseteq L$
Since $l = \map \sup T$ there exists $t \in T$ such that $t > l - \epsilon$.
We have that $\FF$ is a finite subset of $\CC$ such that $\ds \closedint a t \subseteq \bigcup \FF$.
Then $\FF \cup \set L$ is a finite subset of $\CC$ whose union contains $\closedint a {l + \delta}$ for every $\delta \in \openint 0 \epsilon$.
Since $l$ is an upper bound for $T$, it follows that $l + \delta \notin T$.
Thus $l + \delta > b$ for all $\delta > 0$.
That is, $l \ge b$.
But by definition, $l \le b$.
So $l = b$ and so $\CC$ has a finite subcover for $\closedint a b$.
{{qed}}
Theorem: Non-Closed Set of Real Numbers is not Compact

Let $\R$ be the set of real numbers considered as an Euclidean space.
Let $S \subseteq \R$ be non-closed in $\R$.
Then $S$ is not a compact subspace of $\R$.

Proof:
Consider the complement of $S$ in $\R$:
:$S' = \relcomp \R S = \R \setminus S$
As $S$ is not closed, by definition $S'$ is not open.
Thus by definition there exists $x \in S'$ such that:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \notin S'$
where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$.
That is:
:$\forall \epsilon \in \R_{>0}: \exists y \in S: y \in \map {B_\epsilon} x$
That is, for all positive $\epsilon$ there exists a $y$ within $\epsilon$ of $x$.
From Union of Open Sets of Metric Space is Open:
:$O_\epsilon := \openint \gets {x - \epsilon} \cup \openint {x + \epsilon} \to$
is an open set of $\R$.
Let $\CC = \set {O_\epsilon: \epsilon \in \R_{>0} }$.
Then:
:$\ds \bigcup \CC = \R \setminus \set x \supseteq S$
as $x \notin S$.
So $\CC$ is an open cover of $S$.
Let $\FF \subseteq \CC$ be a finite subset of $\CC$.
Then:
:$\ds \bigcup \FF = \openint \gets {x - \epsilon'} \cup \openint {x + \epsilon'} \to$
for some $\epsilon' \in \R_{>0}$.
But, as has been seen, there exists $y \in S$ closer to $x$ than $\epsilon'$.
Thus $\ds \bigcup \FF$ does not cover $S$.
This construction demonstrates an open cover $\CC$ of $S$ which has no finite subcover.
So, by definition, $S$ is not a compact subspace of $\R$.
{{qed}}
Theorem: Non-Closed Set of Real Numbers is not Compact

Let $\R$ be the set of real numbers considered as an Euclidean space.
Let $S \subseteq \R$ be non-closed in $\R$.
Then $S$ is not a compact subspace of $\R$.

Proof:
From:
:Real Number Line is Metric Space
:Metric Space is Hausdorff
:Compact Subspace of Hausdorff Space is Closed
the result follows by the rule of transposition.
{{qed}}
Theorem: Non-Closed Set of Real Numbers is not Compact

Let $\R$ be the set of real numbers considered as an Euclidean space.
Let $S \subseteq \R$ be non-closed in $\R$.
Then $S$ is not a compact subspace of $\R$.

Proof:
By the rule of transposition, it suffices to show that if $S$ is a compact subspace of $\R$, then $S$ is closed.
Consider the relative complement of $S$ in $\R$:
:$T = \relcomp \R S = \R \setminus S$
It remains to be shown that $T$ is open in $\R$.
Let $x \in T$.
For all strictly positive real numbers $\epsilon \in \R_{>0}$, it follows from Union of Open Sets of Metric Space is Open that:
:$O_\epsilon := \openint \gets {x - \epsilon} \cup \openint {x + \epsilon} \to$
is open in $\R$.
Let $\CC = \set {O_\epsilon: \epsilon \in \R_{>0} }$.
Then:
:$\ds \bigcup \CC = \R \setminus \set x \supseteq S$
as $x \notin S$.
So $\CC$ is an open cover for $S$.
Let $\GF$ be a finite subcover of $\CC$ for $S$.
Then:
:$\ds \bigcup \GF = \openint \gets {x - \delta} \cup \openint {x + \delta} \to  \supseteq S$
for some strictly positive real number $\delta \in \R_{>0}$.
By the rule of transposition, it follows that:
:$\openint {x - \delta} {x + \delta} \subseteq T$
Hence, $T$ is open in $\R$.
{{qed}}
Theorem: Compact Subspace of Real Numbers is Closed and Bounded

Let $\R$ be the real number line considered as a Euclidean space.
Let $S \subseteq \R$ be compact subspace of $\R$.
Then $S$ is closed and bounded in $\R$.

Proof:
From:
: Non-Closed Set of Real Numbers is not Compact
: Unbounded Set of Real Numbers is not Compact
the result follows by the Rule of Transposition.
{{qed}}
Theorem: Compact Subspace of Real Numbers is Closed and Bounded

Let $\R$ be the real number line considered as a Euclidean space.
Let $S \subseteq \R$ be compact subspace of $\R$.
Then $S$ is closed and bounded in $\R$.

Proof:
From Real Number Line is Metric Space, $\left({\R, d}\right)$ is a metric space, where $d$ denotes the Euclidean metric on $\R$.
Therefore, the result follows from:
: Metric Space is Hausdorff
: Compact Subspace of Hausdorff Space is Closed
and:
: Compact Metric Space is Totally Bounded
: Totally Bounded Metric Space is Bounded
{{qed}}
Theorem: Category of Subobjects is Category

Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\mathbf{Sub}_{\mathbf C}} C$ be the category of subobjects of $C$.
Then $\map {\mathbf{Sub}_{\mathbf C}} C$ is a metacategory.

Proof:
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
Let $f: m_1 \to m_2$ and $g: m_2 \to m_3$ be morphisms of $\map {\mathbf{Sub}_{\mathbf C}} C$.
That $g \circ f: m_1 \to m_3$ is again a morphism follows from the following commutative diagram in $\mathbf C$:
$\quad\quad \begin{xy}\xymatrix@+2em{
\operatorname{dom} m_1
\ar[r]_*+{f}
\ar[rd]_*+{m_1}
\ar@/^1pc/[rr]^*+{g \circ f}
&
\operatorname{dom} m_2
\ar[r]_*+{g}
\ar[d]^*+{m_2}
&
\operatorname{dom} m_3
\ar[ld]^*+{m_3}
\\
&
C
}\end{xy}$
For every subobject $m$ of $C$, the diagram:
$\quad\quad \begin{xy}\xymatrix{
\operatorname{dom} m
\ar[r]^*+{\operatorname{id}_m}
\ar[rd]_*+{m}
&
\operatorname{dom} m
\ar[d]^*+{m}
\\
&
C
}\end{xy}$
in $\mathbf C$ commutes, for $\operatorname{id}_m = \operatorname{id}_{\operatorname{dom} m}$, and the latter is an identity morphism in $\mathbf C$.
This last property also immediately proves the required behaviour of $\operatorname{id}_m$ with respect to other morphisms of $\map {\mathbf{Sub}_{\mathbf C}} C$.
Since the composition is inherited from $\mathbf C$, it is necessarily associative.
Hence $\map {\mathbf{Sub}_{\mathbf C}} C$ is a metacategory.
{{qed}}
Theorem: Category of Subobjects is Preorder Category

Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\mathbf{Sub}_{\mathbf C} } C$ be the category of subobjects of $C$.
Then $\map {\mathbf{Sub}_{\mathbf C} } C$ is a preorder category.

Proof:
By Category of Subobjects is Category, we know $\map {\mathbf{Sub}_{\mathbf C} } C$ is a metacategory.
By definition of preorder category, it suffices to show that if $f, g: m \to m'$ are morphisms with the same domain and codomain, then $f = g$.
The situation is sketched by the following commutative diagram in $\mathbf C$:
$\quad\quad \begin{xy}\xymatrix@+1em{
\Dom m
\ar[r]<2pt>^*+{f}
\ar[r]<-2pt>_*+{g}
\ar[rd]_*+{m}
&
\Dom {m'}
\ar[d]^*+{m'}
\\
&
C
}\end{xy}$
Thus, we see that $m' \circ f = m = m' \circ g$.
Now $m'$ is a subobject, and {{afortiori}} a monomorphism.
Hence $f = g$, and $\map {\mathbf{Sub}_{\mathbf C} } C$ is a preorder category.
{{qed}}