hcju/proofwikiqa huggingface dataset:
Consider: : $A := \R \setminus \left[{a \,.\,.\, b}\right) = \left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$ Let $A:= \left({-\infty \,.\,.\, a}\right)$ and let $B := \left[{b \,.\,.\, \infty}\right)$. Let $B_\epsilon \left({b}\right)$ be the open $\epsilon$-ball of $b$. We have that $b - \epsilon < b$ and so $B_\epsilon \left({b}\right) = \left({b - \epsilon \,.\,.\, b + \epsilon}\right)$ does not lie entirely in $\left[{b \,.\,.\, \infty}\right)$. Now $b - \epsilon$ may itself lie in $A$. However: : $\left[{a \,.\,.\, b}\right) \cap B_\epsilon \left({b}\right) \ne \varnothing$ and so there are elements of $B_\epsilon \left({b}\right)$ which are not in $\left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$. So $\left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$ is not an open set of $\R$. Thus, by definition, $\left[{a \,.\,.\, b}\right)$ is not a closed set of $\R$. Hence the result. Mutatis mutandis, the argument also shows that $\left({a \,.\,.\, b}\right] \subset \R$ is neither an open set nor a closed set of $\R$. {{qed}}
Consider: : $A := \left({c \,.\,.\, b}\right) \setminus \left[{a \,.\,.\, b}\right) = \left({c \,.\,.\, a}\right)$ Then $A$ is an open interval. By Open Real Interval is Open Set, $A$ is open in $\R$. By the definition of the subspace topology it follows that $A$ is open in $\left({c \,.\,.\, b}\right)$. Thus by the definition of closed set, $\left({c \,.\,.\, b}\right) \setminus A = \left[{a \,.\,.\, b}\right)$ is closed in $\left({c \,.\,.\, b}\right)$. Mutatis mutandis, the argument also shows that $\left({a \,.\,.\, b}\right] \subset \R$ is a closed set of $\left({a \,.\,.\, d}\right)$. {{qed}}
From the definition of topology, $S$ is open in $T$. From Empty Set is Element of Topology, $\varnothing$ is open in $T$. From Underlying Set of Topological Space is Closed $S$ is closed in $T$. Hence the result. {{qed}}
From the definition of closed set, $U$ is open in $T = \struct {S, \tau}$ {{iff}} $S \setminus U$ is closed in $T$. From Underlying Set of Topological Space is Clopen, $S$ is open in $T$. From Set Difference with Self is Empty Set, we have $S \setminus S = \O$, so $\O$ is closed in $T$. {{qed}}
Let $x \in U$, with $U$ an open set of $\tau$. Since $\mathcal B$ is an analytic basis for $\tau$, we have: :$U = \displaystyle \bigcup \mathcal A$ for some $\mathcal A \subseteq \mathcal B$. Since $x \in \displaystyle \bigcup \mathcal A$, there is a $B \in \mathcal A$ such that $x \in B$, by definition of set union. Hence, by definition of $\mathcal B_x$, $B \in \mathcal B_x$. From Set is Subset of Union: General Result, we also have: :$B \subseteq U$ Hence $\mathcal B_x$ is a local basis at $x$. {{qed}}
Follows directly from: : Fort Space is $T_1$ : $T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space {{qed}} Category:T0 Spaces Category:Fort Spaces 3yo8x75pff8wnyiw5wr8rl4jbiwiao4
Let $\alpha: A \to D$ be an object of $\mathbf C \mathbin / D$. Then the identity morphism $\operatorname{id}_\alpha: \alpha \to \alpha$ is by definition $\operatorname{id}_A: A \to A$. Thus $f^* \operatorname{id}_\alpha$ is by definition the unique morphism fitting: ::$\begin{xy}\xymatrix@+1em{  A'   \ar[rr]^*{f_\alpha}   \ar[dd]_*{f^* \alpha}   \ar@{-->}[rd]^*{f^* \mathrm{id}_\alpha} & &  A   \ar[rd]^*{\mathrm{id}_\alpha}   \ar[dd]^(.4)*{\alpha} \\ &  A'   \ar[ld]^*{f^* \alpha}   \ar[rr] |{\hole} ^(.3)*{f_\alpha} & &  A   \ar[ld]^*{\alpha} \\  C   \ar[rr]_*{f} & &  D }\end{xy}$ It is apparent that $\operatorname{id}_{A'}$ satisfies this condition. Since by definition, $\operatorname{id}_{f^* \alpha} = \operatorname{id}_{A'}$, this shows that: :$f^* \operatorname{id}_\alpha = \operatorname{id}_{f^* \alpha}$ Now let $\gamma_1: \alpha_2 \to \alpha_1$ and $\gamma_2: \alpha_3 \to \alpha_2$ be morphisms of $\mathbf C \mathbin / D$. Then $f^* \left({\gamma_1 \circ \gamma_2}\right)$ is the unique morphism fitting the dashed arrow in: ::$\begin{xy}\xymatrix@+1em{  A_3'   \ar[rr]^*{f_{\alpha_3}}   \ar[dd]_*{f^* \alpha_3}   \ar@{-->}[rd] & &  A_3   \ar[rd]^*{\gamma_1 \circ \gamma_2}   \ar[dd]^(.4)*{\alpha_3} \\ &  A_1'   \ar[ld]^*{f^* \alpha_1}   \ar[rr] |{\hole} ^(.3)*{f_{\alpha_1}} & &  A_1   \ar[ld]^*{\alpha_1} \\  C   \ar[rr]_*{f} & &  D }\end{xy}$ Unfolding the composition $\gamma_1 \circ \gamma_2$ yields the expanded diagram: ::$\begin{xy}\xymatrix@+1em{  A_3'   \ar[rr]^*{f_{\alpha_3}}   \ar[ddd]_*{f^* \alpha_3}   \ar[rd]^*{f^* \gamma_2}   \ar@{-->}@/_/[rdd] & &  A_3   \ar[rd]^*{\gamma_2}   \ar[ddd]^*{\alpha_3} \\ &  A_2'   \ar[d]^*{f^* \gamma_1}   \ar[rr] |{\hole} ^(.3)*{f_{\alpha_2}} & &  A_2   \ar[d]^*{\gamma_1} \\ &  A_1'   \ar[ld]^*{f^* \alpha_1}   \ar[rr] |{\hole} ^(.3)*{f_{\alpha_1}} & &  A_1   \ar[ld]^*{\alpha_1} \\  C   \ar[rr]_*{f} & &  D }\end{xy}$ in which $f^* \left({\gamma_1 \circ \gamma_2}\right)$ again fits the dashed arrow. But this means, in particular, that we must have: :$f^* \left({\gamma_1 \circ \gamma_2}\right) = f^* \gamma_1 \circ f^* \gamma_2$ Therefore, $f^*$ is a functor. {{qed}}
Let $\epsilon \in \R_{>0}$ be a strictly positive real number. Let $\map {B_\epsilon} u$ be the open $\epsilon$-ball of $u$ in $\R$. From Distance from Subset of Real Numbers: :$\map d {u, H} = 0$ Thus by definition of distance from subset: :$\exists x \in H: \map d {u, x} < \epsilon$ Thus $x \in \map {B_\epsilon} u$. As $x \in H$ and $x \in \map {B_\epsilon} u$, from the definition of intersection: :$x \in H \cap \map {B_\epsilon} u$ The result follows from Condition for Point being in Closure. {{qed}}
From Underlying Set of Topological Space is Clopen, $S$ is open in $T$. By hypothesis, $H \subseteq S$. The result follows from Open Superset is Open Neighborhood. {{qed}} Category:Topology Category:Neighborhoods 23fxtpbd2ihzx2sxohd59o0o3q2xfuf
By hypothesis, $U$ is open in $T$. Also by hypothesis, $H \subseteq U$. The result follows by definition of open neighborhood. {{qed}} Category:Topology aqh7jnykp72r6rr7jvroil8njg00al0
=== $(1)$ implies $(2)$ === Suppose $\mathbf C$ has all finite products and equalizers. That $\mathbf C$ has pullbacks follows from Pullback as Equalizer. That $\mathbf C$ has a terminal object follows by Empty Product is Terminal Object. {{qed|lemma}} === $(2)$ implies $(1)$ === Suppose $\mathbf C$ has all pullbacks and a terminal object. From Category has Finite Products iff Terminal Object and Binary Products, it suffices to check that $\mathbf C$ has binary products. That this is the case follows from Product as Pullback. That $\mathbf C$ has equalizers follows from Equalizer as Pullback, as it is established that $\mathbf C$ has finite products. {{qed}}
Let $\epsilon \in \R_{>0}$ be a strictly positive real number. Let $\map {B_\epsilon} l$ be the open $\epsilon$-ball of $l$ in $\R$. From Distance from Subset of Real Numbers: :$\map d {l, H} = 0$ Thus by definition of distance from subset: :$\exists x \in H: \map d {l, x} < \epsilon$ Thus $x \in \map {B_\epsilon} l$. As $x \in H$ and $x \in \map {B_\epsilon} l$, from the definition of intersection: :$x \in H \cap \map {B_\epsilon} l$ The result follows from Condition for Point being in Closure. {{qed}}
Proof by Counterexample: Consider the real number line $\struct {\R, \tau_d}$ with the usual (Euclidean) topology $\tau_d$. Let: :$H_n \subseteq \R: H_n = \closedint {\dfrac 1 n} 1$ for $n \ge 2$ where $\closedint {\dfrac 1 n} 1$ denotes the closed real interval from $\dfrac 1 n$ to $1$. From Closed Real Interval is Closed Set, $\closedint {\dfrac 1 n} 1$ is a closed set of $\struct {\R, \tau_d}$. Then from Set is Closed iff Equals Topological Closure: :$\map \cl {H_n} = H_n$ Also: :$\displaystyle \bigcup_{n \mathop \ge 2} \map \cl {H_n} = \bigcup_{n \mathop \ge 2} H_n = \hointl 0 1$ However: :$\displaystyle \map \cl {\bigcup_{n \mathop \ge 2} H_n} = \closedint 0 1$ So: :$\displaystyle \bigcup_{n \mathop \ge 2} \map \cl {H_n} \ne \map \cl {\bigcup_{n \mathop \ge 2} H_n}$
Proof by Counterexample: Let $\struct {\R, \tau_d}$ denote the real number line with the usual (Euclidean) topology $\tau_d$. Let $\struct {\Q, \tau_d}$ be the rational number space, also under the usual (Euclidean) topology $\tau_d$. For a rational number $\alpha \in \Q$, let $B_\alpha$ denote the singleton containing $\alpha$. From Union of Closures of Singleton Rationals is Rational Space: :$\displaystyle \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} = \Q$ From Closure of Union of Singleton Rationals is Real Number Line: :$\displaystyle \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha} = \R$ So it is seen that: :$\displaystyle \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} \subset \displaystyle \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha}$ but it is not the case that: :$\displaystyle \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} = \displaystyle \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha}$ and the result is apparent. {{qed}}
Consider the set $\R \setminus N$, the relative complement of $N$. By definition: :$\displaystyle \R \setminus N = \openint 1 \to \cup \paren {\bigcup_{n \mathop \in \N} \openint {\frac 1 {n + 1} } {\frac 1 n} }$ {{mistake|You forgot $\hointl \gets 0$ for a start}} That is, $\R \setminus N$ is a (countable) union of open real intervals. From Open Real Interval is Open Set and its corollary, each one of these is an open set of $\R$. {{mistake|... except $\hointl \gets 0$ is not -- it's closed.}} From open set axiom $\text O 1$, it follows that $\R \setminus N$ is itself an open set of $\R$. So by definition, $N$ is closed in $\R$. From Closed Set Equals its Closure: :$N = N^-$ where $N^-$ is the closure of $N$ in $\R$. {{mistake|Not quite right -- from Closure of Integer Reciprocal Space we have that $N^- {{=}} N \cup \set 0$}} But $N$ contains no open real intervals and so has no subset which is open in $\R$. Thus by definition $N$ is nowhere dense. {{qed}}
From Closed Set equals its Closure, $H$ is closed in $T$ {{iff}}: : $H = H^-$ where $H^-$ is the closure of $H$. The result then follows directly from Nowhere Dense iff Complement of Closure is Everywhere Dense. {{qed}}
=== Definition 1 iff Definition 2 === {{:Equivalence of Definitions of Topology Generated by Synthetic Basis/Definition 1 iff Definition 2}}{{qed|lemma}} === Definition 1 iff Definition 3 === {{:Equivalence of Definitions of Topology Generated by Synthetic Basis/Definition 1 iff Definition 3}}{{qed}} Category:Topological Bases 56z39n330e0jrnu19q4hfi6yq956j5y
We use Equivalence of Definitions of Topology Generated by Synthetic Basis. We proceed to verify the open set axioms for $\tau$ to be a topology on $X$. === $(\text O 1):$ Union of Open Sets === Let $\AA \subseteq \tau$. It is to be shown that: :$\displaystyle \bigcup \AA \in \tau$ By the definition of $\tau$, it follows from Set is Subset of Union: General Result that: :$\displaystyle \forall x \in \bigcup \AA: \exists U \in \AA: \exists B \in \BB: x \in B \subseteq U \subseteq \bigcup \AA$ By the transitivity of $\subseteq$, the result follows. {{qed|lemma}} === $(\text O 2):$ Pairwise Intersection of Open Sets === Let $U, V \in \tau$. It is to be shown that: :$U \cap V \in \tau$ By the definition of a synthetic basis, we have that: :$\forall A, B \in \BB: A \cap B \in \tau$ Therefore, it follows from Set is Subset of Union: General Result and Set Intersection Preserves Subsets that: :$\displaystyle \forall x \in U \cap V: \exists A, B \in \BB: A \subseteq U, \, B \subseteq V: \exists C \in \BB: x \in C \subseteq A \cap B \subseteq U \cap V$ By the transitivity of $\subseteq$, the result follows. {{qed|lemma}} === $(\text O 3):$ Set Itself === By the definition of a synthetic basis, we have that: :$\forall x \in X: \exists B \in \BB: x \in B \subseteq X$ Hence the result. {{qed}}
Since there are no non-identity morphisms, a cone to $D$ is simply a pair: ::$\begin{xy}\xymatrix@+1em@L+3px{  C_1 &  C   \ar[l]_*+{f_1}   \ar[r]^*+{f_2} &  C_2 }\end{xy}$ of morphisms with common domain $C$. By the UMP of the binary product $C_1 \times C_2$, for such a cone to $D$ there is a unique $u: C \to C_1 \times C_2$ making: ::$\begin{xy}\xymatrix@L+3px@+1em{ &  C   \ar[dl]_*+{f_1}   \ar@{-->}[d]^*{u}   \ar[dr]^*+{f_2} \\  C_1 &  C_1 \times C_2   \ar[l]_*+{p_1}   \ar[r]^*+{p_2} &  C_2 }\end{xy}$ a commutative diagram. The conditions $p_1 \circ u = f_1$ and $p_2 \circ u = f_2$ precisely mean that $u: C \to C_1 \times C_2$ is a morphism of cones. Thus for every cone $\left({C, f_1, f_2}\right)$ to $D$, there is a unique morphism of cones to the cone $\left({C_1 \times C_2, p_1, p_2}\right)$. That is, $\left({C_1 \times C_2, p_1, p_2}\right)$ is a terminal object in $\mathbf{Cone} \left({D}\right)$, the category of cones to $D$. Hence the result, by definition of limit. {{qed}}
By definition: :$\tau_2 = \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$ We examine each of the open set axioms in turn: === $(\text O 1)$: Union of Open Sets === Let $\family {U_i}_{i \matho \in I}$ be an indexed family of elements of $\tau_2$. Let $\ds V = \bigcup_{i \mathop \in I} U_i$ be the union of $\family {U_i}_{i \mathop \in I}$. From Preimage of Union under Mapping: Family of Sets: :$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} U_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {U_i}$ By hypothesis $f^{-1} \sqbrk {U_i} \in \tau_1$. As $\tau_1$ is a topology: :$\ds \bigcup_{i \mathop \in I} f^{-1} \sqbrk {U_i} \in \tau_1$ So $f^{-1} \sqbrk V \in \tau_1$. Thus $V \in \tau_2$ and so $V$ is open by definition. {{qed|lemma}} === $(\text O 2)$: Intersection of Open Sets === Let $U$ and $V$ be elements of $\tau_2$. From Preimage of Intersection under Mapping: :$f^{-1} \sqbrk {U \cap V} = f^{-1} \sqbrk U \cap f^{-1} \sqbrk V$ As $\tau_1$ is a topology: :$f^{-1} \sqbrk U \cap f^{-1} \sqbrk V \in \tau_1$ Hence $U \cap V$ is open by definition. {{qed|lemma}} === $(\text O 3)$: Set Itself === As $f$ is a mapping its domain is $S_1$. That is: :$f^{-1} \sqbrk {S_2} = S_1$ By definition of a topology, $S_1$ is open in $T_1$. Thus by definition $S_2 \in \tau_2$. {{qed|lemma}} All the open set axioms are fulfilled, and the result follows. {{qed}}
By definition of identification topology: :$\tau_2 = \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$ That is: :$V \in \tau_2 \implies f^{-1} \sqbrk V \in \tau_1$ This is precisely the definition of a continuous mapping. {{qed}} Category:Identification Topology p1p49egjbhzpe3wej3gre4srnw99xg3
Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\left({S_1, \tau_1}\right)$. By definition: :$\tau_2 := \left\{{V \in \mathcal P \left({S_2}\right): f^{-1} \left({V}\right) \in \tau_1}\right\} \subseteq \mathcal P \left({S_2}\right)$ where $\mathcal P \left({S_2}\right)$ is the power set of $S_2$. Let $V \subseteq S_2$. Then either: : $f^{-1} \left({V}\right) \in \tau_1 \implies V \in \tau_2$ or: : $f^{-1} \left({V}\right) \notin \tau_1 \implies V \notin \tau_2$ In particular: : $f^{-1} \left({S_2}\right)  = S_1 \in \tau_1 \implies S_2 \in \tau_2$ Thus $\tau_2$ exists. By the same coin, given any $V \subseteq S_2$, either $V \in \tau_2$ or $V \notin \tau_2$, and so the set: :$\tau_2 := \left\{{V \in \mathcal P \left({S_2}\right): f^{-1} \left({V}\right) \in \tau_1}\right\}$ consists of a uniquely defined set of elements. {{qed}} Category:Identification Topology r7e2utcv5y2icdnbuoqigd7zekj6z1o
=== Necessary Condition === By definition, finite products are instances of finite limits. So are equalizers, by Equalizer as Limit. {{qed|lemma}} === Sufficient Condition === Let $C: \mathbf J \to \mathbf C$ be a diagram, with $\mathbf J$ finite. The objective is to construct the limit $D = \varprojlim_j C_j$, that is: :an object $D$ :morphisms $p_j: D \to C_j$ such that: :$C_\alpha p_i = p_j$ for each $\mathbf J$-morphism $\alpha: i \to j$ :for each $E$ and $q_j: E \to C_j$ such that $C_\alpha q_i = q_j$ for each $\alpha$, there is a unique $f: E \to D$ such that $q_j = p_j f$ for each $j$ To this end, consider the finite products: :$\displaystyle \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$ and $\displaystyle \prod_{\alpha \mathop \in \operatorname{mor} \mathbf J} C_{\operatorname{cod} \alpha}$ of the objects of the diagram and the codomains of its morphisms. Next, define the morphisms $\pi, \varepsilon: \displaystyle \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j \to \prod_{\alpha \mathop \in \operatorname{mor} \mathbf J} C_{\operatorname{cod} \alpha}$ by: :$\pi_\alpha = \pr_{C_{\operatorname{cod} \alpha}}$ :$\varepsilon_\alpha = C_\alpha \pr_{C_{\operatorname{dom} \alpha}}$ Now let $e: D \to \displaystyle \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$ be the equalizer of $\pi$ and $\varepsilon$. Also, define $p_j: D \to C_j$ by $p_j = \pr_{C_j} e$ for each $j$. Now for every $\alpha: i \to j$: {{begin-eqn}} {{eqn | l = C_\alpha p_i       | r = C_\alpha \pr_{C_j} e       | c = Definition of $p_i$ }} {{eqn | r = \varepsilon_\alpha e       | c = Definition of $\varepsilon_\alpha$ }} {{eqn | r = \pi_\alpha e       | c = $e$ equalizes $\pi$ and $\varepsilon$ }} {{eqn | r = \pr_{C_j} e       | c = Definition of $\pi_\alpha$ }} {{eqn | r = p_j       | c = Definition of $p_j$ }} {{end-eqn}} Lastly, suppose that $E$ and $q_j$ have the properties stated. By definition of the product $\displaystyle \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$, there is a morphism: :$\displaystyle q: E \to \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$ Now, for every $\alpha: i \to j$: {{begin-eqn}} {{eqn | l = \pi_\alpha q       | r = \pr_j q       | c = Definition of $\pi_\alpha$ }} {{eqn | r = q_j       | c = Definition of $q$ }} {{eqn | r = C_\alpha q_i       | c = Assumption on the $q_i$ }} {{eqn | r = C_\alpha \pr_i q       | c = Definition of $q$ }} {{eqn | r = \varepsilon_\alpha q       | c = Definition of $\varepsilon_\alpha$ }} {{end-eqn}} Hence, since $e$ equalizes $\pi$ and $\varepsilon$, it follows that there exists a unique $f: E \to D$ such that: :$q = ef$ Thus, for each $j$: :$\pr_j q = \pr_j e f$ which, by the definitions of $q_j$ and $p_j$, amounts to: :$q_j = p_j f$ Hence $D$ and the $q_j$ form the desired limit. {{qed}}
It is established in Identification Mapping is Continuous that $f$ is continuous. Let $\tau \subseteq \powerset {S_2}$ be a topology on $S_2$ for which $f$ is $\struct {\tau_1, \tau}$-continuous. Let $U \in \tau$. Then by definition of $\struct {\tau_1, \tau}$-continuous: :$f^{-1} \sqbrk U \in \tau_1$ By definition of $\tau_2$ it then follows that $U \in \tau_2$. By definition of subset it follows that $\tau \subseteq \tau_2$. Thus it has been shown that $\tau_2$ is finer than any topology on $S_2$ for which $f$ is $\struct {\tau_1, \tau}$-continuous. Hence the result, by definition of finest topology. {{qed}}
Let $H'$ denote the derived set of $H$. By Closed Set iff Contains all its Limit Points, $H$ is closed in $T$ {{iff}} $H' \subseteq H$. By Union with Superset is Superset, $H' \subseteq H$ {{iff}} $H = H \cup H'$. The result follows from the definition of closure. {{qed}}
=== 1 implies 2 === Suppose $T_H$ is compact in the sense of Definition 1. Let $\mathcal C$ be a cover of $H$ by open sets of $T$. Then for each $U \in \mathcal C$, $U \cap H$ is open in $T_H$ by definition of the subspace topology. Since $\mathcal C$ is a cover of $H$ it follows that $\mathcal C' = \left\{{U \cap H: U \in \mathcal C}\right\}$ is also a cover of $H$. By hypothesis, there exists a finite subcover $\mathcal F'$ of $\mathcal C'$ for $H$. Thus let $\mathcal F' = \left\{{U_1 \cap H, U_2 \cap H, \ldots, U_n \cap H}\right\}$ where $U_1, U_2, \ldots, U_n \in \mathcal C$. It follows that $\mathcal F = \left\{{U_1, U_2, \ldots, U_n}\right\}$ is a finite subcover of $H$. That is, $T_H$ is compact in the sense of Definition 2. {{qed|lemma}} === 2 implies 1 === Suppose $T_H$ is compact in the sense of Definition 2. Let $\mathcal C$ be a cover of $H$ by open sets of $H$. By definition of the subspace topology, each $V \in \mathcal C$ is of the form $U \cap H$ for some $U \in \tau$. It follows that the collection of all such $U$ is a cover of $H$ by open sets of $T$. By hypothesis, there exists a finite subcover $\mathcal F = \left\{{U_1, U_2, \ldots, U_n}\right\}$ of $\mathcal C$ for $H$. Thus $\mathcal F' = \left\{{U_1 \cap H, U_2 \cap H, \ldots, U_n \cap H}\right\}$ is also a finite subcover of $H$ That is, $T_H$ is compact in the sense of Definition 1. {{qed}}
The composite of $f$ and $g$ is defined as: :$g \circ f := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f \land \tuple {y, z} \in g}$ It is necessary to show that $g \circ f$ is both left-total and many-to-one. === Left-total === From the definition of a mapping: {{begin-eqn}} {{eqn | lo= \forall x \in S_1: \exists y \in S_2:       | l = \map f x       | r = y       | c = {{Defof|Mapping}} }} {{eqn | lo= \forall y \in S_2: \exists z \in S_2:       | l = \map g y       | r = z       | c = {{Defof|Mapping}} }} {{eqn | ll=\leadsto       | lo= \forall x \in S_1: \exists z \in S_3:       | l = \map g {\map f x}       | r = z       | c =  }} {{eqn | ll=\leadsto       | lo= \forall x \in S_1: \exists z \in S_3:       | l = \map {g \circ f} x       | r = z       | c = {{Defof|Composite Mapping}} }} {{end-eqn}} That is, $g \circ f$ is left-total. {{qed|lemma}} === Many-to-one === Suppose $x_1, x_2 \in S_1: x_1 = x_2$. Then as $f$ is a mapping and so by definition many-to-one: :$\map f {x_1} = \map f {x_2}$ and so: :$\map g {\map f {x_1} } = \map g {\map f {x_2} }$ demonstrating that $g \circ f$ is similarly many-to-one. {{qed}}
Let $T = \struct {S, \tau}$ be a normal space. From Normal Space is $T_3$ Space, we have that $T$ is a $T_3$ space. We also have by definition of normal space that $T$ is a $T_1$ (Fréchet) space. From $T_1$ Space is $T_0$ Space we have that $T$ is a $T_0$ (Kolmogorov) space So $T$ is both a $T_3$ space and a $T_0$ (Kolmogorov) space. Hence $T$ is a regular space by definition. {{qed}}
Let $x, y \in S : x \ne y$. Then $x_\alpha \ne y_\alpha$ for some $\alpha \in I$. Since $\struct {S_\alpha, \tau_\alpha}$ is Hausdorff then: :$\exists U, V \in \tau_\alpha: x_\alpha \in U, y_\alpha \in V : U \cap V = \O$ Let $\pr_\alpha: S \to S_\alpha$ be the projection of $S$ to $S_\alpha$. Then: {{begin-eqn}} {{eqn | l = \map {\pr_\alpha^\gets} U \cap \map {\pr_\alpha^\gets} V       | r = \map {\pr_\alpha^\gets} {U \cap V}       | c = Preimage of Intersection under Mapping }} {{eqn | r = \map {\pr_\alpha^\gets} \O }} {{eqn | r = \O }} {{end-eqn}}  By definition of the projection $\pr_\alpha$: :$\map {\pr_\alpha} x = x_\alpha \in U$ By definition of the preimage under $\pr_\alpha$: :$x \in \map {\pr_\alpha^\gets} U$ Similarly: :$y \in \map {\pr_\alpha^\gets} V$ By definition of the Tychonoff topology $\tau$: :$\map {\pr_\alpha^\gets} U, \map {\pr_\alpha^\gets} V \in \tau$ Since $x, y \in S$ were arbitrary, it follows that $T$ is a $T_2$ (Hausdorff) space. {{qed}} Category:Hausdorff Spaces Category:Product Spaces ia1v6mckgd7rylx9ukwbewey9ye5e33
Let $T_\alpha$ and $T_\beta$ be Hausdorff spaces. Let $\tuple {a, b}$ and $\tuple {c, d}$ be two distinct points of the product space $T$. Let $a = c$. Then as $\tuple {a, b} \ne \tuple {c, d}$ it follows that $b \ne d$. As $T_\beta$ is Hausdorff, there exists two disjoint open sets $U, V \subseteq T_\beta$ such that $b \in U, d \in V$. Then $T_\alpha \times U$ and $T_\alpha \times V$ are two open disjoint sets in the product space $T$ containing $\tuple {a, b}$ and $\tuple {c, d}$. Let $a \ne c$. As $T_\alpha$ is Hausdorff, there exists two disjoint open sets $U, V \subseteq T_\alpha$ such that $a \in U, c \in V$. Then $U \times T_\beta$ and $V \times T_\beta$ are two open disjoint sets in the product space $T$ containing $\tuple {a, b}$ and $\tuple {c, d}$. {{qed}}