hcju/proofwikiqa huggingface dataset:
By definition of closure: :$H^- = H' \cup H^i$ where: :$H'$ denotes the set of limit points of $H$ :$H^i$ denotes the set of isolated points of $H$. Let $a \in H$. If $a$ is a limit point of $H$ then $a \in H'$. Suppose $a \notin H'$. Then by definition of limit point: :$\neg \forall \epsilon \in \R_{>0}: \left\{{x \in A: 0 < d \left({x, a}\right) < \epsilon}\right\} \ne \varnothing$ That is: :$\exists \epsilon \in \R_{>0}: \left\{{x \in A: 0 < d \left({x, a}\right) < \epsilon}\right\} = \varnothing$ and so as $d \left({a, a}\right) = 0$: :$\exists \epsilon \in \R_{>0}: \left\{{x \in A: d \left({x, a}\right) < \epsilon}\right\} = \left\{{a}\right\}$ Thus, by definition, $a$ is an isolated point of $H$. So $a \in H'$ or $a \in H^i$. Hence by definition of set union: :$a \in H \subseteq H^-$ By definition of subset: :$H \subseteq H \subseteq H^-$ and hence the result by definition of closure. {{qed}}
From definition of closure, $H^- = H' \cup H^i$. Suppose that $a \in H^-$. If $a \in H^i$, then $a \in H$ and so $\sequence {a, a, \ldots}$ is a sequence in $H$ that converges to $a$. If $a \in H'$, then by Limit Point is Limit of Convergent Sequence there exist a sequence in $H$ that converges to $a$. {{qed|lemma}} For the converse, let $\sequence {x_n}$ be a sequence in $H$ that converges to a point $a \in A$. So let $U$ be an open set that contains $a$. By the definition of an open set, it follows that there exists an open ball centered at $a$ such that $\map {B_\epsilon} a \subseteq U$. But then because $\sequence {x_n}$ converges to $a$ there exists $m \in \N_{\gt 0}$ such that: {{begin-eqn}} {{eqn | l = \map d {x_m, a}       | o = <       | r = \epsilon }} {{eqn | ll= \leadsto       | l = H \cap \map {B_\epsilon} a       | o = \ne       | r = \O       | c =  }} {{eqn | ll= \leadsto       | l = H \cap U       | o = \ne       | r = \O       | c =  }} {{end-eqn}} Because from Set Intersection Preserves Subsets it is seen that $H \cap \map {B_\epsilon} a \subseteq H \cap U$. It has been shown that every open set that contains $a$ also contains a point in $H$. So by Condition for Point being in Closure, it is seen that $a \in H^-$. {{qed}}
From Metric Induces Topology, the topology $\tau$ induced by the metric $d$ is a topology on $M$. From Metric Closure and Topological Closure of Subset are Equivalent, it is sufficient to show that the topological closure of $H$ is contained in $F$. From Set is Closed in Metric Space iff Closed in Induced Topological Space: :$F$ is closed in the topological space $\struct{A, \tau}$ induced by the metric $d$. We have: {{begin-eqn}} {{eqn | l = H       | o = \subseteq       | r = F }} {{eqn | ll= \leadsto       | l = H^-       | o = \subseteq       | r = F^-       | c = Topological Closure of Subset is Subset of Topological Closure }} {{eqn | r = F       | c = Set is Closed iff Equals Topological Closure }} {{end-eqn}} {{qed}}
Let $x \in H^-$. From Point in Closure of Subset of Metric Space iff Limit of Sequence :there exists a sequence $\sequence {a_n}$ of points of $H$ which converges to the limit $x$. By assumption: :$\sequence {a_n}$ is also a sequence of points of $F$ From Subset of Metric Space contains Limits of Sequences iff Closed: :$x \in F$ Thus it has been shown: :$H^- \subseteq F$ {{qed}}
Let $\mathbb K$ be the set of all closed sets $K$ of $M$ such that $H \subseteq K$. From Closure of Subset of Closed Set of Metric Space is Subset: :$\forall K \in \mathbb K: H^- \subseteq K$ From Intersection is Largest Subset: :$\displaystyle H^- \subseteq \bigcap \mathbb K$ {{qed|lemma}} From Closure of Subset of Metric Space is Closed, $H^-$ is closed. From Subset of Metric Space is Subset of its Closure, $H \subseteq H^-$. So $H^-$ is, by definition, a closed set of $M$ which contains $H$. But $\displaystyle \bigcap \mathbb K$ is the intersection of '''all''' closed sets of $M$ which contain $H$. So from Intersection is Subset it follows that: : $\displaystyle \bigcap \mathbb K \subseteq H^-$ {{qed|lemma}} Finally, we have that: : $\displaystyle H^- \subseteq \bigcap \mathbb K$ : $\displaystyle \bigcap \mathbb K \subseteq H^-$ So by definition of set equality: : $\displaystyle H^- = \bigcap \mathbb K$ which is what we needed to prove. {{qed}}
From Rational Numbers form Subfield of Real Numbers: :$\Q \subseteq \R$ By the definition of the Euclidean metric on $\R$: :$\forall x, y \in \R: \map d {x, y} = \size {x - y}$ and so $d_\Q$ is a restriction of $d$: :$d_\Q = d {\restriction}_{\Q \times \Q}$ From Euclidean Metric on Real Number Line is Metric, $d$ is a metric on $\R$. The result follows by definition of metric subspace. {{qed}}
The Euclidean metric on the real number line is a special case of the Euclidean metric on the real vector space $\R^n$. The result follows from Euclidean Metric on Real Vector Space is Metric. {{qed}}
Consider the real number line under the Euclidean metric: :$M = \struct {\R, d}$ where $d$ is the distance function given by: :$\map d {x, y} = \size {x - y}$ === Proof of $\text M 1$ === {{begin-eqn}} {{eqn | l = \map d {x, x}       | r = \size {x - x}       | c = Definition of $d$ }} {{eqn | r = \size 0       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $\text M 1$ holds for $d$. {{qed|lemma}} === Proof of $\text M 2$ === {{begin-eqn}} {{eqn | l = \map d {x, y} + \map d {y, z}       | r = \size {x - y} + \size {y - z}       | c = Definition of $d$ }} {{eqn | o = \ge       | r = \size {\paren {x - y} + \paren {y - z} }       | c = Triangle Inequality for Real Numbers }} {{eqn | r = \size {x - z}       | c =  }} {{eqn | r = \map d {x, z}       | c = Definition of $d$ }} {{end-eqn}} So axiom $M2$ holds for $d$. {{qed|lemma}} === Proof of $\text M 3$ === {{begin-eqn}} {{eqn | l = \map d {x, y}       | r = \size {x - y}       | c = Definition of $d$ }} {{eqn | r = \size {y - x}       | c =  }} {{eqn | r = \map d {y, x}       | c = Definition of $d$ }} {{end-eqn}} So axiom $\text M 3$ holds for $d$. {{qed|lemma}} === Proof of $\text M 4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x - y       | o = \ne       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \size {x - y}       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \map d {x, y}       | o = >       | r = 0       | c = Definition of $d$ }} {{end-eqn}} So axiom $\text M 4$ holds for $d$. {{qed}}
Let $a \in H$. Let $\epsilon \in \R_{>0}$. Let $\delta = \epsilon$. Then: {{begin-eqn}} {{eqn | l = d_H \left({a, y}\right)       | o = <       | r = \delta       | c = for some $y \in H$ }} {{eqn | ll= \implies       | l = d \left({i_H \left({a}\right), i_H \left({y}\right)}\right)       | r = d \left({a, y}\right)       | c =  }} {{eqn | r = d_H \left({a, y}\right)       | c =  }} {{eqn | o = <       | r = \delta       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} So by definition $d_H$ is continuous at $a$. As $a \in H$ is arbitrary, it follows that $d_H$ is continuous on $H$. {{qed}}
We will first prove the theorem for the case $n \le 2047$.  Consider the following sequence of prime numbers: :$2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503$ Each of these prime number is smaller than twice the previous one. Hence every interval $\set {x: n < x \le 2 n}$, with $n \le 2047$, contains one of these prime numbers. === Lemma 1 === {{:Bertrand-Chebyshev Theorem/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Bertrand-Chebyshev Theorem/Lemma 2}}{{qed|lemma}} === Lemma 3 === {{:Bertrand-Chebyshev Theorem/Lemma 3}}{{qed|lemma}} In particular, if $p > \sqrt {2 n}$, then $p$ appears at most once in $\dbinom {2 n} n$. For $n \ge 3$, there is no prime factor $p$ with $\dfrac 2 3 n < p \le n$, for such a prime number divides $n!$ exactly once and $\paren {2 n}!$ exactly twice. Therefore, by Lemma 1: :$\displaystyle \dfrac {2^{2 n} } {2 n + 1} \le \dbinom {2 n} n \le \prod_{p \mathop \le \sqrt {2 n} } 2 n \prod_{\sqrt{2 n} \mathop < p \mathop \le \frac 2 3 n} p \prod_{n \mathop < p \mathop \le 2 n} p$ for $n \ge 3$. Now assume there is no prime number $p$ with $n < p \le 2 n$.  Then we have: {{begin-eqn}} {{eqn | l = \frac {2^{2 n} } {2 n + 1}        | r = \prod_{p \mathop \le \sqrt {2 n} } 2 n \prod_{\sqrt {2 n} \mathop < p \mathop \le \frac 2 3 n} p        | o = \le       | c =  }} {{eqn | r = \paren {2 n}^{\sqrt {2 n} } \prod_{p \mathop \le \frac 2 3 n} p       | o = \le       | c =  }} {{eqn | r = \paren {2 n}^{\sqrt {2 n} } 2^{\frac 4 3 n}       | o = \le       | c = by Lemma 2 }} {{end-eqn}} This is a contradiction if $n$ is large enough. Indeed, we have: :$2^{\frac 2 3 n} \le \paren {2 n + 1} \paren {2 n}^{\sqrt {2 n} }$ Now: :$2 n + 1 \le \paren {2 n}^2 \le \paren {2 n}^{\frac 1 3 \sqrt {2 n} }$ for $n \ge 18$. So: :$2^{2 n} \le \paren {2 n}^{4 \sqrt {2 n} }$ Put $r = \sqrt {2 n}$. Then: :$2^{r^2} \le r^{8 r}$ or equivalently: :$2^r \le r^8$ This fails when $r = 2^6 = 64$. It fails thereafter, since $2^r$ increases faster than $r^8$. So our proof works if: :$n \ge 2^{11} = 2048$ and the examples show it is true for smaller $n$. {{qed}}
When $k \ge 0$, we have: {{begin-eqn}} {{eqn | l = \binom n {k + 1}       | r = \frac {n!} {\left({k + 1}\right)! \left({n - k - 1}\right)!}       | c = Definition of Binomial Coefficient }} {{eqn | r = \frac {n - k} {n - k} \frac {n!} {\left({k + 1}\right)! \left({n - k - 1}\right)!}       | c =  }} {{eqn | r = \frac {n - k} {\left({k + 1}\right) \left({n - k}\right)} \frac {n!} {k! \left({n - k - 1}\right)!}       | c = extracting $k + 1$ from its factorial }} {{eqn | r = \frac {n - k} {k + 1} \frac {n!} {k! \left({n - k}\right)!}       | c = inserting $n - k$ into its factorial }} {{eqn | r = \frac {n - k} {k + 1} \binom n k       | c = Definition of Binomial Coefficient }} {{end-eqn}} In order for $S_n$ to be strictly increasing, it is necessary for $\dfrac {n - k} {k + 1} > 1$. So: {{begin-eqn}} {{eqn | l = \dfrac {n - k} {k + 1}       | o = >       | r = 1       | c =  }} {{eqn | ll= \iff       | l = n - k       | o = >       | r = k + 1       | c =  }} {{eqn | ll= \iff       | l = n       | o = >       | r = 2 k + 1       | c =  }} {{eqn | ll= \iff       | l = n       | o = >       | r = 2 \left({k + 1}\right) - 1       | c =  }} {{end-eqn}} Thus $\dbinom n {k + 1} > \dbinom n k$ {{iff}} $k + 1$ is less than half of $n$. Hence the result. {{Qed}} Category:Binomial Coefficients h1tjmwqkc19n3s0v7w6d6nvdds38hk9
=== $(1)$ implies $(2)$ === Let $\phi: M_1 \to M_2$ be an isometry by definition 1. Then by definition: :$\phi$ is a bijection such that: ::$\forall a, b \in A_1: \map {d_1} {a, b} = \map {d_2} {\map \phi a, \map \phi b}$ By Inverse of Bijection is Bijection, $\phi^{-1}$ is also a bijection. Thus $\forall a, b \in A_2$ there exists $\map {\phi^{-1} } a$ and $\map {\phi^{-1} } b$ in $A_1$. From Inverse of Inverse of Bijection: :$\forall a, b \in A_2: \map {d_2} {\map \phi {\map {\phi^{-1} } a}, \map \phi {\map {\phi^{-1} } b} } = \map {d_1} {\map {\phi^{-1} } a, \map {\phi^{-1} } b}$ and so: :$\forall a, b \in A_2: \map {d_2} {a, b} = \map {d_1} {\map {\phi^{-1} } a, \map {\phi^{-1} } b}$ Thus $M_1$ and $M_2$ are isometric by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $M_1$ and $M_2$ be isometric by definition 2. Then by definition: :there exist inverse mappings $\phi: A_1 \to A_2$ and $\phi^{-1}: A_2 \to A_1$ such that: ::$\forall a, b \in A_1: \map {d_1} {a, b} = \map {d_2} {\map \phi a, \map \phi b}$ :and: ::$\forall u, v \in A_2: \map {d_2} {u, v} = \map {d_1} {\map {\phi^{-1} } u, \map {\phi^{-1} } v}$ By Inverse of Bijection is Bijection, $\phi$ and $\phi^{-1}$ are both bijections. Thus $\phi: M_1 \to M_2$ is a isometry by definition 1. {{qed}}
Let $a \in A_1$. Let $\epsilon \in \R_{>0}$. Let $\delta = \epsilon$. Then: {{begin-eqn}} {{eqn | l = d_1 \left({a, y}\right)       | o = <       | r = \delta       | c = for some $y \in A_1$ }} {{eqn | ll= \implies       | l = d_2 \left({\phi \left({a}\right), \phi \left({y}\right)}\right)       | o = <       | r = \delta       | c = as $d_2 \left({\phi \left({a}\right), \phi \left({y}\right)}\right) = d_1 \left({a, y}\right)$ }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} So by definition $\phi$ is continuous at $a$. As $a \in H$ is arbitrary, it follows that $d_H$ is continuous on $H$. {{qed}}
=== Definition 1 implies Definition 2 === Let $d_1$ and $d_2$ be topologically equivalent by definition 1. Let $\struct {B, d}$ and $\struct {C, d'}$ be metric spaces: Let $f: B \to A$ and $g: A \to C$ be mappings. Then by definition: :$f$ is $\tuple {d, d_1}$-continuous {{iff}} $f$ is $\tuple {d, d_2}$-continuous :$g$ is $\tuple {d_1, d'}$-continuous {{iff}} $g$ is $\tuple {d_2, d'}$-continuous. Suppose $U \subseteq A$ is $d_2$-open. Let $h: A \to A$ be the identity mapping. That is: :$\forall a \in A: \map h a = a$ From Identity Mapping is Continuous, $h$ is $\tuple {d_2, d_2}$-continuous. Hence because $h$ is $\tuple {d_1, d'}$-continuous {{iff}} $h$ is $\tuple {d_2, d'}$-continuous, $h$ is $\tuple {d_1, d_2}$-continuous. From Metric Space Continuity by Open Set, it follows that $h^{-1} \sqbrk U$ is $d_1$-open. But $h^{-1} \sqbrk U = U$, so $U$ is $d_1$-open. A similar argument, still starting with the same proposition that $g$ is $\tuple {d_1, d'}$-continuous {{iff}} $g$ is $\tuple {d_2, d'}$-continuous, shows that if $U$ is $d_1$-open then it is $d_2$-open. Thus $d_1$ and $d_2$ are topologically equivalent by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $d_1$ and $d_2$ be topologically equivalent by definition 2. Then by definition: :$U \subseteq A$ is $d_1$-open {{iff}} $U \subseteq A$ is $d_2$-open. From Metric Space Continuity by Open Set, we have that: :$f$ is $\tuple {d, d_1}$-continuous {{iff}} for every set $U \subseteq A$ which is open in $\struct {A, d_1}$, $f^{-1} \sqbrk U$ is open in $\struct {B, d}$ :$f$ is $\tuple {d, d_2}$-continuous {{iff}} for every set $U \subseteq A$ which is open in $\struct {A, d_2}$, $f^{-1} \sqbrk U$ is open in $\struct {B, d}$. Hence $f$ is $\tuple {d, d_1}$-continuous {{iff}} $f$ is $\tuple {d, d_2}$-continuous. Similarly: :$g$ is $\tuple {d_1, d'}$-continuous {{iff}} for every set $U \subseteq C$ which is open in $\struct {C, d'}$, $g^{-1} \sqbrk U$ is open in $\struct {A, d_1}$ :$g$ is $\tuple {d_2, d'}$-continuous {{iff}} for every set $U \subseteq C$ which is open in $\struct {C, d'}$, $g^{-1} \sqbrk U$ is open in $\struct {A, d_2}$. Hence $g$ is $\tuple {d_1, d'}$-continuous {{iff}} $g$ is $\tuple {d_2, d'}$-continuous. Thus $d_1$ and $d_2$ are topologically equivalent by definition 1. {{qed}}
By the definition of an isometry, there exists a bijection $f: A_1 \to A_2$ such that: : $\forall a, b \in A_1: d_1 \left({a, b}\right) = d_2 \left({\phi \left({a}\right), \phi \left({b}\right)}\right)$ From Isometry of Metric Spaces is Homeomorphism, $f$ is a homeomorphism from $M_1$ to $M_2$. The result follows by definition of homeomorphic metric spaces. {{qed}}
Consider the spaces $\struct {\openint 0 4, d}$ and $\struct {\openint 0 1, d}$, where $d$ is the Euclidean Metric. By Open Real Intervals are Homeomorphic, they are homeomorphic. Let $\phi: \openint 0 4 \to \openint 0 1$ be a mapping. Then: {{begin-eqn}} {{eqn | l = \map d {\map \phi 1, \map \phi 3}       | o = \le       | r = \map d {0, 1} }} {{eqn | r = 1 }} {{eqn | o = <        | r = 2 }} {{eqn | r = \map d {1, 3} }} {{end-eqn}} showing that $\phi$ cannot be an isometry. Therefore the two spaces above are not isometric. {{qed}}
Let $\epsilon \in \R_{>0}$. Let $a \in A$. Set $\delta = \dfrac \epsilon K$. Then: {{begin-eqn}} {{eqn | l = \map {d_1} {x, a}       | o = <       | r = \delta       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_2} {\map {I_A} x, \map {I_A} a}       | o = \le       | r = K \map {d_1} {x, a}       | c =  }} {{eqn | o = <       | r = K \delta       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Hence by definition $I_A$ is continuous at $a$. As $a$ is arbitrary, it follows that this is true for all $a \in A$. Thus $I_A$ is continuous on the whole of $M_1$. {{qed}}
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$. === Definition 1 implies Definition 2 === Let $d_1$ and $d_2$ be Lipschitz equivalent by definition 1: Then by definition: : $\exists h, k \in \R_{>0} \forall x, y \in A: h \map {d_2} {x, y} \le \map {d_1} {x, y} \le k \map {d_2} {x, y}$ Let $K_1 = \dfrac 1 h, K_2 = k$. {{begin-eqn}} {{eqn | lo= \exists h \in \R_{>0}: \forall x, y \in A:       | l = h \map {d_2} {x, y}       | o = \le       | r = \map {d_1} {x, y}       | c = {{Defof|Lipschitz Equivalence|subdef = Metrics|index = 1|Lipschitz Equivalent Metrics}} }} {{eqn | ll= \leadsto       | l = \map {d_2} {x, y}       | o = \le       | r = \dfrac 1 h \map {d_1} {x, y}       | c =  }} {{eqn | r = K_1 \map {d_1} {x, y}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | lo= \exists k \in \R_{>0}: \forall x, y \in A:       | l = \map {d_1} {x, y}       | o = \le       | r = k \map {d_2} {x, y}       | c = {{Defof|Lipschitz Equivalence|subdef = Metrics|index = 1|Lipschitz Equivalent Metrics}} }} {{eqn | r = K_2 \map {d_2} {x, y}       | c =  }} {{end-eqn}} Thus $d_1$ and $d_2$ are Lipschitz equivalent by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $d_1$ and $d_2$ be Lipschitz equivalent by definition 2. Then by definition: : $(1): \quad \exists K_1 \in \R_{>0}: \forall x, y \in A: \map {d_2} {x, y} \le K_1 \map {d_1} {x, y}$ : $(2): \quad \exists K_2 \in \R_{>0}: \forall x, y \in A: \map {d_1} {x, y} \le K_2 \map {d_2} {x, y}$ Let $h = \dfrac 1 {K_1}, k = K_2$. {{begin-eqn}} {{eqn | lo= \exists K_1 \in \R_{>0}: \forall x, y \in A:       | l = \map {d_2} {x, y}       | o = \le       | r = K_1 \map {d_1} {x, y}       | c = {{Defof|Lipschitz Equivalence|subdef = Metrics|index = 2|Lipschitz Equivalent Metrics}} }} {{eqn | ll= \leadsto       | l = \dfrac 1 {K_1} \map {d_2} {x, y}       | o = \le       | r = \map {d_1} {x, y}       | c =  }} {{eqn | r = h \map {d_1} {x, y}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | lo= \exists K_2 \in \R_{>0}: \forall x, y \in A:       | l = \map {d_1} {x, y}       | o = \le       | r = K_2 \map {d_2} {x, y}       | c = {{Defof|Lipschitz Equivalence|subdef = Metrics|index = 2|Lipschitz Equivalent Metrics}} }} {{eqn | r = k \map {d_2} {x, y}       | c =  }} {{end-eqn}} Thus $d_1$ and $d_2$ are Lipschitz equivalent by definition 1. {{qed}} Category:Lipschitz Equivalence q32z449zwrthkf9f1mswdws8g6a7i6k
By definition, $d_1$ and $d_2$ are topologically equivalent on $A$ {{iff}}: :For all metric spaces $\left({B, d}\right)$ and $\left({C, d'}\right)$: :For all mappings $f: B \to A$ and $g: A \to C$: :: $(1): \quad f$ is $\left({d, d_1}\right)$-continuous {{iff}} $f$ is $\left({d, d_2}\right)$-continuous :: $(2): \quad g$ is $\left({d_1, d'}\right)$-continuous {{iff}} $g$ is $\left({d_2, d'}\right)$-continuous. Set $B = C = A$. The result follows by definition of homeomorphic metric spaces. {{qed}} Category:Topologically Equivalent Metrics Category:Homeomorphic Metric Spaces k46uy5cad2030usp0vcgec9fka4wwox
From Relation between $p$-Product Metric and Chebyshev Distance on Real Vector Space: :$\forall x, y \in \R^n: d_\infty \left({x, y}\right) \le d_p \left({x, y}\right) \le n^{1/p} d_\infty \left({x, y}\right)$ The Euclidean metric $d_2$ is a special case of the $p$-product metric $d_p$ for $p = 2$. It follows by definition that $d_\infty$ and $d_2$ are Lipschitz equivalent. It follows from Lipschitz Equivalent Metrics are Topologically Equivalent that $d_\infty$ and $d_2$ are topologically equivalent. The result follows from Metric Spaces on Topologically Equivalent Metrics on same Underlying Set are Homeomorphic. {{qed}}
In order to prove the assertion it is sufficient to prove that the conditions for homeomorphism in definitions $2$ to $4$ are necessary and sufficient conditions for $f$ and $f^{-1}$ to be continuous on $M_1$ and $M_2$ respectively. === Definition 1 implies Definition 2 === Let $f$ and $f^{-1}$ both be continuous by definition $2$. Let $U \subseteq A_1$ be open in $M_1$. As $f^{-1}$ is continuous, $\left({f^{-1}}\right)^{-1} \left[{U}\right] = f \left[{U}\right]$ is open in $M_2$. That is, $f \left[{U}\right] = f \left[{U}\right]$ is open in $M_2$. Let $f \left[{U}\right] \subseteq A_2$ be open in $M_2$. Then as $f$ is continuous, $f^{-1} \left[{f \left[{U}\right]}\right] = U$ is open in $M_1$. Thus: : for all $U \subseteq A_1$, $U$ is an open set of $M_1$ {{iff}} $f \left[{U}\right]$ is an open set of $M_2$. {{qed|lemma}} === Definition 2 implies Definition 4 === Let definition $2$ hold. Let $a \in A_1$. Let $N \subseteq A_1$. Then: :$N$ is a neighborhood of $a$ {{iff}}: :$N$ contains an open set $U$ containing $a$ {{iff}}: :$f \left[{N}\right]$ contains an open set $U' = f \left[{U}\right]$ containing $f \left({a}\right)$ {{iff}}: :$f \left[{N}\right]$ is a neighborhood $f \left({a}\right)$. {{qed|lemma}} === Definition 4 implies Definition 1 === Let definition $4$ hold. Let $a \in A_1$. Let $U \subseteq A_2$ be a neighborhood of $f \left({a}\right)$. Then $f \left[{f^{-1} \left[{U}\right]}\right]$ is a neighborhood $f \left({a}\right)$. Hence $f^{-1} \left[{U}\right]$ is a neighborhood of $a$. Thus $f$ is continuous. Similarly, let $b \in A_2$. Let $V \subseteq A_1$ be a neighborhood of $f^{-1} \left({b}\right)$. Then $f^{-1} \left[{f \left[{V}\right]}\right]$ is a neighborhood $f^{-1} \left({b}\right)$. Hence $f \left[{V}\right]$ is a neighborhood of $f \left({f^{-1} \left({b}\right)}\right) = b$. Thus $f^{-1}$ is continuous. {{qed|lemma}} === Definition 2 iff Definition 3 === This is demonstrated in Continuity of Mapping between Metric Spaces by Closed Sets. {{qed}}
Let $x$ be a Liouville number.  Aiming for a contradiction, suppose $x$ were rational, that is: : $x = \dfrac a b$ with $a, b \in \Z$ and $b > 0$. By definition of a Liouville number, for all $n \in \N$, there exist $p,q \in \Z$ (which may depend on $n$) with $q > 1$ such that: :$0 < \left \lvert x - \dfrac p q \right \rvert < \dfrac 1 {q^n}$ Let $n$ be a positive integer such that $2^{n - 1} > b$. Let $p$ and $q$ be any integers with $q > 1$. We have: :$\left \lvert x - \dfrac p q \right \rvert = \dfrac {\left \lvert a q - b p \right \rvert} {b q}$ If $\left \lvert a q - b p \right \rvert = 0$, this would violate the first inequality. If $\left \lvert a q - b p \right \rvert \ne 0$, then: {{begin-eqn}} {{eqn |l = \left \lvert x - \frac p q \right \rvert       |r = \frac 1 {b q}       |c = as $\left \lvert a q - b p \right \rvert$ is a positive integer       |o = \ge }} {{eqn |r = \frac 1 {2^{n - 1} q}       |c = by our choice of $n$       |o = > }} {{eqn |r = \frac 1 {q^n}       |c = as $q > 1$ by definition       |o = \ge }} {{end-eqn}} which would violate the second inequality. Therefore, in any case, if $n$ is sufficiently large, there cannot exist integers $p$ and $q$ with $q > 1$ satisfying the two inequalities. This is a contradiction. Thus $x$ is irrational. {{qed}} Category:Number Theory 6szpys8wrdmc3pufjkt99dsohygqay5
Let $q = 10^{n!}$ and write: :$\displaystyle L = \frac p q + \sum_{k \mathop = n + 1}^\infty \frac 1 {10^{k!} }$ for some suitable $p \in \Z$. Then: {{begin-eqn}} {{eqn | l = \size {L - \frac p q}       | r = \sum_{k \mathop = n + 1}^\infty \frac 1 {10^{k!} }       | c =   }} {{eqn | r = \frac 2 {10^{\paren {n + 1}!} }       | o = \le       | c =  }} {{eqn | r = \frac 2 {q^{n + 1} }       | c =  }} {{eqn | r = \frac 1 {q^n}       | o = <       | c = as $q \ge 10$ for all $n \in \N$ }} {{end-eqn}} Thus, by definition, $L$ is a Liouville number. Therefore, by Liouville's Theorem, $L$ is transcendental. {{qed}}