hcju/proofwikiqa huggingface dataset:
=== Existence === By Power Set is Sigma-Algebra, there is at least one $\sigma$-algebra containing $\mathcal G$. Next, let $\Bbb E$ be the collection of $\sigma$-algebras containing $\mathcal G$: :$\Bbb E := \set {\Sigma': \mathcal G \subseteq \Sigma', \text{$\Sigma'$ is a $\sigma$-algebra} }$ By Intersection of Sigma-Algebras, $\Sigma := \bigcap \Bbb E$ is a $\sigma$-algebra. Also, by Set Intersection Preserves Subsets: :$\mathcal G \subseteq \Sigma$ Now let $\Sigma'$ be a $\sigma$-algebra containing $\mathcal G$. By construction of $\Sigma$, and Intersection is Subset: General Result: :$\Sigma \subseteq \Sigma'$ {{qed|lemma}} === Uniqueness === Suppose both $\Sigma_1$ and $\Sigma_2$ are $\sigma$-algebras generated by $\mathcal G$. Then property $(2)$ for these $\sigma$-algebras implies both $\Sigma_1 \subseteq \Sigma_2$ and $\Sigma_2 \subseteq \Sigma_1$.  By definition of set equality: :$\Sigma_1 = \Sigma_2$ {{qed}} {{expand|Add proof considering $\Sigma$ as a magma of sets}}
We need to show that conditions '''B1''' and '''B2''' in the definition for basis hold for $\PP$. :'''B1''': Since $A_1 \in \tau_1$ and $A_2 \in \tau_2$, $A_1 \times A_2 \in \PP$. :'''B2''': Suppose $U_1, V_1 \in \tau_1$ and $U_2, V_2 \in \tau_2$. From Cartesian Product of Intersections, $\paren{U_1 \times U_2} \cap \paren{V_1 \times V_2} = \paren{U_1 \cap V_1} \times \paren{U_2 \cap V_2}$. Since $\paren{U_1 \cap V_1} \in \tau_1$ and $\paren{U_2 \cap V_2} \in \tau_2$, we have $\paren{U_1 \cap V_1} \times \paren{U_2 \cap V_2} \in \PP$. So $\paren{U_1 \times U_2} \cap \paren{V_1 \times V_2}$ is the union of (one) set in $\PP$. Hence the result. {{qed}}
From Natural Basis of Tychonoff Topology:Finite Product, a basis for $\tau$ is: :$\BB = \set {U \times V: U \in \tau_1, V \in \tau_2}$ et $U$ be open in $T_1$. Then $\map {\pr_1^{-1} } U = U \times T_2$ is one of the open sets in the basis in the definition of product topology. Thus $\pr_1$ is continuous. The same argument can be applied to $\pr_2$.
=== Necessary Condition === Let $f$ be continuous. Then by: :Projection from Product Topology is Continuous :Composite of Continuous Mappings is Continuous so are $\pr_1 \circ f$ and $\pr_2 \circ f$. {{qed|lemma}} === Sufficient Condition === Suppose $\pr_1 \circ f$ and $\pr_2 \circ f$ are continuous. Let $U_1 \subseteq T_1$ and $U_2 \subseteq T_2$. Then we have: {{begin-eqn}} {{eqn | l = \map {f^{-1} } {U_1 \times U_2}       | r = \map {f^{-1} } {\paren{U_1 \times T_2} \cap \paren{T_1 \times U_2} }       | c = Cartesian Product of Intersections }} {{eqn | r = \map {f^{-1} } {U_1 \times T_2} \cap \map {f^{-1} } {T_1 \times U_2}       | c =  }} {{eqn | r = \map {f^{-1} } {\map {\pr_1^{-1} } {U_1} } \cap \map {f^{-1} } {\map {\pr_2^{-1} } {U_2} }       | c =  }} {{eqn | r = \map {\paren{\pr_1 \circ f}^{-1} } {U_1} \cap \map {\paren{\pr_2 \circ f}^{-1} } {U_2}       | c =  }} {{end-eqn}} Let $U_1$ and $U_2$ be open in $T_1$ and $T_2$ respectively. Then by continuity of $\pr_1 \circ f$ and $\pr_2 \circ f$: :$\map {\paren{\pr_1 \circ f}^{-1} } {U_1}$ and $\map {\paren{\pr_2 \circ f}^{-1} } {U_2}$ are open in $T'$. So $\map {f^{-1} } {U_1 \times U_2}$ is open in $T'$. From Natural Basis of Tychonoff Topology of Finite Product, a basis for the product space $T$ is: :$\BB = \set{U_1 \times U_2 : U_1 \text{ is open in } T_1, U_2 \text{ is open in } T_2}$ Then we have shown that: :$\forall B \in \BB : \map {f^{-1}} B$ is open in $T'$   It follows from Continuity Test using Basis that $f$ is continuous. {{qed}}
From the definition of closed set, $U$ is open in $T$ {{iff}} $S \setminus U$ is closed in $T$. From Underlying Set of Topological Space is Clopen, $S$ is both open and closed in $T$. From Empty Set is Element of Topology, $\O$ is open in $T$. From Empty Set is Closed in Topological Space, we have that $\O$ is closed in $T$. {{qed}}
From Absolutely Continuous Real Function is Uniformly Continuous: :$f$ is uniformly continuous. Therefore, from Uniformly Continuous Real Function is Continuous: :$f$ is continuous. {{qed}} Category:Continuous Real Functions Category:Absolutely Continuous Functions n2cjhawz8cn1ke0wlnevycp12gjkgkt
=== Necessary Condition === Suppose $V \subseteq T'$ is closed in $T'$. Then $T' \setminus V$ is open in $T'$ by definition. So, by definition of subspace topology, $T' \setminus V = T' \cap U$ for some $U$ open in $T$. Then: {{begin-eqn}} {{eqn | l = V       | r = T' \setminus \left({T' \setminus V}\right)       | c = Relative Complement of Relative Complement }} {{eqn | r = T' \setminus \left({T' \cap U}\right)       | c = from above }} {{eqn | r = T' \setminus U       | c = Set Difference with Intersection is Difference }} {{eqn | r = T' \cap \left({T \setminus U}\right)       | c =  }} {{end-eqn}} Thus $T \setminus U$ is closed in $T$. {{qed|lemma}} === Sufficient Condition === Conversely, suppose $V = T' \cap W$ where $W$ closed in $T$. Then $T' \setminus V = T' \setminus \left({T' \cap W}\right) = T' \cap \left({T \setminus W}\right)$ which is open in $T'$. So $V$ is closed in $T'$. {{qed}}
First we show the following. Let $W \in T_2$. We note that $f^{-1} \sqbrk T_2 = T_1$. Hence, from Preimage of Set Difference under Mapping, we have: :$f^{-1} \sqbrk {T_2 \setminus W} = T_1 \setminus f^{-1} \sqbrk W$ === Necessary Condition === Suppose the condition on closed sets holds. Let $U$ be open in $T_2$. By Relative Complement of Relative Complement, $T_2 \setminus \paren { T_2 \setminus U } = U$. Therefore $T_2 \setminus \paren { T_2 \setminus U }$ is open in $T_2$. Then $T_2 \setminus U$ is closed in $T_2$. By hypothesis, $f^{-1} \sqbrk {T_2 \setminus U} = T_1 \setminus f^{-1} \sqbrk U$ is closed in $T_1$. So $f^{-1} \sqbrk U$ is open in $T_1$. This is true for any $U \in T_2$. Hence $f$ is continuous. {{qed|lemma}} === Sufficient Condition === Now let $f$ be continuous. Let $V$ be closed in $T_2$. Then $T_2 \setminus V$ is open in $T_2$. As $f$ is continuous, $f^{-1} \sqbrk {T_2 \setminus V} = T_1 \setminus f^{-1} \sqbrk V$ is open in $T_1$. So $f^{-1} \sqbrk V$ is closed in $T_1$. {{qed}}
We prove the equivalent result that every sequence $\sequence {x_k}_{k \mathop \in \N}$ omits at least one $x \in \R$. Let $\sequence {x_k}_{k \mathop \in \N}$ be a sequence of distinct real numbers. Let a sequence of closed real intervals $\sequence {I_n}$ be defined as follows: Let: :$a_k = \min \set {x_k, x_{k + 1} }$ :$b_k = \max \set {x_k, x_{k + 1} }$ and: :$I_k = \closedint {a_k} {b_k}$ Since the terms of $\sequence {x_k}_{k \mathop \in \N}$ are distinct, $a_k \ne b_k$. Thus $I_k$ is not a singleton. Let: :$I_{n - 1} = \closedint {a_{n - 1} } {b_{n - 1} }$ It can be assumed that infinitely many of the $x_k$ lie inside $I_{n - 1}$. Otherwise the proof is complete. Let $y$ and $z$ be the first two such terms of $\sequence {x_k}_{k \mathop \in \N}$. Let: :$a_n = \min \set {y, z}$ :$b_n = \max \set {y, z}$ :$I_n = \closedint {a_n} {b_n}$ Thus we have sequences: :$\sequence {a_k}_{k \mathop \in \N}$ :$\sequence {b_k}_{k \mathop \in \N}$ with: :$ a_1 < a_2 < \cdots < b_2 < b_1$ So $\sequence {a_k}_{k \mathop \in \N}$ and $\sequence {b_k}_{k \mathop \in \N}$ are monotone, and bounded above and bounded below respectively. Therefore by the Monotone Convergence Theorem (Real Analysis) both are convergent. Let: :$\displaystyle A = \lim_{k \mathop \to \infty} a_k$ :$\displaystyle B = \lim_{k \mathop \to \infty} b_k$ Clearly we have $A \le B$. So: :$\closedint A B \ne \O$ Let $h \in \closedint A B$. Then: :$h \ne a_k, b_k$ for all $k$. We claim that $h \ne x_k$ for all $k$.  Suppose that $h = x_k$ for some $k$.  Then there are only finitely many points in the sequence before $h$ occurs. {{Explain|"the sequence" -- which one in particular? There are a few under discussion here.}} Therefore only finitely many of the $a_k$ precede $h$. Let $a_d$ be the last element of the sequence $\sequence {a_k}_{k \mathop \in \N}$ preceding $h$. We defined $a_{d + 1}$, $b_{d + 1}$ to be interior points of $I_d$, and also $h \in I_{d + 1}$ by the definition of $h$. {{explain|Where are those interior points so defined?}} Therefore $a_{d + 1}$ must precede $h$ in the sequence, for the sequence is monotone increasing. This is a contradiction. {{explain|Specify precisely what assumption this contradiction falsifies.}} {{qed}}
By definition, a perfect set is a set $X$ such that every point $x \in X$ is the limit of a sequence of points of $X$ distinct from $x$. From Real Numbers form Perfect Set, $\R$ is perfect. Therefore it is sufficient to show that a perfect subset of $X \subseteq \R^k$ is uncountable. We prove the equivalent result that every sequence $\sequence {x_k}_{k \mathop \in \N} \subseteq X$ omits at least one point in $X$. Let $y_1 \in X$. Let $B_1:= \map {\BB_r} {y_1}$ be a closed ball centred at $y_1$. Consider a closed ball: :$B_{n - 1} \supseteq B_n := \map {\BB_{\map \delta n} } {y_n}$ such that: :$\map \delta n \le \dfrac {\map \delta {n - 1} } 2$ :$y_n \in X$ :$x_n \notin B_n$ Note that $\map \delta 1 = r$. We can satisfy the condition $x_n \notin B_n$ because $X$ is perfect, so every ball centred at a point of $X$ contains infinitely many points of $X$. Since $\map \delta n \le \dfrac r {2^{n - 1} }$, $y_n$ is Cauchy. Therefore from Perfect Set is Closed, we may let $\displaystyle Y = \lim_{n \mathop \to \infty} y_n \in X$. For $n \in \N$ we have: :$\set {y_m: m > n} \subseteq B_n$ so $Y \in B_n$. But by construction: :$\forall n \in \N: x_n \notin B_n$ Therefore: :$\forall n \in \N: Y \ne x_n$ and the proof is complete. {{qed}}
By Surjection from Natural Numbers iff Countable, a set $A$ is countable {{iff}} there exists a surjection $f: \N \to A$. Suppose there exists a surjection $f: \N \to \R$. Then $\forall x \in \R: \exists n \in \N: f \left({n}\right) = x$ as $f$ is surjective. Let $d_{n, 0}$ be the integer before the decimal point of $f \left({n}\right)$. Similarly, for all $m > 0$, let $d_{n, m}$ be the $m$th digit in the decimal expansion of $f \left({n}\right)$. Let $e_0$ be an integer different from $d_{0, 0}$. Similarly, for all $m > 0$, let $e_m$ be an integer different from $d_{m, m}$. Specifically, we can define $e_0$ to be $d_{0, 0} + 1$, and: :$e_m = \begin{cases} 1 & : d_{m, m} \ne 1 \\ 2 & : d_{m, m} = 1 \end{cases}$ Now consider the real number $\displaystyle x = e_0 + \sum_{n \mathop = 1}^\infty \frac {e_n} {10^n}$. Its decimal expansion is: :$x = \left[{e_0 . e_1 e_2 e_3 \ldots}\right]_{10}$ Since $e_0 \ne d_{0, 0}$, $x \ne f \left({0}\right)$. Similarly, for each $n \in \N$ such that $n \ge 1$, we have that $e_n \ne d_{n, n}$ and so $x \ne f \left({n}\right)$. Thus $x$ is a real number which is not in the set $\left\{{f \left({n}\right): n \in \N}\right\}$. Hence $f$ can not be surjective. {{qed}}
By Cantor's Theorem there is no surjection: : $\N \twoheadrightarrow \mathcal P \left({\N}\right)$ Additionally, we have Power Set of Natural Numbers is not Countable. Therefore, if we can show that $\mathcal P \left({\N}\right)$ injects into $\R$ then there is no injection $\R \hookrightarrow \N$ and $\R$ is uncountable. To prove the theorem we construct an injection $f: \mathcal P \left({\N}\right) \to \R$. For a subset $S \subseteq \N$, let $\chi_S$ be the characteristic function of $S$, and let $d_i = \chi_S \left({i}\right)$ for all $i \in \N$. By the definition of characteristic function, $\left\langle{d_i}\right\rangle_{i \in \N}$ is an infinite sequence of $1$s and $0$s. There are two cases: $\left\langle{d_i}\right\rangle_{i \in \N}$ terminates in an infinite sequence of $1$s, or it does not. Suppose $\left\langle{d_i}\right\rangle_{i \in \N}$ does ''not'' terminate in an infinite sequence of $1$s. Then $f \left({S}\right)$ is the binary expansion of the following number in $\left[{0 \,.\,.\, 1}\right)$: :$0.d_1 d_2 d_3 d_4 \ldots$ Otherwise $\left\langle{d_i}\right\rangle_{i \in \N}$ ''does'' terminate in an infinite sequence of $1$s. Then $f \left({S}\right)$ is the integer expressed in binary as: :$1 d_1 d_2 d_3 \ldots d_k$ where $d_k$ is the last member of the sequence not equal to $1$. In either case, every subset of $\N$, that is, element of $\mathcal P \left({\N}\right)$, is mapped to an element of $\R$. That $f$ is an injection follows from the uniqueness statement of Existence of Base-N Representation. {{qed}}
It is sufficient to show that the real interval $I = \set {x \in \R: 0 < x \le 1}$ is uncountable. Let $x \in I$. From Existence of Base-N Representation, $x$ has a unique representation of the form: :$x = \dfrac {\epsilon_1} 3 + \dfrac {\epsilon_2} {3^2} + \dfrac {\epsilon_3} {3^3} + \cdots$  where $\epsilon_k = 0, 1$ or $2$ and an infinite number of $\epsilon_k$ are different from $0$. Let $S \subseteq I$ be countably infinite. Let $S = \set {x_1, x_2, \ldots}$. Let $\epsilon_{k 1}, \epsilon_{k 2}, \epsilon_{k 3}, \ldots$ be the ternary digits of $x_k$. Let $\epsilon_k = 1 + 2 \epsilon_{k k} - \epsilon_{k k}^2$ so that: :$\epsilon_k = 1$ if $\epsilon_{k k} = 0$ or $\epsilon_{k k} = 2$ :$\epsilon_k = 2$ if $\epsilon_{k k} = 1$ Then: :$(1): \quad \forall k: \epsilon_k \ne 0$ :$(2): \quad \forall k: \epsilon_k \ne \epsilon_{k k}$ We have that $0 < x \le 1$ so $x \in I$. But the real number $\displaystyle x = \sum \epsilon_k 3^{-k}$ is different from every $x_k \in I$. Thus we have found an element of $I$ which is not an element of $S$. Therefore $S$ is a proper subset of $I$. It follows by definition that $I$ is uncountable. {{qed}}
Define a mapping $f: \powerset {\N_{>0} } \to \R$ thus: :$\map f S = 0.d_1 d_2 \ldots$, interpreted as a ternary expansion where $\sequence {d_n}$ is the characteristic function of $S$. That is: :$\displaystyle \map f S = \sum_{i \mathop \in S} 3^{-i}$ By the lemma, $f$ is an injection. {{AimForCont}} that $\R$ is countable. Then there is an injection $g: \R \to \N$. By Composite of Injections is Injection, $g \circ f: \powerset \N \to \N$ is an injection. But this contradicts No Injection from Power Set to Set. Hence, by Proof by Contradiction, $\R$ is not countable. {{qed}}
From the definition of closure, we have that $H^-$ is the union of $H$ and all the limit points of $H$ in $T$. === Necessary Condition === Suppose $x \in H^-$. Then either: :$(1): \quad x \in H$, in which case every open set of $T$ which contains $x$ trivially contains a point in $H$ (that is, $x$ itself); :$(2): \quad x$ is a limit point of $H$ in $T$. If the latter is the case, then it follows directly from the definition of limit point that every open set of $T$ which contains $x$ contains a point in $H$ other than $x$. === Sufficient Condition === Suppose that every open set of $T$ which contains $x$ contains a point in $H$. If $x \in H$, then $x$ is in the union of $H$ and all the limit points of $H$ in $T$. Hence by definition $x \in H^-$. If $x \notin H$ then $x$ must be a limit point of $H$ by definition. So again, $x$ is in the union of $H$ and all the limit points of $H$ in $T$. Hence by definition $x \in H^-$. {{qed}}
Let $H^{\complement}$ denote the relative complement of $H$ in $T$. By definition, we have that $H$ is closed in $T$ {{iff}} $H^{\complement}$ is open in $T$. By Set is Open iff Neighborhood of all its Points, this is equivalent to: :$\forall x \in H^{\complement}: \exists U \in \tau: x \in U \subseteq H^{\complement}$ By Empty Intersection iff Subset of Complement, we have that: :$U \subseteq H^{\complement} \iff U \cap H = \O$ By Condition for Point being in Closure, it follows that $H^{\complement}$ is open in $T$ {{iff}}: :$\forall x \in H^{\complement}: x \notin \map \cl H$ By the Rule of Transposition, this is equivalent to $\map \cl H \subseteq H$. From Set is Subset of its Topological Closure, we have that $H \subseteq \map \cl H$. By definition of set equality, this is equivalent to: :$H = \map \cl H$ {{qed}}
From the definition of closure: :$\map \cl H$ is the union of $H$ and its limit points. Let $x \in \map \cl H$. If $x \in H$ then $x \in K \implies x \in \map \cl K$. Otherwise $x$ is a limit point of $H$. That is, every open set $U$ of $T$ such that $x \in U$ contains $y \in H$ such that $y \ne x$. But as $y \in H$ it follows that $y \in K$. So every open set $U$ of $T$ such that $x \in U$ contains $y \in K$ such that $y \ne x$. This is the definition for a limit point of $K$. Thus $x \in \map \cl K$. {{ExtractTheorem|Extract the above paragraph into something like "Limit Point of Subset is Limit Point" (which proves that $H \subseteq K \implies H' \subseteq K'$). Then use Set Union Preserves Subsets to prove this result.}} {{qed}}
From Topological Closure is Closed, $\map \cl K$ is closed. From Set is Subset of its Topological Closure: :$K \subseteq \map \cl K$ By Subset Relation is Transitive, it follows that: :$H \subseteq \map \cl K$ Hence, by definition of closure as the smallest closed set that contains $H$: :$\map \cl H \subseteq \map \cl K$ {{qed}}
It follows directly from Set is Subset of its Topological Closure that: : $H^- \subseteq \left({H^-}\right)^-$ {{qed|lemma}} Let $x \in \left({H^-}\right)^-$. Then from Condition for Point being in Closure, any $U$ which is open in $T$ such that $x \in U$ contains some $y \in H^-$. If we consider $U$ as an open set containing $y$, it follows that: : $U \cap H \ne \varnothing$ Hence $x \in H^-$. {{qed}}
From Closure of Topological Closure equals Closure: :$\map \cl {\map \cl H} = \map \cl H$ From Set is Closed iff Equals Topological Closure, it follows that $\map \cl H$ is closed. {{qed}}
=== Definition $1$ is equivalent to Definition $2$ === This is proved in Set Closure as Intersection of Closed Sets. {{qed|lemma}} === Definition $2$ is equivalent to Definition $3$ === This is proved in Set Closure is Smallest Closed Set in Topological Space. {{qed|lemma}} === Definition $1$ is equivalent to Definition $4$ === By the definition of the interior, and Set is Subset of its Topological Closure, it easily follows that :$H^\circ \subseteq H \subseteq H^-$ Then: {{begin-eqn}} {{eqn | l = H \cup \partial H       | r = H \cup \paren {H^- \setminus H^\circ}        | c = {{Defof|Boundary (Topology)}} }} {{eqn | r = H \cup \paren {H \setminus H^\circ} \cup \paren {H^- \setminus H}       | c = Union of Relative Complements of Nested Subsets }} {{eqn | r = H \cup \paren {H^- \setminus H}       | c = Set Difference is Subset and Union with Superset is Superset }} {{eqn | r = H^-       | c = Union with Relative Complement }} {{end-eqn}} {{qed|lemma}} === Definition $1$ is equivalent to Definition $5$ === Every isolated point of $H$ is a point of $H$. So by Set Union Preserves Subsets: :$(1): \quad H^i \cup H' \subseteq H \cup H'$ If $S \setminus \paren {H^i \cup H'} = \O$, then $(1)$ yields $H^i \cup H' = H \cup H'$ and so the proof is complete. Otherwise, let $x \in S \setminus \paren {H^i \cup H'} \ne \O$. From De Morgan's Laws: :$S \setminus \paren {H^i \cup H'} = \paren {S \setminus H^i} \cap \paren {S \setminus H'}$ Thus $x$ is a point of $S$ that is neither an isolated point of $H$ nor a limit point of $H$. Then there exists an open set $U$ that contains $x$ such that: :$U \cap H \ne \set x$ and: :$H \cap \paren {U \setminus \set x} = \O$ This implies that :$U \cap H = \O$ By Condition for Point being in Closure: :$x \notin H \cup H'$ It has been shown that: :$x \notin H^i \cup H' \implies x \notin H \cup H'$ From Rule of Transposition: :$x \in H \cup H' \implies x \in H^i \cup H'$ And so :$(2): \quad H \cup H' \subseteq H^i \cup H'$ Combining $(1)$ and $(2)$: :$H \cup H' = H^i \cup H'$ Hence the result. {{qed|lemma}} === Definition $1$ is equivalent to Definition $6$ === By one of the equivalent definitions of an adherent point: :A point $x \in S$ is an '''adherent point of $H$''' {{iff}} every open neighborhood $U$ of $x$ satisfies $H \cap U \ne \O$ From Condition for Point being in Closure: :$x \in H^-$ {{iff}} every open set of $T$ which contains $x$ contains a point in $H$ where: :$H^-$ is the union of $H$ and all the limit points of $H$ in $T$ The equivalence follows. {{qed}}
Since $\displaystyle \bigcap_I H_i^-$ is an intersection of closed sets, it is closed, from Topology Defined by Closed Sets. Also, it contains $\displaystyle \bigcap_I H_i$ and so by the main definition of closure also contains $\displaystyle \left({\bigcap_I H_i}\right)^-$. {{qed}}
From Closure of Union contains Union of Closures we have that: :$\displaystyle \bigcup_{i \mathop = 1}^n \map \cl {H_i} \subseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$ We need now to show that:  :$\displaystyle \bigcup_{i \mathop = 1}^n \map \cl {H_i} \supseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$ Let $\displaystyle K = \bigcup_{i \mathop = 1}^n \map \cl {H_i}$ and $\displaystyle H = \bigcup_{i \mathop = 1}^n H_i$. From Topological Closure is Closed, all of $\map \cl {H_i}$ are closed. So $K$ is the union of a finite number of closed sets. So $K$ is itself closed, from Topology Defined by Closed Sets. From Set is Subset of its Topological Closure: :$\forall i \in \closedint 1 n: H_i \subseteq \map \cl {H_i}$ and so $H \subseteq K$. So from Topological Closure of Subset is Subset of Topological Closure: :$\map \cl H \subseteq \map \cl K$ The result follows. {{qed}}
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $S$. {{AimForCont}} that $\sequence {x_n}_{n \mathop \in \N}$ does not have a convergent subsequence. By Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit, it follows that $\sequence {x_n}_{n \mathop \in \N}$ does not have an accumulation point in $S$. Thus, for each $x \in S$, we can select an open set $U_x$ such that $x \in U_x$ and $U_x$ only contains $x_n$ for finitely many $n \in \N$. The set $\UU = \set {U_x : x \in X}$ is clearly an open cover for $S$. {{wtd|The axiom of choice is implicitly invoked above}} By the definition of a compact space, there exists a finite subcover $\set {U_{x_1}, U_{x_2}, \ldots, U_{x_m} }$ of $\UU$ for $X$. Then $U_{x_1} \cup U_{x_2} \cup \cdots \cup U_{x_m}$ contains all of $S$ since it is a cover. However, each open set in this union only contains $x_n$ for a finite number of $n$, and this is a finite union of sets. So the union only contains $x_n$ for finitely many $n$. This implies that $x_n \in X$ for only finitely many $n \in \N$. This is a contradiction, since $\sequence {x_n}$ is a sequence in $S$, and so $x_n \in S$ for all $n \in \N$. {{qed}} {{Improve|To prove this in ZF, one could simply use Compact Space is Countably Compact and Countably Compact First-Countable Space is Sequentially Compact.}} Category:Sequences Category:Compact Spaces Category:First-Countable Spaces 5slyzzaj1dyro8axdjxiz96vvkwg8hy
Follows directly from: : Infinite Sequence in Countably Compact Space has Accumulation Point : Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit {{qed}}
Let $T = \struct {S, \tau}$ be a countably compact first-countable topological space. By definition of sequentially compact, it is sufficient to show that every infinite sequence in $S$ has a convergent subsequence. Let $\sequence {s_n}$ be any sequence in $S$. By Infinite Sequence in Countably Compact Space has Accumulation Point, $\sequence {s_n}$ has an accumulation point $p \in S$. As $T$ is first-countable, $p$ has a countable local basis, say: :$\set {V_n: V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots}$ Then a subsequence $\sequence {s_{n_i} }$, where $s_{n_i} \in V_i$, converges to $p$. {{handwaving|This last statement is "by intimidation"; justification needed}} {{qed}}
Suppose $\map d {x, y} > \epsilon$. Then $\map {B_{\map d {x, y} - \epsilon} } y$ is an open set containing $y$ and not meeting $\map {B_\epsilon} x$. Hence: :$y \notin \map \cl {\map {B_\epsilon} x}$ {{qed}}
Let: : $H^\circ$ denote the interior of any $H \subseteq S$ : $H^-$ denote the closure of any $H \subseteq S$. From the definition, $H$ is nowhere dense in $T$ {{iff}} $\paren {H^-}^\circ = \varnothing$. From the definition of interior, it follows that $\paren {H^-}^\circ = \varnothing$ {{iff}} every open set of $T$ contains a point of $S \setminus \paren {H^-}$. Thus $S \setminus \paren {H^-}$ is everywhere dense by definition. {{qed}}