hcju/proofwikiqa huggingface dataset:
Let $M_1$ and $M_2$ be Lipschitz equivalent. Then, by definition, $\exists h, k \in \R_{>0}$ such that: : $\forall x, y \in A_1: h d_1 \left({x, y}\right) \le d_2 \left({f \left({x}\right), f \left({y}\right)}\right) \le k d_1 \left({x, y}\right)$ From the definition of open $\epsilon$-ball: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = B_{h \epsilon} \left({f \left({x}\right); d_2}\right)       | c =  }} {{eqn | ll= \implies       | l = d_2 \left({f \left({x}\right), f \left({y}\right)}\right)       | o = <       | r = h \epsilon       | c =  }} {{eqn | ll= \implies       | l = d_1 \left({x, y}\right)       | o = \le       | r = \frac {d_2 \left({f \left({x}\right), f \left({y}\right)}\right)} h < \epsilon       | c =  }} {{eqn | ll= \implies       | l = y       | o = \in       | r = B_\epsilon \left({x; d_1}\right)       | c =  }} {{end-eqn}} ... and: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = B_{\epsilon / k} \left({x; d_1}\right)       | c =  }} {{eqn | ll= \implies       | l = d_1 \left({x, y}\right)       | o = <       | r = \frac \epsilon k       | c =  }} {{eqn | ll= \implies       | l = d_2 \left({f \left({x}\right), f \left({y}\right)}\right)       | o = \le       | r = k {d_1 \left({x, y}\right)} < \epsilon       | c =  }} {{eqn | ll= \implies       | l = y       | o = \in       | r = B_\epsilon \left({f \left({x}\right); d_2}\right)       | c =  }} {{end-eqn}} Thus: : $B_{h \epsilon} \left({f \left({x}\right); d_2}\right) \subseteq B_\epsilon \left({x; d_1}\right)$ : $B_{\epsilon / k} \left({x; d_1}\right) \subseteq B_\epsilon \left({f \left({x}\right); d_2}\right)$ {{qed|lemma}} Now, suppose $U$ is $d_2$-open. Let $x \in U$. Then: : $\exists \epsilon \in \R_{>0}: B_\epsilon \left({f \left({x}\right); d_2}\right) \subseteq U$. Hence: : $B_{\epsilon / k} \left({x; d_1}\right) \subseteq U$ Thus $U$ is $d_1$-open. Similarly, suppose $U$ is $d_1$-open. Let $x \in U$. Then: : $\exists \epsilon \in \R_{>0}: B_\epsilon \left({x; d_1}\right) \subseteq U$ Hence: : $B_{h \epsilon} \left({f \left({x}\right); d_2}\right) \subseteq U$ Thus $U$ is $d_2$-open. The result follows by definition of homeomorphic metric spaces. {{qed}}
Let $r, t \in \R_{\ge 1}$. {{WLOG}}, assume that $r \le t$. For all $x, y \in \R^n$, we are going to show that: :$\displaystyle \map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_r} {x, y}$ Then we can demonstrate Lipschitz equivalence between all of these metrics, from which topological equivalence follows. Let $d_r$ be the metric defined as: :$\displaystyle \map {d_r} {x, y} = \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{1/r}$ === Inequalities for Chebyshev Distance === {{:Relation between P-Product Metric and Chebyshev Distance on Real Vector Space}} {{qed|lemma}} === Inequality for General Case === {{:P-Product Metrics on Real Vector Space are Topologically Equivalent/Inequality for General Case}} When we combine the inequalities, we have: {{improve|In fact, this inequality (for real $r \ge 1$) can be established quite easily without the humongous calculation.}} :$\map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_1} {x, y} \ge n^{-1} \map {d_r} {x, y}$ Therefore, $d_r$ and $d_\infty$ are Lipschitz equivalent for all $r \in \R_{\ge 1}$. {{qed}}
First it is shown that $\map {D_x} {c x} = c$: {{begin-eqn}} {{eqn | l = \map {D_x} {c x}       | r = c \map {D_x} x + x \map {D_x} c       | c = Product Rule }} {{eqn | r = c + x \map {D_x} c       | c = Derivative of Identity Function }} {{eqn | r = c + 0       | c = Derivative of Constant }} {{eqn | r = c }} {{end-eqn}} Next: {{begin-eqn}} {{eqn | l = \map {D_x} {\map f {c x} }       | r = \map {D_x} {c x} \map {D_{c x} } {\map f {c x} }       | c = Chain Rule for Derivatives }} {{eqn | r = c \map {D_{c x} } {\map f {c x} }       | c = from above }} {{end-eqn}} {{qed}} Category:Differential Calculus jt5vlqqn2wg1t6e7w8q2vhlfcae82ro
{{begin-eqn}} {{eqn | l = \cot \left({x + \pi}\right)       | r = \frac {\cos \left({x + \pi}\right)} {\sin \left({x + \pi}\right)}       | c = Definition of Cotangent Function }} {{eqn | r = \frac {-\cos x} {-\sin x}       | c = Cosine of Angle plus Straight Angle, Sine of Angle plus Straight Angle }} {{eqn | r = \cot x       | c =  }} {{end-eqn}} Also, from Derivative of Cotangent Function, we have that: : $D_x \left({\cot x}\right) = -\dfrac 1 {\sin^2 x}$ provided $\sin x \ne 0$. From Shape of Sine Function, we have that $\sin$ is strictly positive on the interval $\left({0 \,.\,.\, \pi}\right)$. From Derivative of Monotone Function, $\cot x$ is strictly decreasing on that interval, and hence can not have a period of ''less'' than $\pi$. Hence the result. {{qed}} Category:Cotangent Function 3lwf13v4mp3rfvrjzzlc3jbgwu3lgnj
* $\cot x$ is continuous and strictly decreasing on $\left({0 \,.\,.\, \pi}\right)$: Continuity follows from the Quotient Rule for Continuous Functions: :$(1): \quad$ Both $\sin x$ and $\cos x$ are continuous on $\left({0 \,.\,.\, \pi}\right)$ from Real Sine Function is Continuous and Cosine Function is Continuous :$(2): \quad \sin x > 0$ on this interval. The fact of $\cot x$ being strictly decreasing on this interval has been demonstrated in the discussion on Cotangent Function is Periodic on Reals. * $\cot x \to + \infty$ as $x \to 0^+$: From Sine and Cosine are Periodic on Reals, we have that both $\sin x > 0$ and $\cos x > 0$ on $\left({0 \,.\,.\, \dfrac \pi 2}\right)$. We have that: :$(1): \quad \cos x \to 1$ as $x \to 0^+$ :$(2): \quad \sin x \to 0$ as $x \to 0^+$ Thus it follows that $\cot x = \dfrac {\cos x} {\sin x} \to + \infty$ as $x \to 0^+$. * $\tan x \to - \infty$ as $x \to \pi^-$: From Sine and Cosine are Periodic on Reals, we have that $\sin x > 0$ and $\cos x < 0$ on $\left({\dfrac \pi 2 \,.\,.\, \pi}\right)$. We have that: :$(1): \quad \cos x \to -1$ as $x \to \pi^-$ :$(2): \quad \sin x \to 0$ as $x \to \pi^-$ Thus it follows that $\cot x = \dfrac {\cos x} {\sin x} \to - \infty$ as $x \to \pi^-$. * $\cot x$ is not defined and discontinuous at $x = n \pi$: From the discussion of Sine and Cosine are Periodic on Reals, it was established that $\forall n \in \Z: x = n \pi \implies \sin x = 0$. As division by zero is not defined, it follows that at these points $\cot x$ is not defined either. Now, from the above, we have: :$(1): \quad \cot x \to + \infty$ as $x \to 0^+$ :$(2): \quad \cot x \to - \infty$ as $x \to \pi^-$ As $\cot \left({x + \pi}\right) = \cot x$ from Cotangent Function is Periodic on Reals, it follows that $\cot x \to + \infty$ as $x \to \pi^+$. Hence the left hand limit and right hand limit at $x = \pi$ are not the same. From the periodic nature of $\cot x$, it follows that the same applies $\forall n \in \Z: x = n \pi$. The fact of its discontinuity at these points follows from the definition of discontinuity. * $\cot \left({n + \dfrac 1 2}\right) \pi = 0$: Follows directly from Sine and Cosine are Periodic on Reals: $\forall n \in \Z: \cos \left({n + \dfrac 1 2}\right) \pi = 0$. === Graph of Cotangent Function === {{:Shape of Cotangent Function/Graph}} {{qed}}
Let: : $y = \log_b x \iff b^y = x$ : $z = \log_a x \iff a^z = x$ Then: {{begin-eqn}} {{eqn | l = z       | r = \map {\log_a} {b^y} }} {{eqn | r = y \log_a b       | c = Logarithms of Powers }} {{eqn | r = \log_b x \log_a b }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {u^v}       | r = \map {\dfrac \d {\d x} } {\map \exp {v \ln u} }       | c = {{Defof|Power to Real Number}} }} {{eqn | r = \map \exp {v \ln u} \map {\dfrac \d {\d x} } {v \ln u}       | c = Chain Rule for Derivatives and Derivative of Exponential Function }} {{eqn | r = \map \exp {v \ln u} \paren {\paren {\ln u} \map {\dfrac \d {\d x} } v + v \map {\dfrac \d {\d x} } {\ln u} }       | c = Product Rule }} {{eqn | r = u^v \paren {\paren {\ln u} \map {\dfrac \d {\d x} } v + \frac v u \map {\dfrac \d {\d x} } u}       | c = Chain Rule for Derivatives }} {{eqn | r = v u^{v - 1} \map {\dfrac \d {\d x} } u + u^v \paren {\ln u} \map {\dfrac \d {\d x} } v       | c = gathering terms }} {{end-eqn}} {{qed}}
By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le  M$ for all $z \in \C$.  For any $R \in \R: R > 0$, consider the function: :$\map {f_R} z := \map f {R z}$ Using the Cauchy Integral Formula, we see that: :$\displaystyle \cmod {\map {f_R'} z} = \frac 1 {2 \pi} \cmod {\int_{\map {C_1} z} \frac {\map f w} {\paren {w - z}^2} \rd w} \le \frac 1 {2 \pi} \int_{\map {C_1} z} M \rd w = M$ where $\map {C_1} z$ denotes the circle of radius $1$ around $z$. Hence: :$\displaystyle \cmod {\map {f'} z} = \cmod {\map {f_R'} z} / R \le M / R$ Since $R$ was arbitrary, it follows that $\cmod {\map {f'} z} = 0$ for all $z \in \C$. Thus $f$ is constant. {{qed}} {{MissingLinks}}
{{WLOG}}, we may assume that: :$U = \Bbb D := \left\{{z \in \C: \left|{z}\right| < 1}\right\}$ and that $z_0 = 0$.  Otherwise, restrict $f$ to a suitable disk centered at $z_0$, and precompose with a suitable affine map. A holomorphic function is continuous. A continuous function is locally bounded. So $(1) \implies (2) \implies (3)$. Also, $(3) \implies (4)$ by definition.  That $(2) \implies (1)$ follows from the proof of the Residue Theorem. (For completeness, we sketch a self-contained argument below.) {{refactor|Extract that proof referred to above, and put it in its own page, if necessary.}} Now assume that $(4)$ holds. That is: :$z \cdot f \left({z}\right) \to 0$ as $z \to 0$. We need to show that (b) holds.  By assumption, the function: :$g \left({z}\right) := \begin{cases} z \cdot f \left({z}\right) & : z \ne 0 \\ 0 & : \text{otherwise} \end{cases}$ is continuous in $\Bbb D$ and holomorphic in $\Bbb D \setminus \left\{ {0}\right\}$. By applying the direction (b) $\implies$ (a) to this function, we see that $g$ is holomorphic in $\Bbb D$. We have: :$\displaystyle g' \left({0}\right) = \lim_{z \to 0} \frac {g \left({z}\right)} z = \lim_{z \to 0} f \left({z}\right)$ so $f$ extends continuously to $\Bbb D$, as claimed. To prove $(2) \implies (1)$, we use Morera's Theorem and the Residue Theorem. By Morera's Theorem, we need to show that: :$\displaystyle \int_C f \left({z}\right) \, \mathrm d z = 0$ for every closed curve $C$ in $\Bbb D$. It follows from the Cauchy integral theorem that we only need to check that: :$\displaystyle \int_C f \left({z}\right) \, \mathrm d z = 0$ for a simple closed loop surrounding $0$, and that this integral is independent of the loop $C$. Letting $C = C_\epsilon \left({0}\right)$ be the circle of radius $\epsilon$ around $0$, we see that: :$\displaystyle \left|{\int_{C_\epsilon} f \left({z}\right) \, \mathrm d z}\right| \le 2 \pi \epsilon \max_{z \in C_\epsilon} \left|{f \left({z}\right)}\right| \to 0$ as $\epsilon\to 0$ (because $f$ is continuous in $0$ by assumption). This completes the proof. {{qed}} {{Namedfor|Georg Friedrich Bernhard Riemann|cat = Riemann}} Category:Complex Analysis Category:Continuous Functions 7l2wp2sscbwqqjmoijs79kwjblfak5j
{{begin-eqn}} {{eqn	| l = \int_{-\infty}^\infty e^{-x^2} \rd x 	| r = 2 \int_0^\infty e^{-x^2} \rd x 	| c = Definite Integral of Even Function }} {{eqn	| r = \frac {2 \sqrt \pi} 2 	| c = Integral to Infinity of Exponential of -t^2 }} {{eqn	| r = \sqrt \pi }} {{end-eqn}} {{qed}} Category:Definite Integrals involving Exponential Function kxvyhs9dfoak7xid7n2a1pf5blmlt52
In order to be metrizable, there needs to be a metric $d$ on $S$, so that $\struct {S, d}$ is a metric space. As $S$ has more than one element, $\exists x, y \in S: x \ne y$. Then: :$\epsilon = \map d {x, y} > 0$ So the open $\epsilon$-ball $\map {B_{\epsilon / 2} } x$ is $d$-open. Hence $\map {B_{\epsilon / 2} } x$ is in the topology which $d$ induces. But $x \in \map {B_{\epsilon / 2} } x$ while $y \notin \map {B_{\epsilon / 2} } x$. Thus $\map {B_{\epsilon / 2} } x \ne \O$ and $\map {B_{\epsilon / 2} } x \ne S$. So the topology induced by $d$ is not the indiscrete topology. {{qed}}
Let $U \in T_3$ be open in $T_3$. As $g$ is continuous, $g^{-1} \sqbrk U \in T_2$ is open in $T_2$. As $f$ is continuous, $f^{-1} \sqbrk {g^{-1} \sqbrk U} \in T_1$ is open in $T_1$. By Inverse of Composite Bijection, $f^{-1} \sqbrk {g^{-1} \sqbrk U} = \paren {g \circ f}^{-1} \sqbrk U$. Hence the result. {{qed}}
Follows directly from: * the open set definition of continuity on a metric space * the definition of continuity on a topological space. {{qed}}
We proceed to verify the open set axioms for $\tau$ to be a topology on $S$. === $(\text O 1):$ Union of Open Sets === Let $\AA \subseteq \tau$. It is to be shown that: :$\displaystyle \bigcup \AA \in \tau$ Define: :$\displaystyle \AA' = \bigcup_{U \mathop \in \AA} \set {B \in \BB: B \subseteq U}$ By Union is Smallest Superset: Family of Sets, it follows that $\AA' \subseteq \BB$. Hence, by Equivalence of Definitions of Topology Generated by Synthetic Basis and General Distributivity of Set Union: :$\displaystyle \bigcup \AA = \bigcup_{U \mathop \in \AA} \bigcup \set {B \in \BB: B \subseteq U} = \bigcup \AA' \in \tau$ {{qed|lemma}} === $(\text O 2):$ Pairwise Intersection of Open Sets === Let $U, V \in \tau$. It is to be shown that: :$U \cap V \in \tau$ Define: :$\OO = \set {A \cap B: A, B \in \BB, \, A \subseteq U, \, B \subseteq V}$ By the definition of a synthetic basis: :$\forall A, B \in \BB: A \cap B \in \tau$ Hence, by the definition of a subset, it follows that $\OO \subseteq \tau$. By open set axiom $(1)$, which has been verified above for $\tau$, we have: :$\displaystyle \bigcup \OO \in \tau$ Since set intersection preserves subsets, we have: :$\displaystyle \forall W \in \OO: \exists A, B \in \BB: W = A \cap B \subseteq U \cap V$ By Union is Smallest Superset: General Result, it follows that: :$\displaystyle \bigcup \OO \subseteq U \cap V$ By the definition of $\tau$, it follows from Set is Subset of Union: General Result that: :$\displaystyle \forall x \in U \cap V: \exists A, B \in \BB: A \subseteq U, \, B \subseteq V: x \in A \cap B \subseteq \bigcup \OO$ That is, by the definition of a subset: :$\displaystyle U \cap V \subseteq \bigcup \OO$ By definition of set equality: :$\displaystyle U \cap V = \bigcup \OO \in \tau$ {{qed|lemma}} === $(\text O 3):$ Set Itself === By the definition of a synthetic basis, $\BB$ is a cover for $S$. By Equivalent Conditions for Cover by Collection of Subsets, it follows that: :$\displaystyle S = \bigcup \BB \in \tau$ {{qed}}
=== Necessary Condition === Suppose that $\BB$ is an analytic basis for $\tau$. We proceed to check the axioms for $\BB$ to be a synthetic basis on $S$. Open set axiom $(\text O 3)$ states that $S \in \tau$. Therefore, by the definition of an analytic basis: :$\displaystyle \exists \SS \subseteq \BB: S = \bigcup \SS$ By Equivalent Conditions for Cover by Collection of Subsets, $\BB$ is a cover for $S$. That is, axiom $(\text B 1)$ for a synthetic basis is satisfied by $\BB$. Suppose that $A, B \in \BB$. By the definition of an analytic basis, $\BB \subseteq \tau$. Therefore $A, B \in \tau$. Open set axiom $(\text O 2)$ states that $A \cap B \in \tau$. Therefore, by the definition of an analytic basis: :$\displaystyle \exists \AA \subseteq \BB: A \cap B = \bigcup \AA$ That is, axiom $(\text B 2)$ for a synthetic basis is satisfied by $\BB$. Therefore, $\BB$ is a synthetic basis on $S$. Let $\tau'$ be the topology on $S$ generated by the synthetic basis $\BB$. It follows from the definition of an analytic basis that $\tau \subseteq \tau'$. Since the subset relation is transitive, we can apply open set axiom $(\text O 1)$ for a topology to conclude that: :$\displaystyle \forall U \in \tau': \exists \AA \subseteq \BB \subseteq \tau: U = \bigcup \AA \in \tau$ That is, $\tau' \subseteq \tau$. By definition of set equality: :$\tau = \tau'$ {{qed|lemma}} === Sufficient Condition === Suppose that $\BB$ is a synthetic basis on $S$, and that $\tau$ is the topology on $S$ generated by $\BB$. Then: :$\displaystyle \BB = \set {\bigcup \set B: \set B \subseteq \BB} \subseteq \tau$ By definition, $\BB$ is an analytic basis for $\tau$. {{qed}}
Let $U \in \tau_2$. By the definition of an analytic basis, it follows that: :$\displaystyle \exists \mathcal A \subseteq \mathcal B: U = \bigcup \mathcal A$ Hence: {{begin-eqn}} {{eqn | l = f^{-1} \left({U}\right)       | r = f^{-1} \left({\bigcup \mathcal A}\right) }} {{eqn | r = \bigcup_{B \mathop \in \mathcal A} f^{-1} \left({B}\right)       | c = Preimage of Union under Mapping: General Result }} {{eqn | o = \in       | r = \tau_1       | c = by hypothesis, and by the definition of a topology }} {{end-eqn}} The result follows from the definition of continuity. {{qed}}
Let $\left({c \,.\,.\, d}\right)$ be an open real interval. Then by definition: :$\left({c \,.\,.\, d}\right) = \left({-\infty \,.\,.\, d}\right) \cap \left({c \,.\,.\, \infty}\right)$ and so $\left({c \,.\,.\, d}\right)$ is the intersection of two elements of $\mathcal B$. From Open Sets in Real Number Line, any open set of $\R$ is the union of countably many open real intervals. The result follows by definition of sub-basis. {{qed}}
We consider $X$ as the universe. Thus, in accordance with Intersection of Empty Set, we take the convention that: :$\displaystyle \bigcap \O = X \in \BB$ By Set is Subset of Union: General Result, it follows that: :$\displaystyle X \subseteq \bigcup \BB$ That is, axiom '''B1''' for a synthetic basis is satisfied. We have that $\BB \subseteq \powerset X$. Let $B_1, B_2 \in \BB$. Then there exist finite $\FF_1, \FF_2 \subseteq \SS$ such that: :$\displaystyle B_1 = \bigcap \FF_1$ :$\displaystyle B_2 = \bigcap \FF_2$ It follows that: :$\displaystyle B_1 \cap B_2 = \bigcap \paren {\FF_1 \cup \FF_2}$ {{explain|proof that $\displaystyle \bigcap_{i \mathop \in I} \bigcap \mathbb S_i$ is equal to $\displaystyle \bigcap \bigcup_{i \mathop \in I} \mathbb S_i$}} By Union is Smallest Superset, $\FF_1 \cup \FF_2 \subseteq \SS$. We have that $\FF_1 \cup \FF_2$ is finite. Hence $B_1 \cap B_2 \in \BB$, so it follows by definition that axiom '''B2''' for a synthetic basis is satisfied. {{qed}}
Throughout the following, we assume the morphism property holds for $\phi$ for both operations. === Left Distributivity === Suppose $\circ_1$ is left distributive over $+_1$. Then: {{begin-eqn}} {{eqn | l = \map \phi x \circ_2 \paren {\map \phi y +_2 \map \phi z}       | r = \map \phi x \circ_2 \map \phi {y +_1 z}       | c =  }} {{eqn | r = \map \phi {x \circ_1 \paren {y +_1 z} }       | c =  }} {{eqn | r = \map \phi {\paren {x \circ_1 y} +_1 \paren {x \circ_1 z} }       | c = as $\circ_1$ is left distributive over $+_1$ }} {{eqn | r = \map \phi {x \circ_1 y} +_2 \phi \paren {x \circ_1 z}       | c =  }} {{eqn | r = \paren {\map \phi x \circ_2 \map \phi y} +_2 \paren {\map \phi x \circ_2 \map \phi z}       | c =  }} {{end-eqn}} So $\circ_2$ is left distributive over $+_2$. {{qed}} === Right Distributivity === Suppose $\circ_1$ is right distributive over $+_1$. Then: {{begin-eqn}} {{eqn | l = \paren {\map \phi x +_2 \map \phi y} \circ_2 \map \phi z       | r = \map \phi {x +_1 y} \circ_2 \map \phi z       | c =  }} {{eqn | r = \map \phi {\paren {x +_1 y} \circ_1 z}       | c =  }} {{eqn | r = \map \phi {\paren {x \circ_1 z} +_1 \paren {y \circ_1 z} }       | c = as $\circ_1$ is right distributive over $+_1$ }} {{eqn | r = \map \phi {x \circ_1 z} +_2 \map \phi {y \circ_1 z}       | c =  }} {{eqn | r = \paren {\map \phi x \circ_2 \map \phi z} +_2 \paren {\map \phi y \circ_2 \map \phi z}       | c =  }} {{end-eqn}} So $\circ_2$ is right distributive over $+_2$. {{qed}} === Distributivity === If $\circ_1$ is distributive over $+_1$, then it is both right and left distributive over $+_1$. Hence from the above, $\circ_2$ is both right and left distributive over $+_2$. That is, $\circ_2$ is distributive over $+_2$. {{qed}} Category:Epimorphisms Category:Distributive Operations ovqap1zaxej3ypd2qv9glmydamx4bfb
From Epimorphism Preserves Semigroups, if $\struct {S, \circ}$ is a semigroup then so is $\struct {T, *}$. From Epimorphism Preserves Identity, if $\struct {S, \circ}$ has an identity $e_S$, then $\map \phi {e_S}$ is the identity for $*$. From Epimorphism Preserves Inverses, if $x^{-1}$ is an inverse of $x$ for $\circ$, then $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$. The result follows from the definition of group. {{qed}}
We verify the open set axioms for $\tau_H$ to be a topology on $H$. === $({O1})$: Union of Open Sets === Let $\mathcal A \subseteq \tau_H$. It is to be shown that: :$\displaystyle \bigcup \mathcal A \in \tau_H$ Define: :$\displaystyle \mathcal A' = \left\{{V \in \tau: V \cap H \subseteq \bigcup \mathcal A}\right\} \subseteq \tau$ Let: :$\displaystyle U = \bigcup \mathcal A'$ By the definition of a topology, we have $U \in \tau$. Then, by the distributivity of intersection over union and by Union is Smallest Superset: Family of Sets: :$\displaystyle U \cap H = \bigcup_{V \mathop \in \mathcal A'} \left({V \cap H}\right) \subseteq \bigcup \mathcal A$ By the definition of $\tau_H$ and by Set is Subset of Union: General Result, we have: :$\displaystyle \forall S \in \mathcal A: \exists V \in \tau: S = V \cap H \subseteq \bigcup \mathcal A$ That is: :$\forall S \in \mathcal A: \exists V \in \mathcal A': S = V \cap H$ By Set is Subset of Union: General Result, we have: :$\displaystyle \forall V \in \mathcal A': V \subseteq U$ Since intersection preserves subsets, it follows that: :$\forall S \in \mathcal A: S \subseteq U \cap H$ By Union is Smallest Superset: General Result, we conclude that: :$\displaystyle \bigcup \mathcal A \subseteq U \cap H$ Hence, by definition of set equality: :$\displaystyle \bigcup \mathcal A = U \cap H \in \tau_H$ {{qed|lemma}} === $({O2})$: Pairwise Intersection of Open Sets === Let $A, B \in \tau_H$. Let $U, V \in \tau$ be such that $A = U \cap H$ and $B = V \cap H$. By the definition of a topology, we have $U \cap V \in \tau$. From Intersection Distributes over Intersection: : $A \cap B = \left({U \cap V}\right) \cap H \in \tau_H$ {{qed|lemma}} === $({O3})$: Set Itself === By the definition of a topology, we have $X \in \tau$. By Intersection with Subset is Subset, it follows that $H = X \cap H \in \tau_H$. {{qed|lemma}} All the open set axioms are fulfilled, and the result follows. {{qed}}
Let $M = \struct {A, d}$ be totally bounded. Then there exist $n \in \N$ and points $x_0, \dots, x_n \in A$ such that: :$\displaystyle \inf_{0 \mathop \le i \mathop \le n} \map d {x_i, x} \le 1$ for all $x \in A$. Let us set: :$a := x_0$ :$\displaystyle D := \max_{0 \mathop \le i \mathop \le n} \map d {x_0, x_i}$ :$K := D + 1$ Now let $x \in A$ be arbitrary. Then by assumption there exists $i$ such that $\map d {x_i, x} \le 1$. Hence: :$\map d {a, x} \le \map d {a, x_i} + \map d {x_i, x} \le 1 + D = K$ So $M$ is bounded, as claimed. {{qed}}
=== Necessary Condition === If $\left\langle{x_n}\right\rangle$ converges to $x$, it trivially follows that $\left\langle{x_n}\right\rangle$ is a subsequence of itself that converges to $x$. {{qed|lemma}} === Sufficient Condition === Suppose that $\left\langle{x_{n_k}}\right\rangle$ is a subsequence of $\left\langle{x_n}\right\rangle$ that converges to $x$. Let $\epsilon$ be a strictly positive real number. By the definition of a Cauchy sequence, there exists a real number $M$ such that: :$\displaystyle \forall i, j \in \N: i, j > M \implies d \left({x_i, x_j}\right) < \frac \epsilon 2$ By the definition of convergence, there exists a real number $N$ such that: :$\displaystyle \forall k \in \N: k > N \implies d \left({x_{n_k}, x}\right) < \frac \epsilon 2$ By the Archimedean principle, there exists a natural number $K > \max \left\{{M, N}\right\}$. Therefore, by the triangle inequality: :$\displaystyle \forall m \in \N: m > K \implies d \left({x_m, x}\right) \le d \left({x_m, x_K}\right) + d \left({x_K, x}\right) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$ That is, $\left\langle{x_n}\right\rangle$ converges to $x$. {{qed}} Category:Cauchy Sequences htq125kdcv946ebxpiasn2vm0n3eovm