hcju/proofwikiqa huggingface dataset:
By Category of Subobjects is Category, we know $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ is a metacategory. By definition of preorder category, it suffices to show that if $f, g: m \to m'$ are morphisms with the same domain and codomain, then $f = g$. The situation is sketched by the following commutative diagram in $\mathbf C$: ::$\begin{xy}\xymatrix@+1em{  \operatorname{dom} m   \ar[r]<2pt>^*+{f}   \ar[r]<-2pt>_*+{g}   \ar[rd]_*+{m} &  \operatorname{dom} m'   \ar[d]^*+{m'} \\ &  C }\end{xy}$ Thus, we see that $m' \circ f = m = m' \circ g$. Now $m'$ is a subobject, and a fortiori a monomorphism. Hence $f = g$, and $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ is a preorder category. {{qed}}
=== Equality of Domains === {{begin-eqn}} {{eqn | l=\operatorname{Dom} \left({f \circ i_A}\right)       | r=\operatorname{Dom} \left({i_A}\right)       | c=Domain of Composite Relation }} {{eqn | r=A       | c=by definition of $i_A$ }} {{eqn | r=\operatorname{Dom} \left({f \restriction_A}\right)       | c=by definition of restriction of mapping }} {{end-eqn}} {{qed|lemma}} === Equality of Codomains === {{begin-eqn}} {{eqn | l=\operatorname{Cdm} \left({f \circ i_A}\right)       | r=\operatorname{Cdm} \left({f}\right)       | c=Codomain of Composite Relation }} {{eqn | r=T       | c=by definition of $f$ }} {{eqn | r=\operatorname{Cdm} \left({f \restriction_A}\right)       | c=by definition of restriction of mapping }} {{end-eqn}} {{qed|lemma}} === Equality of Graph === Let $x \in A$. {{begin-eqn}} {{eqn | l=\left({f \circ i_A}\right) \left({x}\right)       | r=f \left({i_A \left({x}\right)}\right)       | c=by definition of composition of mappings }} {{eqn | r=f \left({x}\right)       | c=by definition of $i_A$ }} {{eqn | r=f \restriction_A \left({x}\right)       | c=by definition of restriction of mapping }} {{end-eqn}} {{qed|lemma}} All three criteria are seen to be fulfilled. The result follows from Equality of Mappings. {{qed}}
By Inclusion Relation on Subobjects is Preordering, $\subseteq$ is a preordering. From Preorder Induces Equivalence, we see that $\sim$ is the equivalence induced by $\subseteq$. In particular, $\sim$ is an equivalence. {{qed}} Category:Subobjects n8ex14wenju6wig6hy69oflijmcpocp
For brevity, let us write $\operatorname{dom} m = B$ and $\operatorname{dom} m' = B'$. Since $m$ and $m'$ are equivalent subobjects of $C$, we have that: :$m \subseteq m'$ and $m' \subseteq m$ where $\subseteq$ denotes the inclusion relation on subobjects. That is, there exist morphisms $f: m \to m'$ and $f': m' \to m$ of $\mathbf{Sub}_{\mathbf C} \left({C}\right)$. Hence, we have morphisms: :$f \circ f': m' \to m'$ :$f' \circ f: m \to m$ By Category of Subobjects is Preorder Category, these are the unique such morphisms. It follows that both are identity morphisms: :$f \circ f' = \operatorname{id}_{m'}$ :$f' \circ f = \operatorname{id}_m$ That is, $f: m \to m'$ is an isomorphism. {{qed}}
Let us check the three conditions for an equivalence in turn. === Reflexivity === To show that $\sim$ is reflexive, note that for $f: m \to m'$: :$f \sim f$ holds iff $m \sim m$ and $m' \sim m'$. By Equivalence of Subobjects is Equivalence, these conditions are satisfied. {{qed|lemma}} === Symmetry === It is to be shown that for $f: m \to n$ and $f': m' \to n'$: :$f \sim f'$ implies $f' \sim f$ By definition of $\sim$, this is equivalent to: :$m \sim m' \land n \sim n'$ implies $m' \sim m \land n' \sim n$ Again by Equivalence of Subobjects is Equivalence, this is satisfied. {{qed|lemma}} === Transitivity === Given $f: m \to n$, $f': m' \to n'$ and $f'': m'' \to n''$, it is to be shown that: :$f \sim f' \land f' \sim f''$ implies $f \sim f''$ By definition of $\sim$, this is equivalent to: :$m \sim m' \land n \sim n' \land m' \sim m'' \land n' \sim n''$ implies $m \sim m'' \land n \sim n''$ which also follows from Equivalence of Subobjects is Equivalence. {{qed|lemma}} Thence $\sim$ is proven an equivalence. {{qed}} Category:Subobjects 9urack6agtmknhli7c7y523gsh5snk8
{{begin-eqn}} {{eqn | l = \size {x + y}^2       | r = \paren {x + y}^2       | c =  }} {{eqn | r = x^2 + 2 x y + y^2       | c =  }} {{eqn | r = \size x^2 + 2 x y + \size y^2       | c =  }} {{eqn | o = \le       | r = \size x^2 + 2 \size {x y} + \size y^2       | c = Negative of Absolute Value }} {{eqn | r = \size x^2 + 2 \size x \cdot \size y + \size y^2       | c = Absolute Value Function is Completely Multiplicative }} {{eqn | r = \paren {\size x + \size y}^2       | c =  }} {{end-eqn}} Then by Order is Preserved on Positive Reals by Squaring: :$\size {x + y} \le \size x + y$ {{qed}}
This can be seen to be a special case of Minkowski's Inequality, with $n = 1$. {{qed}}
We have that Real Numbers form Ordered Integral Domain. Therefore Sum of Absolute Values on Ordered Integral Domain applies directly. {{qed}}
Let $\mathbf x, \mathbf y \in \R^n$. We have: {{begin-eqn}} {{eqn | l = \norm {\mathbf x + \mathbf y}^2       | r = \paren {\mathbf x + \mathbf y} \cdot \paren {\mathbf x + \mathbf y}       | c = Dot Product of Vector with Itself }} {{eqn | r = \mathbf x \cdot \mathbf x + \mathbf x \cdot \mathbf y + \mathbf y \cdot \mathbf x  + \mathbf y \cdot \mathbf y       | c = Dot Product Distributes over Addition }} {{eqn | r = \mathbf x \cdot \mathbf x + 2 \paren {\mathbf x \cdot \mathbf y} + \mathbf y \cdot \mathbf y       | c = Dot Product Operator is Commutative }} {{eqn | r = \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2       | c = Dot Product of Vector with Itself }} {{end-eqn}} From the Cauchy-Bunyakovsky-Schwarz Inequality: {{begin-eqn}} {{eqn | l = \size {\mathbf x \cdot \mathbf y}       | o = \le        | r = \norm {\mathbf x} \norm {\mathbf y} }} {{eqn | ll= \leadsto       | l = \mathbf x \cdot \mathbf y       | o = \le       | r = \norm {\mathbf x} \norm {\mathbf y}       | c = Negative of Absolute Value }} {{eqn | l = \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2       | o = \le       | r = \norm {\mathbf x}^2 + 2 \paren {\norm {\mathbf x} \norm {\mathbf y} } + \norm {\mathbf y}^2 }} {{eqn | r = \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2 }} {{eqn | ll= \leadsto       | l = \norm {\mathbf x + \mathbf y}^2       | o = \le       | r = \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2 }} {{eqn | ll= \leadsto       | l = \norm {\mathbf x + \mathbf y}       | o = \le       | r = \norm {\mathbf x} + \norm {\mathbf y}       | c = taking the square root of both sides }} {{end-eqn}} {{qed}} To prove that the equality holds if the vectors are scalar multiples of each other, assume: :$\exists \lambda \in \R, \lambda \ge 0: \mathbf v = \lambda \mathbf w$ === Sufficient Condition === {{begin-eqn}} {{eqn | l = \norm {\mathbf v + \mathbf w}       | r = \norm {\lambda \mathbf w + \mathbf w} }} {{eqn | r = \norm {\paren {\lambda + 1} \mathbf w} }} {{eqn | r = \paren {\lambda + 1} \norm {\mathbf w} }} {{eqn | r = \lambda \norm {\mathbf w} + 1 \norm {\mathbf w} }} {{eqn | r = \norm {\lambda \mathbf w} + \norm {1 \mathbf w} }} {{eqn | r = \norm {\mathbf v} + \norm {\mathbf w} }} {{end-eqn}} {{qed|lemma}} === Necessary Condition === {{proof wanted}}
Let $X$ denote either $\R$ or $\C$ as appropriate. From Real Number Line is Metric Space and Complex Plane is Metric Space the distance function $d: X \times X \to \R$ can be defined as: :$\map d {x, y} = \size {x - y}$ From the Reverse Triangle Inequality as applied to metric spaces: :$(1): \quad \forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$ Let $z = 0$. Then $(1)$ translates to: :$\forall x, y, z \in X: \size {\size {x - 0} - \size {y - 0} } \le \size {x - y}$ Hence the result. {{qed}}
From the proof 2 of the corollary to this result, which is derived independently: :$\size {x - y} \ge \size x - \size y$ There are two cases: $(1): \quad \size x \ge \size y$ We have : :$\size {\size x - \size y} = \size x - \size y$ and the proof is finished. {{qed|lemma}} $(2): \quad \size y \ge \size x$ We have: :$\size {y - x} \ge \size y - \size x = \size {\size y - \size x}$ But: :$\size {y - x} = \size {x - y}$ and: :$\size {\size y - \size x} = \size {\size x - \size y}$ From this we have: :$-\size {\size x - \size y} \ge -\size {x - y}$ Since, by Negative of Absolute Value, we have that: :$\size x - \size y \ge -\size {\size x - \size y}$ it follows that: :$-\size {x - y} \le \size x - \size y \le \size {x - y}$ The result follows. {{qed}}
From the Reverse Triangle Inequality: :$\cmod {x - y} \ge \cmod {\cmod x - \cmod y}$ By the definition of both absolute value and complex modulus: :$\cmod {\cmod x - \cmod y} \ge 0$ As: :$\cmod x - \cmod y = \pm \cmod {\cmod x - \cmod y}$ it follows that: :$\cmod {\cmod x - \cmod y} \ge \cmod x - \cmod y$ Hence the result. {{qed}}
By the Triangle Inequality: :$\cmod {x + y} - \cmod y \le \cmod x$ Let $z = x + y$. Then $x = z - y$ and so: :$\cmod z - \cmod y \le \cmod {z - y}$ Renaming variables as appropriate gives: :$\cmod {x - y} \ge \cmod x - \cmod y$ {{qed}}
Let $z = x + i y$. By definition of complex modulus: :$\cmod z = \sqrt {x^2 + y^2}$ and so: :$\cmod z \ge 0$ Thus: :$-\cmod z\le 0$ Hence the result. {{qed}}
Follows directly from Quotient Rule for Real Sequences, setting :$\sequence {y_n} := \sequence {x_n}$ and: :$\sequence {x_n} := \tuple {1, 1, 1, \ldots}$ {{qed}}
We have that $\struct {\R, d}$ is a metric space, where $d$ is the Euclidean metric on $\R$. Thus the theorem follows immediately from Limit of Function by Convergent Sequences. {{qed}} {{ACC|Limit of Function by Convergent Sequences}}
{{AimForCont}} $A$ has no $\omega$-accumulation points in $X$. Then for any $x \in X$, there exists some open set $U_x$ such that $U_x$ only contains a finite number of points in $A$. The collection of all such open sets $\CC = \set {U_x: x \in X}$ is an open cover for $X$. Since $X$ is compact, $\CC$ has a finite subcover $\CC'$. $\CC'$ contains a finite number of sets in $\CC$, and each of those sets contain a finite number of points in $A$. Therefore $\bigcup \CC'$ contains a finite number of points in $A$. However by definition of a subcover, $A \subseteq X \subseteq \bigcup \CC'$. Thus $\bigcup \CC'$ contains all points in $A$, of which there are infinite. This is a contradiction. Therefore $A$ must have an $\omega$-accumulation point in $X$. {{qed}} Category:Compact Spaces Category:Omega-Accumulation Points kg2bxqe4nthappgc1mzu9tkh6hfzjkl
Let $\epsilon \in \R_{>0}$. Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball of $a$. We have that $a - \epsilon < a$ and so $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ does not lie entirely in $\hointr a b$. Thus $\hointr a b$ is not a neighborhood $a$. It follows that $\hointr a b$ is not an open set of $\R$. Mutatis mutandis, the argument also shows that $\hointl a b \subset \R$ is not an open set of $\R$. {{qed}}
By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set. That is, there are no points in $S$ which have an open ball some of whose elements are not in $S$. As there are no elements in $\varnothing$, the result follows vacuously. {{qed}}
By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set. Let $x \in A$. An open ball of $x$ in $M$ is by definition a subset of $A$. Hence the result. {{qed}}
From the definition of standard discrete metric: :$\forall x, y \in A: d \left({x, y}\right) = \begin{cases} 0 & : x = y \\ 1 & : x \ne y \end{cases}$ Let $\epsilon \in \R_{>0}$ be such that $0 < \epsilon \le 1$. Let $x \in S$. Let $B_\epsilon \left({x}\right)$ be the open $\epsilon$-ball of $x$. Then by definition of $\epsilon$ and $d$: :$B_\epsilon \left({x}\right) = \left\{{x}\right\}$ Thus: : $\forall x \in S: B_\epsilon \left({x}\right) \subseteq S$ Hence the result by definition of open set. {{qed}}
Consider the constant mapping $f_0: \R^2 \to \R$ defined as: :$\forall \left({x, y}\right) \in \R^2: f \left({x, y}\right) = 0$ Then by Constant Mapping is Continuous $f$ is a continuous mapping. But consider any open set $U \subseteq A_1$ of $M_1$ such that $U \ne \varnothing$. Then $f \left({U}\right) = \left\{{0}\right\} = \left[{0 \,.\,.\, 0}\right]$ which is a closed interval of $\R$. The result follows from Closed Real Interval is not Open Set. {{qed}}
Consider the open real interval $\openint {-\dfrac 1 n} {\dfrac 1 n} \subseteq \R$. From Open Real Interval is Open Set, $\openint {-\dfrac 1 n} {\dfrac 1 n}$  is open in $\R$ for all $n \in \N$. But: :$\displaystyle \bigcap_{n \mathop \in \N} \openint {-\dfrac 1 n} {\dfrac 1 n} = \set 0 = \closedint 0 0$ which is a closed interval of $\R$. The result follows from Closed Real Interval is not Open Set. {{qed}}
=== Definition from Open Neighborhood is equivalent to Definition from Closure === {{:Equivalence of Definitions of Adherent Point/Definition 1 iff Definition 2}}{{qed|lemma}} ===  Definition from Open Neighborhood is equivalent to Definition from Neighborhood === {{:Equivalence of Definitions of Adherent Point/Definition 1 iff Definition 3}}{{qed}} Category:Set Closures Category:Equivalence of Definitions of Adherent Point 82ykqg71wrlnlh3p49zzvwysi1pyw8m
Follows directly from: :Compact Space is Countably Compact :Countably Compact Metric Space is Compact {{qed}} {{ACC|Countably Compact Metric Space is Compact}}
This follows directly from the results: :Metric Space is First-Countable :Countably Compact First-Countable Space is Sequentially Compact {{qed}}
Proof by Contradiction. {{AimForCont}} that $A$ does not have an $\omega$-accumulation point in $X$. Let $\SS \subseteq \powerset A$ be the set of all finite subsets of $A$. By Set of Finite Subsets of Countable Set is Countable, we have that $\SS$ is countable. For all (finite) $F \in \SS$, define: :$U_F = \paren {F \cup \paren {X \setminus A} }^\circ$ where ${}^\circ$ denotes the interior. By Image of Countable Set under Mapping is Countable, we have that $\CC = \set {U_F: F \in \SS}$ is countable. By the definition of an $\omega$-accumulation point, it follows that: :$\forall x \in X: \exists U \in \tau: x \in U: \exists F \in \SS: U \cap A = F$ By Set Difference Union Intersection, we have: :$U = \paren {U \cap A} \cup \paren {U \setminus A} \subseteq F \cup \paren {X \setminus A}$ By Set Interior is Largest Open Set, it follows that $\CC$ is a countable open cover for $X$. Hence, by the definition of a countably compact space, there exists a finite subcover $\CC'$ of $\CC$ for $X$. By the Principle of Finite Choice, there exists an indexed family $\family {G_V}_{V \mathop \in \CC'}$ of elements of $\SS$ such that: :$\forall V \in \CC': V = U_{G_V} = \paren {G_V \cup \paren {X \setminus A} }^\circ$ Define: :$\displaystyle B = \bigcup_{V \mathop \in \CC'} G_V \subseteq A$ By Union of Finite Sets is Finite, it follows that $B$ is finite. We can use the definition of a cover to conclude that: :$X \subseteq B \cup \paren {X \setminus A}$ By Relative Complement inverts Subsets, it follows that: :$\paren {X \setminus B} \cap A = \O$ Hence, $A \subseteq B$. But then by Subset of Finite Set is Finite, it follows that $A$ is finite, a contradiction.
Follows directly from: :Sequentially Compact Space is Countably Compact :Countably Compact First-Countable Space is Sequentially Compact {{qed}} {{ACC|Sequentially Compact Space is Countably Compact}}
By the definition of a first-countable space, there exists a countable local basis $\BB$ at $x$. By Surjection from Natural Numbers iff Countable, there exists a surjection $\phi: \N \to \BB$. For all $n \in \N$, define the set: :$\displaystyle U_n = \bigcap_{k \mathop = 0}^n \map \phi k$ By General Intersection Property of Topological Space, it follows that $U_n$ is an open neighborhood of $x$. {{explain|What is the exact connection with PoRD?}} Using the Principle of Recursive Definition, we construct a strictly increasing sequence $\sequence {n_k}_{k \mathop \in \N}$ in $\N$. By the definition of an accumulation point, we can choose $n_0 \in \N$ such that $x_{n_0} \in U_0$. For all $k \in \N$, let $n_{k + 1} > n_k$ be the (unique) smallest natural number such that $x_{n_{k + 1} } \in U_{k + 1}$. Such an $n_{k + 1}$ exists by the definition of an accumulation point, and by the well-ordering principle. We now show that $x$ is a limit point of $\sequence {x_{n_k} }$. Let $U$ be an open neighborhood of $x$. By the definition of a local basis, there exists an $H \in \BB$ such that $H \subseteq U$. By the definition of a surjection, there exists a natural number $m$ such that $H = \map \phi m$. By construction, we have: :$\forall k \in \N: k > m \implies x_{n_k} \in U_k \subseteq \map \phi m = H$ Hence the result, by the definition of a limit point. {{qed}} Category:First-Countable Spaces Category:Sequences ldlam0wvs63p4v87p2v8zkb910vd0b1
By the definition of separability, we can choose a subset $S \subseteq X$ that is countable and everywhere dense. Define: :$\BB = \set {\map {B_{1/n} } x: x \in S, \, n \in \N_{>0} }$ where $\map {B_\epsilon } x$ denotes the open $\epsilon$-ball of $x$ in $M$. We have that Cartesian Product of Countable Sets is Countable. Hence, by Image of Countable Set under Mapping is Countable, it follows that $\BB$ is countable. Let $\tau$ denote the topology on $X$ induced by the metric $d$. It suffices to show that $\BB$ is an analytic basis for $\tau$. From Open Ball of Metric Space is Open Set, we have that $\BB \subseteq \tau$. We use Equivalence of Definitions of Analytic Basis. Let $y \in U \in \tau$. By the definition of an open set, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} y \subseteq U$. By the Archimedean Principle, there exists a natural number $n > \dfrac 2 \epsilon$. That is: :$\dfrac 2 n < \epsilon$ and so: :$\map {B_{2/n} } y \subseteq \map {B_\epsilon} y$. From Subset Relation is Transitive, we have $\map {B_{2/n} } y \subseteq U$. By the definition of everywhere denseness, and by Equivalence of Definitions of Adherent Point, there exists an $x \in S \cap \map {B_{1/n} } y$. By axiom $\left({M3}\right)$ for a metric, it follows that $y \in \map {B_{1/n} } x$. For all $z \in \map {B_{1/n} } x$, we have: {{begin-eqn}} {{eqn | l = \map d {z, y}       | o = \le       | r = \map d {z, x} + \map d {x, y}       | c = Metric Space Axiom $(\text M 2)$ }} {{eqn | r = \map d {z, x} + \map d {y, x}       | c = Metric Space Axiom $(\text M 3)$ }} {{eqn | o = <       | r = \frac 2 n }} {{end-eqn}} That is: :$\map {B_{1/n} } x \subseteq \map {B_{2/n} } y$ From Subset Relation is Transitive, we have: :$y \in \map {B_{1/n} } x \subseteq U$ Hence the result. {{qed}}
Firstly, note that by Equalizer is Monomorphism, local membership of $e$ is defined. === Necessary Condition === Suppose that $z \in_C e$. By definition of local membership, there is an $h: Z \to E$ such that $z = e \circ h$. Then since $e$ is the equalizer of $f$ and $g$: :$f \circ e = g \circ e$ from which we deduce: :$f \circ z = f \circ e \circ h = g \circ e \circ h = g \circ z$ by using metacategory axiom $(C3)$, i.e. associativity of $\circ$. {{qed|lemma}} === Sufficient Condition === Suppose that $f \circ z = g \circ z$. Since $e$ is the equalizer of $f$ and $g$, we find $h: Z \to E$ such that: :$z = e \circ h$ Hence $z \in_C e$, by definition of local membership. {{qed}}
This follows directly from: * Sequentially Compact Metric Space is Totally Bounded * Totally Bounded Metric Space is Separable. {{qed}} {{ACC|Totally Bounded Metric Space is Separable}} Category:Separable Spaces Category:Metric Spaces Category:Sequentially Compact Spaces 5ngsevn0ih6kfj87bntzl2pno8t3wip
This follows directly from: : Sequentially Compact Metric Space is Separable : Separable Metric Space is Second-Countable. {{qed}} {{ACC|Sequentially Compact Metric Space is Separable}} Category:Second-Countable Spaces Category:Metric Spaces Category:Sequentially Compact Spaces au7mekk5yhl715pl6nvwpjyl5tw6cpe
Let $A$ be a set and let $\mathcal D$ be the set of all metrics on $A$. In the following, let $d_1, d_2, d_3 \subseteq \mathcal D$ be arbitrary. Checking in turn each of the criteria for equivalence: === Reflexivity === Let $d_1$ be a metric on $A$. Then trivially: :$U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_1$-open. That is, $d_1 \sim d_1$ and so $\sim$ has been shown to be reflexive. {{qed|lemma}} === Symmetry === Let $d_1 \sim d_2$. That is, let $d_1, d_2$ be topologically equivalent metrics on $A$. Then by definition: :$U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_2$-open. It follows from Biconditional is Commutative that: :$U \subseteq A$ is $d_2$-open $\iff$ $U \subseteq A$ is $d_1$-open. That is, $d_2 \sim d_1$ and so $\sim$ has been shown to be symmetric. {{qed|lemma}} === Transitivity === Let $d_1 \sim d_2$ and $d_2 \sim d_3$. Then by definition: :$U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_2$-open. :$U \subseteq A$ is $d_2$-open $\iff$ $U \subseteq A$ is $d_3$-open. Then by Biconditional is Transitive: :$U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_3$-open. That is, $d_1 \sim d_3$ and so $\sim$ has been shown to be transitive. {{qed|lemma}} $\sim$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
Consider the identity mapping: : $f: A \to A: \forall x \in A: f \left({x}\right) = x$ Then $f: \left({A, d_1}\right) \to \left({A, d_2}\right)$ can be considered as a Lipschitz equivalence. The result then follows from Lipschitz Equivalent Metric Spaces are Homeomorphic. {{qed}}
By definition of Lipschitz equivalence: : $\forall x, y \in A: h \map {d_2} {x, y} \le \map {d_1} {x, y} \le k \map {d_2} {x, y}$ for some $h, k \in \R_{>0}$. Let $x \in A$. Let $\epsilon \in \R_{>0}$. Let $\map {B_{h \epsilon} } {x; d_1}$ denote the open $h \epsilon$-ball with respect to $d_1$ of $x \in A$. Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \map {B_{h \epsilon} } {x; d_1}       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_1} {x, y}       | o = <       | r = h \epsilon       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_2} {x, y}       | o = \le       | r = \frac {\map {d_1} {x, y} } h       | c =  }} {{eqn | o = <       | r = \epsilon       | c =  }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \map {B_\epsilon} {x; d_2}       | c =  }} {{eqn | ll= \leadsto       | l = \map {B_{h \epsilon} } {x; d_1}       | o = \subseteq       | r = \map {B_\epsilon} {x; d_2}       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \map {B_{\epsilon / k} } {x; d_2}       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_2} {x, y}       | o = <       | r = \frac \epsilon k       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_1} {x, y}       | o = \le       | r = k \frac {\map {d_2} {x, y} } h       | c =  }} {{eqn | o = <       | r = \epsilon       | c =  }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \map {B_\epsilon} {x; d_1}       | c =  }} {{eqn | ll= \leadsto       | l = \map {B_{\epsilon / k} } {x; d_2}       | o = \subseteq       | r = \map {B_\epsilon} {x; d_1}       | c =  }} {{end-eqn}} Now suppose $U \subseteq A$ is $d_1$-open. Let $x \in U$. Then: :$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {x; d_1} \subseteq U$ Thus: :$\map {B_{\epsilon / k} } {x; d_2} \subseteq \map {B_\epsilon} {x; d_1} \subseteq U$ and so $U$ is $d_2$-open. Mutatis mutandis, if $U \subseteq A$ is $d_2$-open, it follows that $U$ is $d_1$-open. {{qed}}
By definition of Lipschitz equivalence: $\exists K_1, K_2 \in \R_{>0}$ such that: : $(1): \quad \forall x, y \in A: d_2 \left({x, y}\right) \le K_1 d_1 \left({x, y}\right)$ : $(2): \quad \forall x, y \in A: d_1 \left({x, y}\right) \le K_2 d_2 \left({x, y}\right)$ By Identity Mapping between Metrics separated by Scale Factor is Continuous: : the identity mapping from $M_1$ to $M_2$ is continuous : the identity mapping from $M_2$ to $M_1$ is continuous. Hence the result by definition of topological equivalence. {{qed}}
Follows directly from: :Compact Space is Countably Compact :Countably Compact Metric Space is Sequentially Compact :Sequentially Compact Metric Space is Complete {{qed}}