hcju/proofwikiqa huggingface dataset:
That there can be as many as $14$ will be demonstrated by example. === Complement === {{:Kuratowski's Closure-Complement Problem/Complement}} === Interior === {{:Kuratowski's Closure-Complement Problem/Interior}} === Closure === {{:Kuratowski's Closure-Complement Problem/Closure}} === Exterior === {{:Kuratowski's Closure-Complement Problem/Exterior}} === Closure of Complement === {{:Kuratowski's Closure-Complement Problem/Closure of Complement}} === Closure of Interior === {{:Kuratowski's Closure-Complement Problem/Closure of Interior}} === Interior of Closure === {{:Kuratowski's Closure-Complement Problem/Interior of Closure}} === Interior of Closure of Interior === {{:Kuratowski's Closure-Complement Problem/Interior of Closure of Interior}} === Interior of Complement of Interior === {{:Kuratowski's Closure-Complement Problem/Interior of Complement of Interior}} === Closure of Interior of Complement === {{:Kuratowski's Closure-Complement Problem/Closure of Interior of Complement}} === Closure of Interior of Closure === {{:Kuratowski's Closure-Complement Problem/Closure of Interior of Closure}} === Interior of Complement of Interior of Closure === {{:Kuratowski's Closure-Complement Problem/Interior of Complement of Interior of Closure}} === Complement of Interior of Closure of Interior === {{:Kuratowski's Closure-Complement Problem/Complement of Interior of Closure of Interior}} === Proof of Maximum === It remains to be shown that there can be no more than $14$. {{:Kuratowski's Closure-Complement Problem/Proof of Maximum}}
Consider an arbitrary subset $A$ of a topological space $T = \left({S, \tau}\right)$. To simplify the presentation: :let $a$ be used to denote the operation of taking the complement of $A$ relative to $S$: $a \left({A}\right) = S \setminus A$ :let $b$ be used to denote the operation of taking the closure of $A$ in $T$: $b \left({A}\right) = A^-$ :let $I$ be used to denote the identity operation on $A$, that is: $I \left({A}\right) = A$. :let the parentheses and the reference to $A$ be removed, so as to present, for example: ::$a \left({b \left({a \left({A}\right)}\right)}\right)$ :as: ::$a b a$ From Relative Complement of Relative Complement: :$a \left({a \left({A}\right)}\right) = A$ or, using the compact notation defined above: :$(1): \quad a a = I$ and from Closure of Topological Closure equals Closure: :$b \left({b \left({A}\right)}\right) = b \left({A}\right) = A^-$ or, using the compact notation defined above: :$(2): \quad b b = b$ Let $s$ be a finite sequence of elements of $\left\{{a, b}\right\}$. By successive applications of $(1)$ and $(2)$, it is possible to eliminate all multiple consecutive instances of $a$ and $b$ in $s$, and so reduce $s$ to one of the following forms: :$\text{a)}: \quad a b a b \ldots a$ :$\text{b)}: \quad b a b a \ldots a$ :$\text{c)}: \quad a b a b \ldots b$ :$\text{d)}: \quad b a b a \ldots b$ From Closure of Complement of Closure is Regular Closed: :$b a b$ is regular closed. By Interior equals Complement of Closure of Complement, the interior of $A$ is: :$a b a$ Recall the definition of regular closed: :a set $A$ is regular closed {{iff}} it equals the closure of its interior. And so as $b a b$ is regular closed: :$b a b = b a b a \left({b a b}\right)$ So, adding an extra $b$ to either of $a b a b a b a$ or $b a b a b a$ will generate a string containing $b a b a b a b$ which can be reduced immediately to $b a b$. It follows that the possible different subsets of $S$ that can be obtained from $A$ by applying $a$ and $b$ can be generated by none other than: :$I$ :$a$ :$a b$ :$a b a$ :$a b a b$ :$a b a b a$ :$a b a b a b$ :$a b a b a b a$ :$b$ :$b a$ :$b a b$ :$b a b a$ :$b a b a b$ :$b a b a b a$ ... a total of $14$. Hence the result. {{qed}}
For ease of analysis, let: :$A_1 := \left({0 \,.\,.\, 1}\right)$ :$A_2 := \left({1 \,.\,.\, 2}\right)$ :$A_3 := \left\{ {3} \right\}$ :$A_4 := \left({\Q \cap \left({4 \,.\,.\, 5}\right)}\right)$ Thus: :$\displaystyle A = \bigcup_{i \mathop = 1}^4 A_i$ By De Morgan's Laws: :$\displaystyle A' := \R \setminus A = \bigcap_{i \mathop = 1}^4 \left({\R \setminus A_i}\right)$ {{begin-eqn}} {{eqn | l = \R \setminus A_1       | r = \left({\gets \,.\,.\, 0}\right] \cup \left[{1 \,.\,.\, \to}\right)       | c =  }} {{eqn | l = \R \setminus A_2       | r = \left({\gets \,.\,.\, 1}\right] \cup \left[{2 \,.\,.\, \to}\right)       | c =  }} {{eqn | l = \R \setminus A_3       | r = \left({\gets \,.\,.\, 3}\right) \cup \left({3 \,.\,.\, \to}\right)       | c =  }} {{eqn | l = \R \setminus A_4       | r = \left({\gets \,.\,.\, 4}\right] \cup \R \setminus \Q \cup \left[{5 \,.\,.\, \to}\right)       | c = De Morgan's Laws }} {{end-eqn}} from which the result follows by inspection. {{qed}} Category:Kuratowski's Closure-Complement Problem d9oxzyl9w658n6mgvygx6400yaq9p0r
From Interior equals Complement of Closure of Complement: :$A^\circ = A^{\prime \, - \, \prime}$ From Kuratowski's Closure-Complement Problem: Closure of Complement: {{begin-eqn}} {{eqn | l = A^{\prime \, -}       | r = \left({\gets \,.\,.\, 0}\right]       | c = Unbounded Closed Real Interval }} {{eqn | o =       | ro= \cup       | r = \left\{ {1} \right\}       | c = Singleton }} {{eqn | o =       | ro= \cup       | r = \left[{2 \,.\,.\, \to}\right)       | c = Unbounded Closed Real Interval }} {{end-eqn}} It can be determined by inspection that: :$A^{\prime \, - \, \prime} = \left({0 \,.\,.\, 1}\right) \cup \left({1 \,.\,.\, 2}\right)$ Hence the result. {{qed}} Category:Kuratowski's Closure-Complement Problem Category:Set Interiors aohordjb80shikn1mcq1ohtd7rzl2q4
From Closure of Union of Adjacent Open Intervals: :$\paren {\openint 0 1 \cup \openint 1 2}^- = \closedint 0 2$ From Real Number is Closed in Real Number Line: :$\set 3$ is closed in $\R$ From Set is Closed iff Equals Topological Closure: :$\set 3^- = \set 3$ From Closure of Rational Interval is Closed Real Interval: :$\paren {\Q \cap \openint 4 5 }^- = \closedint 4 5$ The result follows from Closure of Finite Union equals Union of Closures. {{qed}} Category:Kuratowski's Closure-Complement Problem Category:Set Closures l9svs3wen8l6wzv0kj3p5en5vi0iw2t
By definition, the exterior of $A$ in $\R$ can be defined either as: :the complement of the closure of $A$ in $\R$: $A^{- \, \prime}$ or as: :the interior of the complement of $A$ in $\R$: $A^{\prime \, \circ}$ From Kuratowski's Closure-Complement Problem: Closure: {{begin-eqn}} {{eqn | l = A^-       | r = \closedint 0 2       | c = {{Defof|Closed Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \set 3       | c = {{Defof|Singleton}} }} {{eqn | o =       | ro= \cup       | r = \closedint 4 5       | c = {{Defof|Closed Real Interval}} }} {{end-eqn}} It follows by inspection that: {{begin-eqn}} {{eqn | l = A^e = A^{- \, \prime}       | r = \openint \gets 0       | c = {{Defof|Unbounded Open Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \openint 2 3 \cup \openint 3 4       | c = {{Defof|Union of Adjacent Open Intervals}} }} {{eqn | o =       | ro= \cup       | r = \openint 5 \to       | c = {{Defof|Unbounded Open Real Interval}} }} {{end-eqn}} {{qed}} Category:Kuratowski's Closure-Complement Problem Category:Set Exteriors 6l2oj9k014ngv3qrsqeg832zewp7oal
From Kuratowski's Closure-Complement Problem: Complement: {{begin-eqn}} {{eqn | l = A'       | r = \hointl \gets 0       | c = {{Defof|Unbounded Closed Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \set 1       | c = {{Defof|Singleton}} }} {{eqn | o =       | ro= \cup       | r = \hointr 2 3       | c = {{Defof|Half-Open Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \hointl 3 4       | c = adjacent to Half-Open Real Interval }} {{eqn | o =       | ro= \cup       | r = \paren {\R \setminus \Q \cap \closedint 4 5}       | c = Irrational Numbers from $4$ to $5$ }} {{eqn | o =       | ro= \cup       | r = \hointr 5 \to       | c = {{Defof|Unbounded Closed Real Interval}} }} {{end-eqn}} From Real Number is Closed in Real Number Line: :$\set 3$ is closed in $\R$ and: :$\hointl \gets 0$ is closed in $\R$ and: :$\hointr 5 \to$ is closed in $\R$ Then from Set is Closed iff Equals Topological Closure: :$\set 3^- = \set 3$ :$\hointl \gets 0^- = \hointl \gets 0$ :$\hointr 5 \to^- = \hointr 5 \to$ From Closure of Half-Open Real Interval is Closed Real Interval: :$\hointr 2 3 = \closedint 2 3$ and: :$\hointl 3 4 = \closedint 3 4$ From Closure of Irrational Interval is Closed Real Interval: :$\paren {\R \setminus \Q \cap \closedint 4 5}^- = \closedint 4 5$ From Closure of Finite Union equals Union of Closures: {{begin-eqn}} {{eqn | l = A^{\prime \, -}       | r = \hointl \gets 0       | c = {{Defof|Unbounded Closed Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \set 1       | c = {{Defof|Singleton}} }} {{eqn | o =       | ro= \cup       | r = \closedint 2 3       | c = {{Defof|Closed Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \closedint 3 4       | c = {{Defof|Closed Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \closedint 4 5       | c = {{Defof|Closed Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \hointr 5 \to       | c = {{Defof|Unbounded Closed Real Interval}} }} {{end-eqn}} The result follows. {{qed}} Category:Kuratowski's Closure-Complement Problem 5j3uqggmkqu6tdln6ihka6gej6chaz3
From Kuratowski's Closure-Complement Problem: Interior: :$A^\circ = \openint 0 1 \cup \openint 1 2$ From Closure of Union of Adjacent Open Intervals: :$A^{\circ \, -} = \closedint 0 2$ {{qed}} Category:Kuratowski's Closure-Complement Problem Category:Set Interiors Category:Set Closures 1tlrjvqjiw9pxy8edfnepwds3ks1j02
From Kuratowski's Closure-Complement Problem: Closure: {{begin-eqn}} {{eqn | l = A^-       | r = \closedint 0 2       | c = {{Defof|Closed Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \set 3       | c = {{Defof|Singleton}} }} {{eqn | o =       | ro= \cup       | r = \closedint 4 5       | c = {{Defof|Closed Real Interval}} }} {{end-eqn}} From Interior of Closed Real Interval is Open Real Interval: :$\closedint 0 2^\circ = \openint 0 2$ and: :$\closedint 4 5^\circ = \openint 4 5$ From Interior of Singleton in Real Number Line is Empty: :$\set 3^\circ = \O$ {{Finish|Get past this: Interior of Union is not necessarily Union of Interiors}} {{qed}} Category:Kuratowski's Closure-Complement Problem Category:Set Interiors Category:Set Closures ewc4ug707iz5wmjr2f1zooy69ikydy6
From Kuratowski's Closure-Complement Problem: Closure of Interior: {{begin-eqn}} {{eqn | l = A^{\circ \, -}       | r = \left[{0 \,.\,.\, 2}\right]       | c = {{Defof|Closed Real Interval}} }} {{end-eqn}} From Interior of Closed Real Interval is Open Real Interval: :$\left[{0 \,.\,.\, 2}\right]^\circ = \left({0 \,.\,.\, 2}\right)$ Hence the result. {{qed}} Category:Kuratowski's Closure-Complement Problem Category:Set Interiors Category:Set Closures 46s5m5z3ka0ul2mo5rjbcwvhbilhqig
From Complement of Interior equals Closure of Complement: :$A^{\circ \, \prime} = A^{\prime \, -}$ From Kuratowski's Closure-Complement Problem: Closure of Complement: {{begin-eqn}} {{eqn | l = A^{\prime \, -}       | r = \hointl \gets 0       | c = {{Defof|Unbounded Closed Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \set 1       | c = {{Defof|Singleton}} }} {{eqn | o =       | ro= \cup       | r = \hointr 2 \to       | c = {{Defof|Unbounded Closed Real Interval}} }} {{end-eqn}} From Interior of Closed Real Interval is Open Real Interval: :$\hointl \gets 0^\circ = \openint \gets 0$ and: :$\hointr 2 \to^\circ = \openint 2 \to$ From Interior of Singleton in Real Number Line is Empty: :$\set 1^\circ = \O$ {{Finish|Get past this: Interior of Union is not necessarily Union of Interiors}} {{qed}} Category:Kuratowski's Closure-Complement Problem Category:Set Interiors Category:Set Complement c2o9wneuuzcfipk8cgw6lb7l4ry38bx
From Kuratowski's Closure-Complement Problem: Exterior: {{begin-eqn}} {{eqn | l = A^e       | r = \openint \gets 0       | c = {{Defof|Unbounded Open Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \openint 2 3 \cup \openint 3 4       | c = {{Defof|Union of Adjacent Open Intervals}} }} {{eqn | o =       | ro= \cup       | r = \openint 5 \to       | c = {{Defof|Unbounded Open Real Interval}} }} {{end-eqn}} From Closure of Open Real Interval is Closed Real Interval: :$\openint \gets 0^- = \hointl \gets 0$ and: :$\openint 5 \to^- = \hointr 5 \to$ From Closure of Union of Adjacent Open Intervals: :$\paren {\openint 2 3 \cup \openint 3 4}^- = \closedint 2 4$ The result follows from Closure of Finite Union equals Union of Closures. {{qed}} Category:Kuratowski's Closure-Complement Problem Category:Set Interiors Category:Set Closures Category:Set Complement 1eih3f697wm4pv5pbxpfiymynq5qazp
From Kuratowski's Closure-Complement Problem: Interior of Closure: {{begin-eqn}} {{eqn | l = A^{- \, \circ}       | r = \left({0 \,.\,.\, 2}\right)       | c = {{Defof|Open Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \left({4 \,.\,.\, 5}\right)       | c = {{Defof|Open Real Interval}} }} {{end-eqn}} From Closure of Open Real Interval is Closed Real Interval: :$\left({0 \,.\,.\, 2}\right)^- = \left[{0 \,.\,.\, 2}\right]$ and: :$\left({4 \,.\,.\, 5}\right)^- = \left[{4 \,.\,.\, 5}\right]$ The result follows from Closure of Finite Union equals Union of Closures. {{qed}} Category:Kuratowski's Closure-Complement Problem Category:Set Interiors Category:Set Closures 8m8djt5kb04phfl9rc9qu0tpwfb0ehx
From Kuratowski's Closure-Complement Problem: Interior of Closure: {{begin-eqn}} {{eqn | l = A^{- \, \circ}       | r = \left({0 \,.\,.\, 2}\right)       | c = {{Defof|Open Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \left({4 \,.\,.\, 5}\right)       | c = {{Defof|Open Real Interval}} }} {{end-eqn}} By inspection: {{begin-eqn}} {{eqn | l = A^{- \, \circ \, '}       | r = \left({\gets \,.\,.\, 0}\right]       | c = {{Defof|Unbounded Closed Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \left[{2 \,.\,.\, 4}\right]       | c = {{Defof|Closed Real Interval}} }} {{eqn | o =       | ro= \cup       | r = \left[{5 \,.\,.\, \to}\right)       | c = {{Defof|Unbounded Closed Real Interval}} }} {{end-eqn}} From Interior of Closed Real Interval is Open Real Interval: :$\left({\gets \,.\,.\, 0}\right]^\circ = \left({\gets \,.\,.\, 0}\right)$ and: :$\left[{2 \,.\,.\, 4}\right]^\circ = \left({2 \,.\,.\, 4}\right)$ and: :$\left[{5 \,.\,.\, \to}\right)^\circ = \left({5 \,.\,.\, \to}\right)$ {{Finish|Get past this: Interior of Union is not necessarily Union of Interiors}} {{qed}} Category:Kuratowski's Closure-Complement Problem r98z9ozwhyltndvcnakqeb8bxhaa8d6
From Kuratowski's Closure-Complement Problem: Interior of Closure of Interior: {{begin-eqn}} {{eqn | l = A^{\circ \, - \, \circ}       | r = \left({0 \,.\,.\, 2}\right)       | c = {{Defof|Open Real Interval}} }} {{end-eqn}} The result follows by inspection. {{qed}} Category:Kuratowski's Closure-Complement Problem 4paw14owtms8gr8v0mcwb31g4ih419q
From Irrationals are Everywhere Dense in Topological Space of Reals, $\R \setminus \Q$ is everywhere dense in $\R$. It follows by definition of everywhere dense that $\paren {\R \setminus \Q}^- = \R$. {{qed}} Category:Irrational Number Space Category:Real Number Line with Euclidean Topology Category:Set Closures 43wj1draqpd0swslfbj7dprl6gi9vm8
{{begin-eqn}} {{eqn | l = A^-       | r = \left({\left({a \,.\,.\, b}\right) \cup \left({b \,.\,.\, c}\right)}\right)^-       | c =  }} {{eqn | r = \left({a \,.\,.\, b}\right)^- \cup \left({b \,.\,.\, c}\right)^-       | c = Closure of Finite Union equals Union of Closures }} {{eqn | r = \left[{a \,.\,.\, b}\right] \cup \left[{b \,.\,.\, c}\right]       | c = Closure of Open Ball in Metric Space }} {{eqn | r = \left[{a \,.\,.\, c}\right]       | c = {{Defof|Closed Real Interval}} }} {{end-eqn}} {{qed}} Category:Set Closures Category:Union of Adjacent Open Intervals p1ts9xym8w2qx2gim0b0f2bqkl6dw7t
{{begin-eqn}} {{eqn | l = H^{\circ \, \prime}       | r = H^{\prime \, -}       | c = Complement of Interior equals Closure of Complement }} {{eqn | ll= \leadsto       | l = \paren {H^{\circ \, \prime} }^\prime       | r = \paren {H^{\prime \, -} }^\prime       | c =  }} {{eqn | ll= \leadsto       | l = H^{\circ \, \prime \, \prime}       | r = H^{\prime \, - \, \prime}       | c = Composition of Mappings is Associative }} {{eqn | ll= \leadsto       | l = H^\circ       | r = H^{\prime \, - \, \prime}       | c = Relative Complement of Relative Complement }} {{end-eqn}} {{qed}} Category:Set Closures Category:Set Interiors 289qxln45jqaqop7asiju08vtc3a97r
For an arbitrary $H \subseteq \R$, let $H^-$ denote the closure of $H$ in $\R$. By the definition of closure: :$H^-$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $H$ as a subset. Let $A := \openint a b$ be the open interval between $a$ and $b$. By definition of $H_1$ and $H_2$: :$A \subseteq H_1$ and: :$A \subseteq H_2$ From Closure of Open Real Interval is Closed Real Interval: :$\openint a b^- = \closedint a b$ Thus as $\closedint a b$ is the closure of $\openint a b$, it follows that: :$\closedint a b$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $\openint a b$. But $\closedint a b$ also contains as subsets both $H_1$ and $H_2$. Thus we have: :$A \subseteq H_1 \subseteq \closedint a b$ and: :$A \subseteq H_2 \subseteq \closedint a b$ Thus it follows that $\closedint a b$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $H_1$ and $H_2$. Hence the result. {{qed}} Category:Real Intervals Category:Set Closures e2fcgq4qmbpjesvpx91oj0d7xf7w5jz
Let $\Bbb I$ be an open real interval. From Closure of Real Interval is Closed Real Interval: :$\Bbb I^- = \closedint a b$ From Closure of Irrational Numbers is Real Numbers: :$\paren {\R \setminus \Q}^- = \R$ From Closure of Intersection is Subset of Intersection of Closures: :$\paren{\Bbb I \cap \paren{\R \setminus \Q}}^- \subseteq \Bbb I^- \cap \paren {\R \setminus \Q}^-$ From Intersection with Subset is Subset: :$\Bbb I^- \cap \paren {\R \setminus \Q}^- = \closedint a b$ and so: :$\paren {\Bbb I \cap \paren {\R \setminus \Q} }^- \subseteq \closedint a b$ From Irrationals are Everywhere Dense in Topological Space of Reals: :$\closedint a b \subseteq \paren {\Bbb I \cap \paren {\R \setminus \Q} }^-$ and the result follows. {{qed}} Category:Real Intervals Category:Irrational Number Space Category:Set Closures 2sgds1491k5cm9vkp7ei81psyz1rljc
Let $\Bbb I$ be an open real interval. From Closure of Real Interval is Closed Real Interval: :$\Bbb I^- = \closedint a b$ From Closure of Rational Numbers is Real Numbers: :$\Q^- = \R$ From Closure of Intersection is Subset of Intersection of Closures: :$\paren{\Bbb I \cap \Q}^- \subseteq \Bbb I^- \cap \Q^-$ From Intersection with Subset is Subset: :$\Bbb I^- \cap \Q^- = \closedint a b$ and so: :$\paren {\Bbb I \cap \Q}^- \subseteq \closedint a b$ From Rationals are Everywhere Dense in Topological Space of Reals: :$\closedint a b \subseteq \paren {\Bbb I \cap \Q}^-$ and the result follows. {{qed}} Category:Real Intervals Category:Irrational Number Space m64xrdzlhmu195l3hgeur2klv6xxdnb
{{begin-eqn}} {{eqn | l = \set a^\circ       | r = \closedint a a^\circ       | c = {{Defof|Closed Real Interval}} }} {{eqn | r = \openint a a       | c = Interior of Closed Real Interval is Open Real Interval }} {{eqn | r = \set {x \in \R: a < x < a}       | c = {{Defof|Open Real Interval}} }} {{eqn | r = \O       | c = {{Defof|Empty Real Interval}} }} {{end-eqn}} {{qed}} Category:Set Interiors Category:Singletons Category:Real Number Line with Euclidean Topology 9eq8s0dt9sqtlvbodatp8avtdc6tknh
Let $\tau_d$ denote the metric topology for $d$. Let $\epsilon \in \R_{>0}$ such that $\epsilon < 1$. Let $a \in \Z_{>0}$. Recall the definition of the open $\epsilon$-ball of $a$ in $\struct {\Z_{>0}, d}$: :$\map {B_\epsilon} a := \set {x \in A: \map d {x, a} < \epsilon}$ But we have: :$\forall x \in \Z_{>0}, x \ne a: \map d {x, a} \ge 1$ and so: :$\forall x \in \Z_{>0}, x \ne a: x \notin \map {B_\epsilon} a$ It follows that: :$\map {B_\epsilon} a := \set a$ Thus by definition of $\tau_d$: :$\forall a \in \Z_{>0}: \set a \in \tau_d$ It follows from Basis for Discrete Topology that $\struct {\Z_{>0}, \tau_d}$ is a discrete topological space. {{qed}}
Let $\tau_\delta$ denote the metric topology for $\delta$. In Scaled Euclidean Metric is Metric it is demonstrated that $\delta$ is indeed a metric on $\Z_{>0}$. Let $a \in \Z_{>0}$. Recall the definition of the open $\epsilon$-ball of $a$ in $\struct {\Z_{>0}, \delta}$: :$\map {B_\epsilon} a := \set {x \in A: \map \delta {x, a} < \epsilon}$ Let $x \in \R_{>0}$. Let $\epsilon \in \R_{>0}$ such that $\epsilon < \dfrac 1 {a \paren {a + 1} }$. But we have: {{begin-eqn}} {{eqn | ll= \forall x \in \Z_{>0}, x \ne a:       | l = \map \delta {x, a}       | r = \frac {\size {x - a} } {x a}       | c =  }} {{eqn | r = \size {\frac 1 x - \frac 1 a}       | c =  }} {{eqn | o = \ge       | r = \size {\frac 1 {a + 1} - \frac 1 a}       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} and so: :$\forall x \in \Z_{>0}, x \ne a: x \notin \map {B_\epsilon} a$ It follows that: :$\map {B_\epsilon} a := \set a$ Thus by definition of $\tau_d$: :$\forall a \in \Z_{>0}: \set a \in \tau_\delta$ It follows from Basis for Discrete Topology that $\struct {\Z_{>0}, \tau_\delta}$ is a discrete topological space. {{qed}}
=== Proof of $M1$ === {{begin-eqn}} {{eqn | l = \delta \left({x, x}\right)       | r = \dfrac {\left\lvert{x - x}\right\rvert} {x^2}       | c = Definition of $\delta$ }} {{eqn | r = 0       | c = as $\left\lvert{x - x}\right\rvert = 0$ }} {{end-eqn}} So axiom $M1$ holds for $\delta$. {{qed|lemma}} === Proof of $M2$ === {{begin-eqn}} {{eqn | l = \delta \left({x, y}\right) + \delta \left({y, z}\right)       | r = \frac {\left\lvert{x - y}\right\rvert} {x y} + \dfrac {\left\lvert{y - z}\right\rvert} {y z}       | c = Definition of $\delta$ }} {{eqn | r = \frac {z \left\lvert{x - y}\right\rvert + x \left\lvert{y - z}\right\rvert} {x y z}       | c = Sum of Quotients of Real Numbers }} {{eqn | r = \frac {\left\lvert{x z - y z}\right\rvert + \left\lvert{x y - x z}\right\rvert} {x y z}       | c = Valid, as $x, z > 0$ }} {{eqn | o = \ge       | r = \frac {\left\lvert{x z - y z + x y - x z}\right\rvert} {x y z}       | c = Triangle Inequality }} {{eqn | r = \frac {\left\lvert{x y - y z}\right\rvert} {x y z}       | c = simplifying }} {{eqn | r = \frac {\left\lvert{x - z}\right\rvert} {x z}       | c = simplifying further }} {{eqn | r = \delta \left({x, z}\right)       | c = Definition of $\delta$ }} {{end-eqn}} So axiom $M2$ holds for $\delta$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = \delta \left({x, y}\right)       | r = \frac {\left\lvert{x - y}\right\rvert} {x y}       | c = Definition of $\delta$ }} {{eqn | r = \frac {\left\lvert{y - x}\right\rvert} {y x}       | c = Definition of Absolute Value and Real Multiplication is Commutative }} {{eqn | r = \delta \left({y, x}\right)       | c = Definition of $\delta$ }} {{end-eqn}} So axiom $M3$ holds for $\delta$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \implies       | l = \left\lvert{x - y}\right\rvert       | o = >       | r = 0       | c = Definition of Absolute Value }} {{eqn | ll= \implies       | l = \frac {\left\lvert{x - y}\right\rvert} {x y}       | o = >       | r = 0       | c = as $x y > 0$ }} {{eqn | ll= \implies       | l = \delta \left({x, y}\right)       | o = >       | r = 0       | c = Definition of $\delta$ }} {{end-eqn}} So axiom $M4$ holds for $\delta$. {{qed}} Category:Scaled Euclidean Metric kpi92ur6wjx2s5ij2gy6b1hlycyrq3x
From Topology induced by Usual Metric on Positive Integers is Discrete: :$\struct {\Z_{>0}, \tau_d}$ is a discrete space. From Topology induced by Scaled Euclidean Metric on Positive Integers is Discrete: :$\struct {\Z_{>0}, \tau_\delta}$ is a discrete space. Let $I_{\Z_{>0} }$ be the identity mapping from $\Z_{>0}$ to itself. From Mapping from Discrete Space is Continuous: :$I_{\Z_{>0} }: \struct {\Z_{>0}, \tau_d} \to \struct {\Z_{>0}, \tau_\delta}$ is continuous and: :$I_{\Z_{>0} }: \struct {\Z_{>0}, \tau_\delta} \to \struct {\Z_{>0}, \tau_d}$ is continuous. Hence the result by definition of homeomorphic. {{qed}}