hcju/proofwikiqa huggingface dataset:
By Closed Sets of Double Pointed Topology, $\map \cl {\map {\pr_1} X} \times D$ is closed in $\tau$. Furthermore, for $\tuple {s, d} \in X$, one has: :$s \in \map {\pr_1} X \subseteq \map \cl {\map {\pr_1} X}$ by definition of closure. Since also $d \in D$, we conclude that: :$X \subseteq \map \cl {\map {\pr_1} X} \times D$ By $(3)$ of Equivalence of Definitions of Topological Closure, it now suffices to prove that: {{explain|Check the link above -- probably better to link directly to the specific definition of Closure, and also below}} :$\map \cl {\map {\pr_1} X} \times D \subseteq C$ for any closed set $C$ with $X \subseteq C$. By Closed Sets of Double Pointed Topology, for some $C'$ closed in $\tau$, we have: :$C = C' \times D$ so that $C' = \map {\pr_1} C$. By Image Preserves Subsets, it follows that: :$\map {\pr_1} X \subseteq C'$ and by $(3)$ of Equivalence of Definitions of Topological Closure, this means: :$\map \cl {\map {\pr_1} X} \subseteq C'$ whence by Cartesian Product of Subsets: :$\map \cl {\map {\pr_1} X} \times D \subseteq C' \times D = C$ Hence the result. {{qed}}
By Open Sets of Double Pointed Topology, $X^\circ$ must be of the form: :$X^\circ = U \times D$ with $U$ open in $\tau_S$. By Set Interior is Largest Open Set, we have for any open set $U' \times D$ of $\tau$ that: :$U' \times D \subseteq X \iff U' \times D \subseteq X^\circ = U \times D$ By Cartesian Product of Subsets, we have for open sets $V, V'$: :$V \subseteq V' \iff V \times D \subseteq V' \times D$ since $D$ is non-empty. Combining these two equivalences, we have: :$U' \times D \subseteq X \iff U' \subseteq U$ The condition that $U' \times D \subseteq X$ can be expressed by: :$\forall s \in U': \forall d \in D: \tuple {s, d} \in X$ By definition of the set $A$, this is equivalent to: :$U' \subseteq A$ Combining this with the above, it follows that: :$U' \subseteq A \iff U' \subseteq U$ Recall that $U$ is open in $\tau_S$. By Set Interior is Largest Open Set, we conclude: :$U = A^\circ$ and so: :$X^\circ = A^\circ \times D$ as desired. {{qed}}
By definition of relative complement we have: :$\relcomp S A = S \setminus A$ where $S \setminus A$ denotes set difference. By Cartesian Product Distributes over Set Difference, we have: :$\paren {S \setminus A} \times T = \paren {S \times T} \setminus \paren {A \times T}$ and the latter equals $\relcomp {S \times T} {A \times T}$. In conclusion, we obtain: :$\relcomp S A \times T = \relcomp {S \times T} {A \times T}$ as desired. Mutatis mutandis, the other statement follows from this argument as well. {{qed}}   Category:Cartesian Product Category:Set Complement dh1vzamll5e3p7yr6kpomzg7vyo5xh3
=== Necessary Condition === Let $M' = \struct {Y, d_Y}$ be bounded in $M$ in the sense that: :$\exists a \in X, K \in \R: \forall x \in Y: \map d {x, a} \le K$ Although not specified, $K \le 0$ for the definition to make sense, as $\map d {x, a} \ge 0$. Let $\map {B_{K + 1} } a$ be the open $K + 1$-ball of $x$. By definition of open ball: :$\map {B_{K + 1} } a := \set {x \in M: \map d {x, a} < K + 1}$ Let $y \in S$. Then by definition: :$\map d {y, a} \le K < K + 1$ and so: :$y \in \map {B_{K + 1} } a$ It follows by the definition of subset that: :$Y \subseteq \map {B_{K + 1} } a$ and so $Y$ can be fitted inside an open ball. {{qed|lemma}} === Sufficient Condition === Suppose: :$\exists x \in M, \epsilon \in \R_{>0}: Y \subseteq \map {B_\epsilon} x$ where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$. Then by definition: :$\forall y \in Y: \map d {x, y} \le \epsilon$ and so the condition for boundedness in $M$ is fulfilled. {{qed}}
Let $s \in S$. Then: {{begin-eqn}} {{eqn|l = s      |o = \in      |r = \complement \left({\operatorname{pr}_1 \left({X}\right)}\right) }} {{eqn|ll= \iff      |l = s      |o = \notin      |r = \operatorname{pr}_1 \left({X}\right)      |c = Definition of set complement }} {{eqn|ll= \iff      |l = \forall t \in T: \left({s, t}\right)      |o = \notin      |r = X      |c = Definition of first projection }} {{eqn|ll= \implies      |l = \exists t \in T: \left({s, t}\right)      |o = \notin      |r = X      |c = Universal Instantiation, Existential Generalisation }} {{eqn|ll= \iff      |l = \exists t \in T: \left({s, t}\right)      |o = \in      |r = \complement \left({X}\right)      |c = Definition of set complement }} {{eqn|ll= \iff      |l = s      |o = \in      |r = \operatorname{pr}_1 \left({\complement \left({X}\right)}\right)      |c = Definition of first projection }} {{end-eqn}} In conclusion: :$s \in \complement \left({\operatorname{pr}_1 \left({X}\right)}\right) \implies s \in \operatorname{pr}_1 \left({\complement \left({X}\right)}\right)$ and by definition of subset, the first relation follows. Mutatis mutandis the other relation can be established. {{qed}} Category:Projections Category:Set Complement i08z4t5bvw4gd35eho07xembiphzhlr
By definition of interior, $X^\circ$ is open in $\tau$, and: :$Y^\circ \subseteq Y$ Hence, by Subset Relation is Transitive: :$X^\circ \subseteq Y$ is an open set contained in $Y$. The result follows by Set Interior is Largest Open Set. {{qed}} Category:Subsets Category:Set Interiors rh05xi8umelkm4d6icp3t0sg58yibbu
Consider the open $\epsilon$-ball $B_\epsilon \paren a$ where $\epsilon = b + 1 - a$. As $b \ge a$ we have that $b + 1 > a$ and so $\epsilon > 0$. Let $x \in \mathcal I$. Then, whatever type of real interval $\mathcal I$ actually is, $z \ge a$ and $x \le b$. As $\epsilon > 0$ it follows that $x > a - \epsilon$. Also: {{begin-eqn}} {{eqn | l = x       | o = \le       | r = b       | c =  }} {{eqn | r = a + b - a       | c =  }} {{eqn | o = <       | r = a + \left({b  + 1 - a}\right)       | c =  }} {{eqn | r = a + \epsilon       | c =  }} {{end-eqn}} That is: : $a - \epsilon < x < a + \epsilon$ and so: : $x \in B_\epsilon \paren a$ The result follows by definition of bounded space. {{qed}}
Let $a \in \R$. Let $K \in \R_{>0}$. Consider the open $K$-ball $B_K \left({a}\right)$. By the Archimedean Principle there exists $n \in \N$ such that $n > a + K$. As $\N \subseteq \Z$: :$\exists n \in \Z: a + K < n$ and so: : $n \notin B_K \left({a}\right)$ As this applies whatever $a$ and $K$ are, it follows that there is no $B_K \left({a}\right)$ which contains all the integers. Hence the result, by definition of bounded space. {{qed}}
Consider the relative complement of $I_a$ in $\R$: :$J = \relcomp \R I = \R \setminus I = \hointr a \to$ 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 $I_a$ is not a closed set of $\R$. Mutatis mutandis, the same proof shows that neither is $I_b$ a closed set of $\R$. {{qed}} Category:Real Intervals Category:Real Number Line with Euclidean Metric 29glcnf6iq1fmqb0kswaa274dyowxzc
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
From Open Real Interval is not Closed Set, $I$ is not a closed set of $\R$. The result follows by definition of compact. {{qed}}
From Set of Integers is not Bounded, $\Z$ is not bounded in $\R$. The result follows by definition of compact. {{qed}}
=== Necessary Condition === {{:Heine-Borel Theorem/Euclidean Space/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === Let $C \subseteq \R^n$ be compact. From Compact Subspace of Metric Space is Bounded, it follows that $C$ is bounded. From Metric Space is Hausdorff, it follows that $\R^n$ is a Hausdorff space. Then Compact Subspace of Hausdorff Space is Closed shows that $C$ is closed. {{qed}} {{Namedfor|Heinrich Eduard Heine|name2 = Émile Borel}}
{{AimForCont}} $f$ is not bounded. Let $A_n = \set {x \in X: \norm {\map f x} < n}$ for every $n \in \N$. Then each $A_n$ is open, since $A_n = \map {\norm f^{-1} } {-n, n}$. {{explain|the notation $\map {\norm f^{-1} } {-n, n}$}} Moreover, $X = \bigcup A_n$. Since $X$ is compact, there are $n_1, n_2, \dots, n_m \in \N$ such that $X = \bigcup A_{n_k}$. However, since $f$ is not bounded, there exists $x \in X$ such that $\norm {\map f X} \ge \max \set {n_1, n_2, \dots, n_m}$, which is a contradiction. {{qed}} Category:Continuous Functions Category:Compact Spaces 1umavq3f7zzciycwmr8m6qorc6hz40c
=== Definition 1 implies Definition 2 === Let $T = \left({S, \tau}\right)$ be hereditarily compact by definition 1. Let $\left\langle{U_i}\right\rangle_{i \mathop \in I}$ be an indexed family of open sets of $T$. We have that: :$\displaystyle \bigcup_{i \mathop \in I} U_i \subset S$ By hypothesis, $\displaystyle \bigcup_{i \mathop \in I} U_i$ is compact when considered as a subspace of $T$. Furthermore, $\displaystyle U_i = U_i \cap \bigcup_{i \mathop \in I} U_i$ is open in $\displaystyle \bigcup_{i \mathop \in I} U_i$, by definition of the subspace topology. Therefore $\left\langle{U_i}\right\rangle_{i \mathop \in I}$ is an open cover for $\displaystyle \bigcup_{i \mathop \in I} U_i$. Since $\displaystyle \bigcup_{i \mathop \in I} U_i$ is compact, there exists a finite subcover of $\left\langle{U_i}\right\rangle_{i \mathop \in I}$ for $\displaystyle \bigcup_{i \mathop \in I} U_i$, say $\left\langle{U_j}\right\rangle_{j \mathop \in J}$. Then by definition of open cover: :$\displaystyle \bigcup_{j \mathop \in J} U_j = \bigcup_{i \mathop \in I} U_i$ Thus $T = \left({S, \tau}\right)$ is hereditarily compact by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $T = \left({S, \tau}\right)$ be hereditarily compact by definition 2. Let $Y \subset S$ be a subspace of $T$. Let $\left\langle{V_i}\right\rangle_{i \mathop \in I}$ be an open cover for $Y$: :$\displaystyle \bigcup_{i \mathop \in I} V_i = Y$ Then by definition of the subspace topology: :$V_i = U_i \cap Y$ for a certain $V_i \in \tau$ But then $\left\langle{U_i}\right\rangle_{i \mathop \in I}$ is an indexed family of open sets of $T$. By hypothesis, there exists a finite subset $J \subset I$ such that: :$\displaystyle \bigcup_{j \mathop \in J} U_i = \bigcup_{i \mathop \in I} U_i$ But then: {{begin-eqn}} {{eqn | l = \bigcup_{i \mathop \in I} V_i       | r = \bigcup_{i \mathop \in I} \left({U_i \cap Y}\right)       | c =  }} {{eqn | r = \left({\bigcup_{i \mathop \in I} U_i}\right) \cap Y       | c =  }} {{eqn | r = \left({\bigcup_{j \mathop \in J} U_j}\right) \cap Y       | c =  }} {{eqn | r = \bigcup_{j \mathop \in J} \left({U_j \cap Y}\right)       | c =  }} {{eqn | r = \bigcup_{j \mathop \in J} V_j       | c =  }} {{end-eqn}} Thus $\left\langle{V_j}\right\rangle_{j \mathop \in J}$ is a finite subcover of $\left\langle{V_i}\right\rangle_{i \mathop \in I}$ for $Y$.  Thus $Y$ is a compact subspace of $T$. Thus $T = \left({S, \tau}\right)$ is hereditarily compact by definition 1. {{qed}} Category:Hereditarily Compact Spaces 2umbvuqo8zx89g51nl151h7o5wqmfy3
=== Necessary Condition === Let $\left({S, \tau}\right)$ be hereditarily compact. That is, every subspace of $\left({S, \tau}\right)$ is compact. Let $H \subseteq S$ be a subspace of $\left({S, \tau}\right)$. 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}} === Sufficient Condition === Let $S$ be finite. From $T_2$ (Hausdorff) Space is $T_1$ Space, $\left({S, \tau}\right)$ is a $T_1$ (Fréchet) space. From Finite $T_1$ Space is Discrete, it follows that $\left({S, \tau}\right)$ has the discrete topology. As $\left({S, \tau}\right)$ (and every subspace) is finite and discrete, and Discrete Space is Compact iff Finite, then $S$ (and every subspace) is compact. Thus $\left({S, \tau}\right)$ is hereditarily compact. {{qed}}
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 hereditarily compact. {{qed}}
Let $M$ be both complete and totally bounded. Let $\left \langle{a_k}\right \rangle$ 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: :$\displaystyle A = \bigcup_{i \mathop = 1}^n B_\epsilon \left({x_i}\right)$ where $B_\epsilon \left({x_i}\right)$ 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 $j_k \in \left\{{0, \dots, n}\right\}$ such that $d \left({a_k, x_{j_k}}\right) \le \epsilon$. For some $j \in \left\{{0, \dots, n}\right\}$, 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 $d \left({a_k, x}\right) \le \epsilon$ for infinitely many $k$. Now let $\left \langle{a_k}\right \rangle$ be any infinite sequence in $A$. By $(1)$, there is some $x_1 \in X$ such that $d \left({a_k, x_1}\right) \le 1/2$ 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 $\left \langle{a_k}\right \rangle$ which consisting of those elements for which $d \left({a_k, x_1}\right) \le 1/2$. Thus we can find $x_2 \in A$ such that infinitely many $k$ satisfy both $d \left({a_k, x_2}\right) \le 1/4$ and $d \left({a_k, x_1}\right) \le 1/2$. Now we proceed inductively, to obtain a sequence $\left \langle {x_m}\right \rangle$ with the property that there exist infinitely many $k$ such that, for $1 \le j \le m$: :$(2) \quad d \left({a_k, x_j}\right) \le 2^{-j}$ Now define a subsequence $\left \langle {a_{k_m}}\right \rangle$ 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 $1/2^{n-1} < \epsilon$. Then: :$d \left({a_{k_r}, a_{k_s}}\right) \le d \left({a_{k_r}, x_n}\right) + d \left({a_{k_s}, x_n}\right) \le 2 \cdot 2^{-n} < \epsilon$ whenever $r, s \ge n$. So this subsequence is a Cauchy sequence and hence, because $M = \left({A, d}\right)$ is complete by assumption, it is convergent. Thus we see that $\left \langle{a_k}\right \rangle$ has a convergent subsequence. Hence, by definition, $M$ is sequentially compact. {{qed}}
Let $C \subseteq \R^n$ be closed and bounded. Since $C$ is bounded, $C \subseteq \left[{a \,.\,.\, b}\right]^n = B$ 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 Tychonoff Topology, it follows that $B$ is compact in the usual Euclidean topology induced by the Euclidean metric. By the Corollary to 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}}
The proof holds for $n = 1$, as follows. Suppose $C$ is a closed, bounded subspace of $\R$. Then $C \subseteq \left[{a \,.\,.\, b}\right]$ for some $a, b \in \R$. Moreover, $C$ is closed in $\left[{a \,.\,.\, b}\right]$ by the Corollary to 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, $C \subseteq \left[{a \,.\,.\, b}\right] \times \left[{a \,.\,.\, b}\right] \times \cdots \times \left[{a \,.\,.\, b}\right] = B$ for some $a, b \in \R$. Now $B$ is compact by Topological Product of Compact Spaces. Also, $C$ is closed in $B$ by the Corollary to Closed Set in Topological Subspace. Hence $C$ is compact, by Closed Subspace of Compact Space is Compact. {{qed}}
Let $S$ be closed and bounded. As $S$ is bounded, there exist some $a, b \in \R$ such that: : $S \subseteq \left({a \,.\,.\, b}\right)$ where $\left({a \,.\,.\, b}\right)$ is the open interval between $a$ and $b$. It follows that $S \subseteq \left[{a \,.\,.\, b}\right]$. Consider the set: : $U = \complement_\R \left({S}\right) \cap \left({a - 1 \,.\,.\, b + 1}\right)$ By inspection it can be seen that: : $U = \left({a - 1 \,.\,.\, a}\right) \cup \left({b \,.\,.\, b + 1}\right)$ By Union of Open Sets of Metric Space is Open, it follows that $U$ is open in $\R$. Let $\mathcal C$ be an open cover for $S$. Then we can construct the open cover $\mathcal C' = \mathcal C \cup \left\{{U}\right\}$ for $\left[{a \,.\,.\, b}\right]$. Let $\mathcal F' \subseteq \mathcal C'$ be a finite subcover of $\mathcal C'$ for $\left[{a \,.\,.\, b}\right]$. Then $\mathcal F = \mathcal F' \setminus \left\{{U}\right\}$ is the desired finite subcover for $S$. Hence it is sufficient to prove that any open cover for $\left[{a \,.\,.\, b}\right]$, has a finite subcover. So, suppose $S = \left[{a \,.\,.\, b}\right]$ and create the set $T \subseteq \left[{a \,.\,.\, b}\right]$ as follows: Let $a \le x \le b$. Then $x \in T$ {{iff}} $\left[{a \,.\,.\, x}\right]$ has a finite subcover of $\mathcal C$. We have that $a \in T$ and that $b$ is an upper bound for $T$. Let $l = \sup \left({T}\right)$ be the supremum of $T$. Let $L \in \mathcal C$ such that $l \in L$. Since $L$ is open: :$\exists \epsilon \in \R_{>0}: \left({l - \epsilon \,.\,.\, l + \epsilon}\right) \subseteq L$ Since $l = \sup \left({T}\right)$ there exists $t \in T$ such that $t > l - \epsilon$. We have that $\mathcal F$ is a finite subset of $\mathcal C$ such that $\displaystyle \left[{a \,.\,.\, t}\right] \subseteq \bigcup \mathcal F$. Then $\mathcal F \cup \left\{{L}\right\}$ is a finite subset of $\mathcal C$ whose union contains $\left[{a \,.\,.\, l + \delta}\right]$ for every $\delta \in \left({0 \,.\,.\, \epsilon}\right)$. 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 $\mathcal C$ has a finite subcover for $\left[{a \,.\,.\, b}\right]$. {{qed}}
=== Definition 1 implies Definition 2 === Let $\mathcal B$ be an analytic basis for $\tau$ by definition 1. Let $U \in \tau$. By definition 1 of an analytic basis, we can choose $\mathcal A \subseteq \mathcal B$ such that: :$\displaystyle U = \bigcup \mathcal A$ By the definition of union: :$\forall x \in U: \exists B \in \mathcal A: x \in B$ By Union is Smallest Superset: :$\forall B \in \mathcal A: B \subseteq U$ Since $\mathcal A \subseteq \mathcal B$, the result follows. Thus $\mathcal B$ is an analytic basis for $\tau$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\mathcal B$ be an analytic basis for $\tau$ by definition 2. Let $U \in \tau$. Let $\mathcal A = \left\{{B \in \mathcal B: B \subseteq U}\right\}$. Then $\mathcal A \subseteq \mathcal B$. Let $x \in U$ be arbitrary. Since $\mathcal B$ is an analytic basis for $\tau$ by definition 2, there is some $B_x \in \mathcal B$ such that: :$x \in B_x \subseteq U$ Hence, by construction of $\mathcal A$, $B_x \in \mathcal A$. Thus: :$\displaystyle x \in \bigcup \mathcal A$ and it follows that: :$\displaystyle U \subseteq \bigcup \mathcal A$ By Union is Smallest Superset applied to $\mathcal A$ and $U$: :$\displaystyle \bigcup \mathcal A \subseteq U$ By definition of set equality and definition 1 of an analytic basis, the result follows. Thus $\mathcal B$ is an analytic basis for $\tau$ by definition 1. {{qed}} Category:Topological Bases 19b01qj8o31r1o49cogo8h7hch7b4ir
By the rule of transposition, it suffices to show that if $S$ is a compact subspace of $\R$, then $S$ is bounded. Let $\mathcal C$ be the set of all open $\epsilon$-balls of $0$ in $\R$: :$\mathcal C = \set {\map {B_\epsilon} 0: \epsilon \in \R_{>0}}$ We have that: :$\displaystyle \bigcup \mathcal C = \R \supseteq S$ From Open Ball of Metric Space is Open Set, it follows that $\mathcal C$ is an open cover for $S$. Let $\mathcal F$ be a finite subcover of $\mathcal C$ for $S$. Then $\displaystyle \bigcup \mathcal F$ is the largest open $\epsilon$-ball in $\mathcal F$. Thus: :$S \subseteq \displaystyle \bigcup \mathcal F \in \mathcal C$ Hence, $S$ is bounded. {{qed}}
Consider the complement of $S$ in $\R$: :$S' = \complement_\R \left({S}\right) = \R \setminus S$ As $S$ is not closed, by definition $S'$ is not open. Thus by definition there exists $x \in S'$ such that: :$\forall \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \notin S'$ where $B_\epsilon \left({x}\right)$ denotes the open $\epsilon$-ball of $x$. That is: :$\forall \epsilon \in \R_{>0}: \exists y \in S: y \in B_\epsilon \left({x}\right)$ That is, for all positive $\epsilon$ there exists a $y$ within $\epsilon$ of $x$. From Union of Open Sets of Metric Space is Open: :$O_\epsilon := \left({-\infty \,.\,.\, x - \epsilon}\right) \cup \left({x + \epsilon \,.\,.\, \infty}\right)$ is an open set of $\R$. Let $\mathcal C = \left\{{O_\epsilon: \epsilon \in \R_{>0}}\right\}$. Then: :$\displaystyle \bigcup \mathcal C = \R \setminus \left\{{x}\right\} \supseteq S$ as $x \notin S$. So $\mathcal C$ is an open cover of $S$. Let $\mathcal F \subseteq \mathcal C$ be a finite subset of $\mathcal C$. Then: :$\displaystyle \bigcup \mathcal F = \left({-\infty \,.\,.\, x - \epsilon'}\right) \cup \left({x + \epsilon' \,.\,.\, \infty}\right)$ for some $\epsilon' \in \R_{>0}$. But, as has been seen, there exists $y \in S$ closer to $x$ than $\epsilon'$. Thus $\displaystyle \bigcup \mathcal F$ does not cover $S$. This construction demonstrates an open cover $\mathcal C$ of $S$ which has no finite subcover. So, by definition, $S$ is not a compact subspace of $\R$. {{qed}}
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}}
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}}
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}}
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory. Let $f: m \to m'$ and $g: m' \to m''$ be morphisms of $\mathbf{Sub}_{\mathbf C} \left({C}\right)$. That $g \circ f: m \to m''$ is again a morphism follows from the following commutative diagram in $\mathbf C$: ::$\begin{xy}\xymatrix@+2em{  \operatorname{dom} m   \ar[r]_*+{f}   \ar[rd]_*+{m}   \ar@/^1pc/[rr]^*+{g \circ f} &  \operatorname{dom} m'   \ar[r]_*+{g}   \ar[d]^*+{m'} &  \operatorname{dom} m''   \ar[ld]^*+{m''} \\ &  C }\end{xy}$ For every subobject $m$ of $C$, the diagram: ::$\begin{xy}\xymatrix{  \operatorname{dom} m   \ar[r]^*+{\operatorname{id}_m}   \ar[rd]_*+{m} &  \operatorname{dom} m   \ar[d]^*+{m} \\ &  C }\end{xy}$ 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 $\mathbf{Sub}_{\mathbf C} \left({C}\right)$. Since the composition is inherited from $\mathbf C$, it is necessarily associative. Hence $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ is a metacategory. {{qed}}