hcju/proofwikiqa huggingface dataset:
Let $M = \struct {A, d}$ be a metric space. Let $\sequence {C_\alpha}$ be an open cover of $M$ indexed by ordinals. Let $\map {B_r} x$ be the open $r$-ball in $M$ around $x$. For $n \in \N$, let $D_{\alpha n}$ denote the union of all the open $r$-balls $\map {B_{2^{-n} } } x$ such that: :$(1): \quad$ $\alpha$ is the smallest ordinal such that $x \in C_\alpha$ :$(2): \quad$ $x \notin D_{\beta j}$ for $j < n$ :$(3): \quad$ $\map {B_{3 \times 2^{-n} } } x \subseteq C_\alpha$ It is to be shown that $\sequence {D_{\alpha n} }$ is a locally finite open refinement of $\sequence {C_\alpha}$ which covers $M$. By construction, $D_{\alpha n}$ is a refinement of $\sequence {C_\alpha}$. Let $x \in A$. Then there exists a smallest ordinal $\alpha$ such that $x \in C_\alpha$. Then there exists $n$ sufficiently large such that $(3)$ holds. Then by $(2)$: :$x \in D_{\beta j}$ for some $j \le n$. Thus $\sequence {D_{\alpha n} }$ covers $M$. Now assume $x \in A$. Let $\alpha$ be the smallest ordinal such that $x \in D_{\alpha n}$ for some $n$. Choose $j$ such that $\map {B_{2^{-j} } } x \subseteq D_{\alpha n}$. Let $i \ge n$. As $i > n$, every one of the open balls of radius $2^{-i}$ used to define $D_{\beta i}$ has its center $y$ outside $D_{\alpha n}$ And because $\map {B_{2^{-j} } } x \subseteq D_{\alpha n}$, we have that $\map d {x, y} \ge 2^{-j}$. But $i \ge j + 1$ and $n + j \ge j + 1$. So: :$\map {B_{2^{-n - j} } } x \cap \map {B_{2^{-j} } } y = \varnothing$ Thus if $i \ge n$, then $\map {B_{2^{-n - j} } } x$ intersects no $D_{\beta i}$. Let $i < n$. Let us choose: :$p \in D_{\beta i}$ :$q \in D_{\gamma i}$ :$\beta < \gamma$ There exist points $y$ and $z$ such that: :$p \in \map {B_{2^{-j} } } y \subseteq D_{\beta i}$ :$q \in \map {B_{2^{-j} } } z \subseteq D_{\gamma i}$ By $(3)$: :$\map {B_{3 \times 2^{-n} } } y \subseteq C_\beta$ But by $(2)$: :$z \notin C_\beta$ So: :$\map d {y, z} \ge 3 \times 2^{-i}$ and: :$\map d {p, q} > 2^{-i} \ge 2^{-n - j + 1}$ Thus if $i < n$, then $\map {B_{2^{-n - j} } } x$ intersects $D_{\beta i}$ for at most one $\beta$. Thus $D_{\alpha n}$ is locally finite. Then $D_{\alpha n}$ is a locally finite open refinement of $\sequence {C_\alpha}$ which covers $M$. Hence by definition $M$ is paracompact. {{qed}}
Let $x \in A$. Let: :$\BB = \set {\map {B_{1/n} } x: n \in \N_{>0} }$ where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$ in $M$. By Surjection from Natural Numbers iff Countable, we have that $\BB$ is countable. By the definition of a first-countable space, it suffices to show that $\BB$ is a local basis at $x$. By Open Ball of Metric Space is Open Set, every element of $\BB$ is an open neighborhood of $x$. Let $U$ be an open neighborhood of $x$. By the definition of an open set, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} x \subseteq U$. By the Archimedean Principle, there exists a natural number $n > \dfrac 1 \epsilon$. That is, $\dfrac 1 n < \epsilon$, and so: :$\map {B_{1/n} } x \subseteq \map {B_\epsilon} x \subseteq U$ The result follows from Subset Relation is Transitive. {{qed}}
Follows directly from: :Separable Metric Space is Second-Countable :Second-Countable Space is Separable {{qed}} {{ACC|Second-Countable Space is Separable}}
=== Sufficient Condition === We have from Second-Countable Space is Lindelöf that second-countability implies Lindelöf in all topological spaces, regardless of whether they are metric spaces or not. {{qed|lemma}} === Necessary Condition === Suppose $M = \struct {A, d}$ is Lindelöf. Let us define the open covers on $A$: :$\CC_k = \set {\map {N_{1/k} } x: x \in S}$ for all $k \in \N_{>0}$. As $M$ is Lindelöf, each one of these has a countable subcover. The union of all these subcovers is a countable basis for the topology on $A$. Hence the result, by definition of second-countable space. {{qed}}
This follows directly from the results: :Countably Compact Metric Space is Sequentially Compact :Sequentially Compact Space is Countably Compact {{qed}} {{ACC|Sequentially Compact Space is Countably Compact}}
From Metric Space fulfils all Separation Axioms, a metric space is a $T_1$ (Fréchet) space. The result follows from $T_1$ Space is Weakly Countably Compact iff Countably Compact. {{qed}}
This follows directly from: : Metric Space is First-Countable : Countably Compact First-Countable Space is Sequentially Compact : Sequentially Compact Metric Space is Second-Countable : Second-Countable Space is Compact iff Countably Compact {{qed}}
We have that a Metric Space is Countably Compact iff Sequentially Compact. Then we have that a Sequentially Compact Metric Space is Separable. For each $n$, a metric space which is countably compact can be covered by finitely many open $\paren {1/n}$-balls: $\map {B_{1/n} } {x_i}$. {{explain|why?}} So $\set {x_i}$ is a dense subset of $A$ which is countable. So if a metric space is countably compact it is by definition second-countable. {{explain|by definition? See unresolved issues in Talk page.}} The result follows from Second-Countable Space is Compact iff Countably Compact. {{qed}}
We have from Strongly Locally Compact Space is Weakly Locally Compact that strong local compactness implies weak local compactness in all topological spaces, regardless of whether they are metric spaces or not. So all we need to do is demonstrate that if $M$ is weakly locally compact then it is strongly locally compact. We have that a metric space is a $T_2$ (Hausdorff) space. The result follows from Weakly Locally Compact Hausdorff Space is Strongly Locally Compact. {{qed}}
The relevant justifications are listed as follows: :Metric Space is Compact iff Countably Compact. :Metric Space is Countably Compact iff Sequentially Compact. :Metric Space is Weakly Countably Compact iff Countably Compact. :Compact Space is Weakly $\sigma$-Locally Compact. :By definition, a weakly $\sigma$-locally compact space is both weakly locally compact and $\sigma$-compact. :$\sigma$-Compact Space is Lindelöf. :Metric Space is Lindelöf iff Second-Countable. :Metric Space is Separable iff Second-Countable. :Metric Space is Weakly Locally Compact iff Strongly Locally Compact. {{qed}} {{ACC}} {{explain|Indicate which of the above theorems uses ACC in its proof}}
Let $M = \struct {A, d}$ be extremally disconnected. Let $p \in A$. As $M$ is a metric space, $\set p$ can be expressed as: :$\set p = \displaystyle \bigcap_{n \mathop \in \N_{>0} } \paren {\map {B_{1 / n} } p}^-$ where: :$\map {B_{1 / n} } p$ denotes the open $1 / n$-ball of $p$ :$\paren {\map {B_{1 / n} } p}^-$ denotes the closure of $\map {B_{1 / n} } p$ That is, as the intersection of the closures of the open $1 / n$-ball of $p$ for all non-zero natural numbers. Now let: :$\displaystyle U = \bigcup_{n \mathop \in \N_{>0} } \map {B_{1 / 2 n} } p \setminus \paren {\map {B_{1 / \paren {2 n + 1} } } p}^-$ Then either $U$ or the complementary set of annuli is an open set which has $p$ as a non-interior limit point provided $\set p$ is not open. {{explain|Take another run at that last sentence, it needs amplification.}} So if $M$ is not the discrete topology, it cannot be extremally disconnected. {{qed}}
The claim that $\map \chi e = 1$ is shown by Group Homomorphism Preserves Identity. Let $g \in G$ be arbitrary. Let $k$ be the order of $g$. By Order of Element Divides Order of Finite Group: :$\exists l \in \Z: m = k l$ Therefore: :$g^m = \paren {g^k}^l = e^l = e$ By the homomorphism property: :$1 = \map \chi e = \map \chi {g^m} = \map \chi g^m$ Hence the result. {{Qed}} Category:Analytic Number Theory 0ze9r5yuv8ntst78141gfauk9wn6afm
If $a \in H$, then for $n = 1$, $a^n \in H$. Suppose that $a \notin H$. By Identity of Subgroup, we have that the identity $e$ of $G$ is in $H$. Moreover, by Element of Finite Group is of Finite Order, there is $n \in \N$ such that $a^n = e \in H$. Therefore, for any $g \in G$, there is ''some'' $n \in \N$ such that $a^n \in H$. That is, the set of positive integers $n$ such that $a^n \in H$ is non-empty. So by the Well-Ordering Principle, this set has a least element, and the indicator is well defined. {{qed}} Category:Subgroups 9ellxqergqaoxuoyn896eccdklzha7o
Let $M = \struct {A, d}$ be compact. Let $\epsilon > 0$. Then the family $\set {\map {B_\epsilon} x: x \in A}$ of open $\epsilon$-balls forms an open cover of $A$. By the definition of compact, there exists a finite subcover. That is, there are points $x_0, \ldots, x_n$ such that: :$\displaystyle A = \bigcup_{0 \mathop \le i \mathop \le n} \map {B_\epsilon} {x_i}$ as required.  {{qed}}
=== $K$ is Subgroup of $G$ === We first show that $K$ is a subgroup of $G$ using the Two-Step Subgroup Test. $(1): \quad K \ne \varnothing$: From Identity of Subgroup: :$e \in H$ From Indicator is Well-Defined, $n > 0$ and so $n - 1 \ge 0$. Thus $k = 0$ fulfils the condition that $0 \le k < n$, and so: :$e = e a^0 \in K$ Thus $K \ne \varnothing$. $(2): \quad K$ is closed: Let $r = x a^k \in K$ and $s = y a^l \in K$ where $x, y \in H$ and $0 \le k < n, 0 \le l < n$. $G$ is abelian, so $r$, $s$ and $a$ commute. So: :$r s = x y a^{k + l}$. Since $H$ is a group and $x, y \in H$ then also $x y \in H$. Let $k + l \le n - 1$. Then $r s \in K$ by definition of $K$. Let $n \le k + l \le 2 n$. Then: :$r s = x y a^n a^{k + l - n}$ By definition of indicator, $a^n \in H$. Thus as $x, y, a^n \in H$ it follows that $x y a^n \in H$. As $0 \le k + l - n \le n - 1$ it follows that $\left({x y a^n}\right) a^{k + l - n} \in K$ by definition of $K$.  So again $r s \in K$. Thus $K$ is closed. $(3): \quad K$ is closed under inversion: Let $r = x a^k \in K$ where $x \in H$ and $0 \le k < n$. Then $x^{-1} \in H$. Let $k = 0$. Then  :$r^{-1} = x^{-1} e = x^{-1} a^0 \in K$ Let $k > 0$. Then: :$r^{-1} = x^{-1} a^{-k} = x^{-1} a^{-n} a^{n - k}$ By definition of indicator, $a^n \in H$. We have that $x^{-1}, a^n \in H$. As $G$ is abelian, $x$ and $a$ commute. Thus from Inverse of Commuting Pair: : $x^{-1} a^{-n} = \left({x a^n}\right)^{-1} \in H$ Also: :$0 \le n - k < n -1$ so: :$r^{-1} = \left({x a^n}\right)^{-1} a^{n - k} \in K$ Therefore by the Two-Step Subgroup Test we have shown that $K$ is a subgroup of $G$. === $H$ is Subgroup of $K$ === As: :$\forall x \in H: x = x a^0 \in K$ and so $H \subseteq K$. === Order of $K$ === It remains to be shown that: :$\left\vert{K}\right\vert = n q$ where $q = \left\vert{H}\right\vert$. Let $x a^k = y a^l \in K$. WLOG, let $k \ge l$ (if not, just relabel the two). We have: : $a^{k - l} = x^{-1} y$ Since $H$ is a subgroup of $G$: :$x^{-1} y \in H$ and therefore: :$a^{k - l} \in H$ By definition of indicator, $n$ is the least (strictly) positive integer such that $a^n \in H$ Hence because $n > k \ge k - l$: :$k - l = 0$ Therefore $k = l$, and $x = y$. Therefore the elements $x a^k$ with $x \in H$ and $0 \le k < n$ are distinct. So for each $x \in H$ there are $n$ elements of the form $x a^k$ in $K$. We have that there are $q$ elements in $H$. Thus it follows that there are $n q$ such elements. By definition of $K$, these form all of $K$. Thus: :$\left\vert{K}\right\vert = n q$ as required {{qed}} Category:Subgroups ccxml4gy512g7balk0j604fvssw1m8y
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $A$. As $M$ is sequentially compact, $\sequence {x_n}$ has a convergent subsequence. By Convergent Subsequence of Cauchy Sequence in Metric Space, this implies that the entire sequence $\sequence {x_n}$ is convergent. Hence, $M$ is complete. {{qed}}
Let $\tau_1$ be the topology on $A_1$ induced by $d_1$. Let $\tau_2$ be the topology on $A_2$ induced by $d_2$. Let $\left\langle{x_n}\right\rangle$ be a Cauchy sequence in $A_2$. From Inverse of Isometry of Metric Spaces is Isometry, $\phi^{-1}$ is an isometry. Since Isometric Image of Cauchy Sequence is Cauchy Sequence, $\left\langle{\phi^{-1} \left({x_n}\right)}\right\rangle$ is a Cauchy sequence. Since $M_1$ is a complete metric space, $\left\langle{\phi^{-1} \left({x_n}\right)}\right\rangle$ converges. Since Isometry Preserves Sequence Convergence, $\left\langle{\phi \left({\phi^{-1} \left({x_n}\right)}\right)}\right\rangle$ converges. But: : $\left\langle{\phi \left({\phi^{-1} \left({x_n}\right)}\right)}\right\rangle = \left\langle{x_n}\right\rangle$ so $\left\langle{x_n}\right\rangle$ converges. Thus each Cauchy sequence in $M_2$ converges. It follows by definition that so $M_2$ is a complete metric space. {{qed}}
Let $\epsilon \in \R_{>0}$. Let $\left\langle{b_n}\right\rangle$ be a Cauchy sequence in $A_2$. Thus: : $\exists N_1 \in \N: d_2 \left({b_n, b_m}\right) < \epsilon$ whenever $n, m \ge N_1$ and $b_n, b_m \in A_2$. We have that $M_1$ is isometric to $M_2$. Isometry is Equivalence Relation and so in particular symmetric. Hence $M_2$ is isometric to $M_1$, via $\phi^{-1}$. Thus: : $d_1 \left({\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b_m}\right)}\right) = d_2 \left({b_n, b_m}\right) < \epsilon$ whenever $n, m \ge N_1$ and $\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b_m}\right) \in A$. So $\left\langle{\phi^{-1} \left({b_n}\right)}\right\rangle$ is Cauchy in $A_1$. Since $A_1$ is complete, $\left\langle{\phi^{-1} \left({b_n}\right)}\right\rangle$ converges in $A_1$ to, say, $a$. By definition of isometry, $\phi^{-1}$ is a bijection, and in particular surjective Thus there exists some $b \in A_2$ such that $\phi^{-1} \left({b}\right) = a$. Since $\left\langle{\phi^{-1} \left({b_n}\right)}\right\rangle$ converges to $\phi^{-1} \left({b}\right)$, there exists some $N_2 \in \N$ such that: :$d_1 \left({\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b}\right)}\right) < \epsilon$ whenever $n \ge N_2$ and $\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b}\right) \in A_1$. Since $M_2$ is  isometric to $M_1$, we have: :$d_1 \left({\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b}\right)}\right) = d_2 \left({b_n, b}\right)$ and so: :$d_2 \left({b_n, b}\right) < \epsilon$ whenever $n \ge N_2$ and $b_n, b \in A_2$. Thus $\left\langle{b_n}\right\rangle$ converges in $A_2$. Since $\left\langle{b_n}\right\rangle$ was an arbitrary Cauchy sequence, we have that $M_2$ is complete, as required. {{qed}}
Let $\UU$ be a quasiuniformity on $S$ which has a symmetric filter basis $\BB$. From the definition of filter basis, all the elements of $\UU$ can be formed from intersections of elements of $\BB$. But from Intersection of Symmetric Relations is Symmetric, it follows that all elements of $\UU$ are symmetric. Now suppose $\UU$ is a uniformity. If $\BB$ is a filter basis of $\UU$ then all the elements of $\BB$ are also elements of $\UU$. Hence $\BB$ is a symmetric filter basis of $\UU$. {{qed}}
Let $T = \struct {S, \tau}$ be a topological space. Let $\BB$ be defined as: :$\BB := \set {u_G: u_G = \paren {G \times G} \cup \paren {\paren {S \setminus G} \times G}, G \in \tau}$ Then $\BB$ is a filter sub-basis for a quasiuniformity on $S$ such that $\struct {\struct {S, \UU}, \tau}$ is a quasiuniform space. {{finish|Prove that above assertion.}}
We check whether the Uniformity Axioms are satisfied. === $\text U 1$ === Let $x \in A$. Then $\tuple {x, x} \in \Delta_A$. Let $u \in \UU$. From definition: :$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$ By Metric Space Axiom M1: :$\map d {x, x} = 0 < \epsilon$ Hence we have $\tuple {x, x} \in u_\epsilon$. By definition of subset, $\Delta_A \subseteq u_\epsilon \subseteq u$. {{qed|lemma}} === $\text U 2$ === Let $u, v \in \UU$. From definition: :$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$ :$\exists \delta \in \R_{>0}: u_\delta \subseteq v$ {{WLOG}} assume that $\delta \ge \epsilon$. Let $\tuple {x, y} \in u_\epsilon$. Then $\map d {x, y} < \epsilon \le \delta$. Thus $\tuple {x, y} \in u_\delta$. By definition of subset, $u_\epsilon \subseteq u_\delta$. Hence: {{begin-eqn}} {{eqn | l = u_\epsilon       | r = u_\epsilon \cap u_\delta       | c = Intersection with Subset is Subset }} {{eqn | o = \subseteq       | r = u \cap v       | c = Set Intersection Preserves Subsets }} {{eqn | ll = \leadsto       | l = u \cap v       | o = \in       | r = \UU }} {{end-eqn}} {{qed|lemma}} === $\text U 3$ === Let $v \in A \times A$. Suppose $u \in \UU$ and $u \subseteq v$. From definition: :$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$ Thus $u_\epsilon \subseteq v$. Therefore $v \in \UU$ as well. {{qed|lemma}} === $\text U 4$ === Let $u \in \UU$. From definition: :$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$ Let $v = u_{\epsilon / 2}$. Since $u_{\epsilon / 2} \subseteq v$ and $\dfrac \epsilon 2 > 0$, $v \in \UU$. From definition: {{begin-eqn}} {{eqn | l = v \circ v       | r = \set {\tuple {x, z}: \exists y \in A: \tuple {x, y} \in v, \tuple {y, z} \in v} }} {{eqn | r = \set {\tuple {x, z}: \exists y \in A: \map d {x, y} < \frac \epsilon 2, \map d {y, z} < \frac \epsilon 2} }} {{end-eqn}} Let $\tuple {x, z} \in v \circ v$. Then: :$\exists y \in A: \map d {x, y} < \frac \epsilon 2, \map d {y, z} < \frac \epsilon 2$ By Metric Space Axiom M2: {{begin-eqn}} {{eqn | l = \map d {x, z}       | o = \le       | r = \set {\tuple {x, z}: \exists y \in A: \tuple {x, y} \in v, \tuple {y, z} \in v} }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2 }} {{eqn | r = \epsilon }} {{end-eqn}} Hence $\tuple {x, z} \in u_\epsilon \subseteq u$. By definition of subset, $v \circ v \subseteq u$. {{qed|lemma}} === $\text U 5$ === Let $u \in \UU$. From definition: :$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$ Suppose $\tuple {b, a} \in u_\epsilon$. Then $\map d {b, a} < \epsilon$. By Metric Space Axiom M3: :$\map d {a, b} = \map d {b, a} < \epsilon$ Hence $\tuple {a, b} \in u_\epsilon \subseteq u$. Consider the set $u^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in u}$. From the above, $\tuple {b, a} \in u^{-1}$. By definition of subset, $u_\epsilon \subseteq u^{-1}$. Hence $u^{-1} \in \UU$. {{qed|lemma}} {{finish|It remains to show that the topology induced by the uniformity $\UU$ is the same as the one induced by the pseudometric $d$}}
From definition it is clear that a metric space is a pseudometric space. The result then follows from Pseudometric Space generates Uniformity. {{qed}}
Define $h: \C \to \C$ by $\map h z = z + c$. Then: {{begin-eqn}} {{eqn | l = \map h {\map f z}       | r = \map h {z \times c}       | c =  }} {{eqn | r = z \times c + c       | c =  }} {{eqn | r = \paren {z + 1} \times c       | c =  }} {{eqn | r = \map f {z + 1}       | c =  }} {{end-eqn}} Thus $\map f z = z \times c$ is a superfunction and $\map h z = z + c$ is the corresponding transfer function. {{qed}} Category:Superfunctions cab3ee5qz40q2bqeshg3oe1373qvkwn
Let $T = \struct {S, \tau}$ be the discrete space on $S$. Then by definition $\tau = \powerset S$, that is, is the power set of $S$. We confirm the criteria for $T$ to be a topology: :$(1): \quad$ By definition of power set, $\O \in \powerset S$ and $S \in \powerset S$. :$(2): \quad$ From Power Set with Union is Monoid, $\powerset S$ is closed under set union. :$(3): \quad$ From Power Set with Intersection is Monoid, $\powerset S$ is closed under set intersection. {{qed}}
Let $\phi$ be any topology on $S$. Let $U \in \phi$. Then, by the definition of topology, $U \subseteq S$. Then, by the definition of discrete topological space, $U \in \tau$. Hence by definition of subset, $\phi \subseteq \tau$. Hence by definition of finer topology, $\tau$ is finer than $\phi$. {{qed}}
Let $U \subseteq S$. By definition of discrete topological space, $U \in \tau$. By definition of closed set, $\relcomp S U$ is closed in $T$, where $\relcomp S U$ is the relative complement of $U$ in $S$.  But from Set Difference is Subset: :$\relcomp S U = S \setminus U \subseteq S$ and so: :$\relcomp S U \in \tau$ That is, $\relcomp S U$ is both closed and open in $T$.  Then by Relative Complement of Relative Complement: :$\relcomp S {\relcomp S U} = U$ which is seen to be both closed and open in $T$. {{qed}}
=== Necessary Condition === Let $T = \struct {S, \tau}$ be the discrete space on $S$. Then by definition $\tau = \powerset S$, that is, $\tau$ is the power set of $S$. Let $x \in S$. Then from Set in Discrete Topology is Clopen it follows that $\set x$ is open in $T$. Thus by definition $x \in S$ is an isolated point of $T$. {{qed|lemma}} === Sufficient Condition === Let $T$ be such that $x \in S$ is an isolated point of $T$ for all $x \in T$. {{AimForCont}} $T = \struct {S, \tau}$ is not the discrete space on $S$. Then from Basis for Discrete Topology it follows that: :$\BB := \set {\set x: x \in S}$ is not a basis for $T$. Thus: :$\exists x \in S: \set x \notin \tau$ Let $U \in \tau$ such that $x \in U$. Then as $U \ne \set x$ it follows that $\exists y \in U: y \ne x$. That is, there are no open sets of $T$ which contain only $x$. So $x$ is not an isolated point. It follows by Proof by Contradiction that $T$ is the discrete space on $S$. {{qed|lemma}} Hence the result. {{qed}}
Let $U \subseteq S$. From Set in Discrete Topology is Clopen it follows that $U$ is open in $T$. Let $x \in U$. Then $U$ is an open neighborhood of $x$ containing (trivially) an element of $U$, that is, $x$. So, by definition, $x$ is an adherent point of $U$. Now suppose $x \notin U$. Then $\set x$ is an open neighborhood of $x$ such that $\set x \cap U = \O$. So $x$ can not be an adherent point of $U$. {{qed}}
Let $A \subseteq S$. Then from Set in Discrete Topology is Clopen it follows that $A$ is both open and closed in $T$. From Closed Set Equals its Closure we have that $A = A^-$. From Set Interior is Largest Open Set, we have that $A^\circ = A$. {{qed}}
Let $A \subseteq S$. Then from Set in Discrete Topology is Clopen it follows that $A$ is both open and closed in $T$. The result follows from Set Clopen iff Boundary is Empty. {{qed}}
From the definition of continuous: :$U \in \tau_2 \implies \phi^{-1} \sqbrk U \in \tau_1$ But as $\phi^{-1} \sqbrk U \subseteq S_1$ it follows from the definition of discrete space that $\phi^{-1} \sqbrk U \in \tau_1$. {{qed}}
Define $h : \C \to \C$ by $h \left({ z }\right) = z \times c$.  Then {{begin-eqn}} {{eqn | l=h \left({ f \left({ z }\right) }\right)       | r=h \left({ c^z }\right)       | c= }} {{eqn | r=c^z \times c       | c= }} {{eqn | r=c^{z + 1}       | c= }} {{eqn | r=f \left({ z + 1 }\right)       | c= }} {{end-eqn}} Thus  $f \left({z}\right) = c^z$ is a superfunction and $h \left({z}\right) = z \times c$ is the corresponding transfer function. {{qed}} Category:Superfunctions 2z342fbx3ysenws7x3br3acdq21fska
Let $a \in A$. From Subset of Standard Discrete Metric Space is Open, a set $U \subseteq A$ is open in $M$. So, in particular, $\set a$ is open in $\struct {A, d}$. This holds for all $a \in A$. From Metric Induces Topology it follows that $\set a$ is an open set in $\struct {A, \tau_{A, d} }$. The result follows from Basis for Discrete Topology. {{qed}}
We have that a Standard Discrete Metric induces Discrete Topology. Then we use Metric Space fulfils all Separation Axioms. {{qed}}
By definition, a neighborhood $N_x$ of $x$ is any subset of $S$ containing an open set which itself contains $x$. That is: :$\exists U \in \tau: x \in U \subseteq N_x \subseteq S$ From Set in Discrete Topology is Clopen we have that $\set x$ is open set in $S$. So by Set is Subset of Itself, $\set x$ is a subset of $S$ containing an open set $\set x$ which itself contains $x$. Hence the result. {{qed}}
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$ of $T$. Then from Interior Equals Closure of Subset of Discrete Space we have that $\set x$ equals its closure in $T$. From Singleton Set in Discrete Space is Compact, we have that $\set x$ is compact in $T$. Hence the result by definition of strongly locally compact. {{qed}}
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$. From the definition of local basis, it is clear that $\set {\set x}$ is (trivially) a local basis at $x$. That is, that every open set of $S$ containing $x$ also contains at least one of the sets of $\set {\set x}$. Equally trivially, we have that $\set {\set x}$ is countable. Hence the result by definition of first-countable. {{qed}}