hcju/proofwikiqa huggingface dataset:
First to establish uniqueness of $f$. Suppose that $f, g: P \to P'$ both satisfy the conditions. Define $A \subseteq P$ as: :$A := \left\{{n \in P: f \left({n}\right) = g \left({n}\right)}\right\}$ Then the first condition implies that $0 \in A$. Now suppose that $n \in A$. Then: {{begin-eqn}} {{eqn|l = f \left({s \left({n}\right)}\right)      |r = s' \left({f \left({n}\right)}\right)      |c = Assumption on $f$ }} {{eqn|r = s' \left({g \left({n}\right)}\right)      |c = By hypothesis }} {{eqn|r = g \left({s \left({n}\right)}\right)      |c = Assumption on $g$ }} {{end-eqn}} Hence $s \left({n}\right) \in A$. Since $P$ is a Peano structure, it follows that $A = P$. Hence by Equality of Mappings, $f = g$. For the existence of such an $f$, we apply the Principle of Recursive Definition with $0' \in P'$ and $s': P' \to P'$. Lastly, it is to be established that the $f$ so obtained is a bijection. Let $A' \subseteq P$ be defined as: :$A' := \left\{{n \in P: \forall m \in P: f \left({n}\right) = f \left({m}\right) \implies n = m}\right\}$ It follows from Axiom $(P4)$ that $f \left({n}\right) = 0'$ implies $n = 0$. Hence $0 \in A'$. Suppose now that $n \in A'$, but $s \left({n}\right) \notin A'$. Then, since $f \left({s \left({n}\right)}\right) \ne 0'$, there exists some $m \in P, m \ne n$ such that: :$f \left({s \left({n}\right)}\right) = f \left({s \left({m}\right)}\right)$ That is, such that: :$s' \left({f \left({n}\right)}\right) = s' \left({f \left({m}\right)}\right)$ But by Axiom $(P3)$, $s'$ is injective. Hence $f \left({n}\right) = f \left({m}\right)$, contradicting the assumption that $n \in A'$. Therefore $s \left({n}\right) \in A'$, so that $A' = P$ by Axiom $(P5)$. That is, $f$ is an injection. Next, define $A'' \subseteq P'$ as: :$A'' := \left\{{n' \in P': \exists n \in P: f \left({n}\right) = n'}\right\}$ Since $f \left({0}\right) = 0'$, we have $0' \in A''$. Suppose now that $n' \in A''$, and let $f \left({n}\right) = n'$. Then: {{begin-eqn}} {{eqn|l = f \left({s \left({n}\right)}\right)      |r = s' \left({f \left({n}\right)}\right)      |c = Definition of $f$ }} {{eqn|r = s' \left({n'}\right)      |c = Definition of $n$ }} {{end-eqn}} meaning $s' \left({n'}\right) \in A''$. Hence by Axiom $(P5)$, $A'' = P'$. That is, $f$ is surjective. It follows that $f$ is the sought bijection. {{qed}} Category:Abstract Algebra 19j09orgblsnntox1mcxhjb7duzqpc5
The $p$-adic metric on $\Q$ is defined as the metric induced by the $p$-adic norm on $\Q$. It follows from Metric Induced by Norm is Metric that $d_p$ is a metric. {{qed}} Category:P-adic Metrics fkm4lqmfi69cxlte36doik06uiubwpw
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$. Let $\epsilon \in \R_{>0}$. Let $x \in A_1$. Pick any $\delta \in \R_{>0}$. Let $y \in A_1$ such that $d_1 \left({x, y}\right) < \delta$. Now we have: :$f_c \left({x}\right) = c = f_c \left({y}\right)$ Hence: :$d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) = d_2 \left({c, c}\right)$ By definition of a metric: :$d_2 \left({c, c}\right) = 0 < \epsilon$ Thus we have that: :$\forall \epsilon > 0: \forall x \in A_1: \exists \delta \in \R: d_1 \left({x, y}\right) < \delta \implies d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) < \epsilon$ Hence by definition of continuity on a metric space, $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$. {{qed}}
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$. From Constant Function is Uniformly Continuous, $f_c$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$. The result follows from Uniformly Continuous Function is Continuous. {{qed}}
Follows directly from: :Constant Function is Uniformly Continuous: Real Function :Uniformly Continuous Function is Continuous: Real Function. {{qed}}
Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \R^n$. Let $\epsilon \in \R_{>0}$. Let $\delta = \dfrac {\epsilon} {\sqrt n}$. Let $x = \tuple {x_1, x_2, \ldots, x_n}$ be such that $\map {d_\infty} {x, a} < \delta$. That is: :$\ds \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} } < \delta$ Then: {{begin-eqn}} {{eqn | l = \map {d_2} {x, a}       | r = \sqrt {\sum_{i \mathop = 1}^n \paren {a_i - x_i} }       | c =  }} {{eqn | o = \le       | r = \sqrt {n \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} } }       | c =  }} {{eqn | o = <       | r = \sqrt {n \delta^2}       | c =  }} {{eqn | r = \sqrt {\epsilon^2}       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} The result follows by definition of $\tuple {d_\infty, d_2}$-continuity. {{qed}}
Let $a = \left({a_1, a_2, \ldots, a_n}\right) \in \R^n$. Let $\epsilon \in \R_{>0}$. Let $\delta = \epsilon$. Let $x = \left({x_1, x_2, \ldots, x_n}\right)$ be such that $d_2 \left({x, a}\right) < \delta$. That is: :$\displaystyle \sqrt{\sum_{i \mathop = i}^n \left({a_i - x_i}\right)} < \delta$ Then: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = i}^n \left({a_i - x_i}\right)       | o = <       | r = \delta^2       | c =  }} {{eqn | ll= \implies       | lo= \forall i \in \left\{ {1, 2, \ldots, n}\right\}:       | l = \left({a_i - x_i}\right)^2       | o = <       | r = \delta^2       | c =  }} {{eqn | ll= \implies       | lo= \forall i \in \left\{ {1, 2, \ldots, n}\right\}:       | l = \left\vert{a_i - x_i}\right\vert       | o = <       | r = \delta       | c =  }} {{eqn | ll= \implies       | l = \max_{i \mathop \le i \mathop \le n} \left\{ {\left\vert{a_i - x_i}\right\vert}\right\}       | o = <       | r = \delta       | c =  }} {{eqn | ll= \implies       | l = d_\infty \left({x, a}\right)       | o = <       | r = \delta       | c =  }} {{end-eqn}} The result follows by definition of $\left({d_2, d_\infty}\right)$-continuity. {{qed}}
From Metric Induces Topology, the metric spaces described are topological spaces. The result follows from Composite of Continuous Mappings is Continuous. {{qed}}
Let $f$ and $g$ be continuous mappings. By definition: :$f$ is continuous at $a \in X_1$ for all $a \in X_1$ :$g$ is continuous at $\map f a \in X_2$ for all $\map f a \in X_2$. The result follows from Composite of Continuous Mappings at Point between Metric Spaces is Continuous at Point {{qed}}
Let $\epsilon \in \R_{>0}$. The strategy is to find a $\delta \in \R_{>0}$ such that: : $d_1 \left({x, a}\right) < \delta \implies d_3 \left({g \left({f \left({x}\right)}\right), g \left({f \left({a}\right)}\right)}\right) < \epsilon$ As $g$ is continuous at $f \left({a}\right)$: :$\exists \eta \in \R_{>0}: \forall y \in X_2: d_2 \left({y, f \left({a}\right)}\right) < \eta \implies d_3 \left({g \left({y}\right), g \left({f \left({a}\right)}\right)}\right) < \epsilon$ As $f$ is continuous at $a$: :$\forall \eta \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in X_1: d_1 \left({x, a}\right) < \delta \implies d_2 \left({f \left({x}\right), f \left({a}\right)}\right) < \eta$ Hence: :$d_3 \left({g \left({f \left({x}\right)}\right), g \left({f \left({a}\right)}\right)}\right) < \epsilon$ {{qed}}
Let $\Z_{>0}$ be the set of (strictly) positive integers. Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$. Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the metric on $\Z_{>0}$ defined as: :$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$ Let $\tau_d$ denote the metric topology for $d$. Let $\tau_\delta$ denote the metric topology for $\delta$. From Topologies induced by Usual Metric and Scaled Euclidean Metric on Positive Integers are Homeomorphic: :$\tau_d$ and $\tau_\delta$ are homeomorphic. From Positive Integers under Usual Metric is Complete Metric Space: :$\struct {\Z_{>0}, d}$ is a complete metric space. However, from Positive Integers under Scaled Euclidean Metric is not Complete Metric Space: :$\struct {\Z_{>0}, \delta}$ is not a complete metric space. Hence the result. {{qed}}
The $L^1$ metric on $X$ is defined as: :$\displaystyle \forall f, g \in S: d_1 \left({f, g}\right) := \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t$ Let $\epsilon \in \R_{>0}$. Let $f \in X$. Let $\delta = \epsilon$. Then: {{begin-eqn}} {{eqn | lo= \forall g \in X:       | l = d_1 \left({f, g}\right)       | o = <       | r = \delta       | c =  }} {{eqn | ll= \implies       | l = \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t       | o = <       | r = \delta       | c = Definition of $L^1$ Metric }} {{eqn | ll= \implies       | l = \left\vert{\int_a^b \left({f \left({t}\right) - g \left({t}\right)}\right) \ \mathrm d t}\right\vert       | o = <       | r = \delta       | c = Absolute Value of Definite Integral }} {{eqn | ll= \implies       | l = \left\vert{\int_a^b f \left({t}\right) \ \mathrm d t - \int_a^b g \left({t}\right) \ \mathrm d t}\right\vert       | o = <       | r = \delta       | c = Linear Combination of Integrals }} {{eqn | ll= \implies       | l = \left\vert{I \left({f}\right) - I \left({g}\right)}\right\vert       | o = <       | r = \delta       | c = Definition of $I$ }} {{eqn | ll= \implies       | l = d \left({I \left({f}\right), I \left({g}\right)}\right)       | o = <       | r = \delta       | c = Definition of $d$ }} {{eqn | ll= \implies       | l = d \left({I \left({f}\right), I \left({g}\right)}\right)       | o = <       | r = \epsilon       | c = Definition of $\delta$ }} {{end-eqn}} Thus it has been demonstrated that: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall g \in X: d_1 \left({f, g}\right) < \delta \implies d \left({I \left({f}\right), I \left({g}\right)}\right) < \epsilon$ Hence by definition of continuity at a point, $I$ is continuous at $f$. As $f$ is chosen arbitrarily, it follows that $I$ is continuous for all $f \in X$. The result follows by definition of continuous mapping. {{qed}}
Let $\epsilon \in \R_{>0}$. Let $x \in \mathcal A$. Let $k \in \left\{{1, 2, \ldots, n}\right\}$. Then as $f_k$ is continuous: :$(1): \quad \exists \delta_k \in \R_{>0}: \forall y_k \in A_k: d_k \left({x_k, y_k}\right) < \delta_k \implies d' \left({f_k \left({x_k}\right), f_k \left({y_k}\right)}\right) < \epsilon$ Let $\delta = \max \left\{ {\delta_k: k \in  \left\{{1, 2, \ldots, n}\right\} }\right\}$. Then: {{begin-eqn}} {{eqn | lo= \forall y \in \mathcal A:       | l = d_\infty \left({x, y}\right)       | o = <       | r = \delta       | c =  }} {{eqn | ll= \implies       | l = \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}       | o = <       | r = \delta       | c = Definition of Chebyshev Distance }} {{eqn | ll= \implies       | lo= \forall k \in \left\{ {1, 2, \ldots, n}\right\}:       | l = d_k \left({x_k, y_k}\right)       | o = <       | r = \delta       | c =  }} {{eqn | ll= \implies       | l = d'_k \left({f_k \left({x_k}\right), f_k \left({y_k}\right)}\right)       | o = <       | r = \epsilon       | c = from $(1)$ }} {{eqn | ll= \implies       | l = \max_{i \mathop = 1}^n \left\{ {d'_i \left({f_i \left({x_i}\right), f_i \left({y_i}\right)}\right)}\right\}       | o = <       | r = \epsilon       | c =  }} {{eqn | ll= \implies       | l = d_\infty \left({F \left({x}\right), F \left({y}\right)}\right)       | o = <       | r = \epsilon       | c = Definition of Chebyshev Distance }} {{end-eqn}} Thus it has been demonstrated that: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in X: d_\infty \left({x, y}\right) < \delta \implies d_\infty \left({F \left({x}\right), F \left({y}\right)}\right) < \epsilon$ Hence by definition of continuity at a point, $F$ is continuous at $x$. As $x$ is chosen arbitrarily, it follows that $F$ is continuous for all $x \in X$. The result follows by definition of continuous mapping. {{qed}}
First we note that: {{begin-eqn}} {{eqn | l = \left\vert{\left({x_1 + x_2}\right) - \left({y_1 + y_2}\right)}\right\vert       | r = \left\vert{\left({x_1 - y_1}\right) + \left({x_2 - y_2}\right)}\right\vert       | c =  }} {{eqn | o = \le       | r = \left\vert{x_1 - y_1}\right\vert + \left\vert{x_2 - y_2}\right\vert       | c = Triangle Inequality for Real Numbers }} {{eqn | n = 1       | o = \le       | r = 2 \max \left\{ {\left\vert{x_1 - y_1}\right\vert, \left\vert{x_2 - y_2}\right\vert}\right\}       | c =  }} {{end-eqn}} Let $\epsilon \in \R_{>0}$. Let $x = \left({x_1, x_2}\right) \in \R^2$. Let $\delta = \dfrac \epsilon 2$. Then: {{begin-eqn}} {{eqn | lo= \forall y =  \left({y_1, y_2}\right) \in \R^2:       | l = d_\infty \left({x, y}\right)       | o = <       | r = \delta       | c =  }} {{eqn | ll= \implies       | l = \max \left\{ {\left\vert{x_1 - y_1}\right\vert, \left\vert{x_2 - y_2}\right\vert}\right\}       | o = <       | r = \delta       | c = Definition of Chebyshev Distance }} {{eqn | ll= \implies       | l = 2 \max \left\{ {\left\vert{x_1 - y_1}\right\vert, \left\vert{x_2 - y_2}\right\vert}\right\}       | o = <       | r = \epsilon       | c = Definition of $\epsilon$ }} {{eqn | ll= \implies       | l = \left\vert{\left({x_1 + x_2}\right) - \left({y_1 + y_2}\right)}\right\vert       | o = <       | r = \epsilon       | c = from $(1)$ }} {{eqn | ll= \implies       | l = \left\vert{f \left({x}\right) -  f \left({y}\right)}\right\vert       | o = <       | r = \epsilon       | c = Definition of $f$ }} {{eqn | ll= \implies       | l = d \left({f \left({x}\right), f \left({y}\right)}\right)       | o = <       | r = \epsilon       | c = Definition of Usual Metric on $\R$ }} {{end-eqn}} Thus it has been demonstrated that: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in \R^2: d_\infty \left({x, y}\right) < \delta \implies d \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$ Hence by definition of continuity at a point, $f$ is continuous at $x$. As $x$ was chosen arbitrarily, it follows that $f$ is continuous for all $x \in \R^2$. The result follows by definition of continuous mapping. {{qed}}
First we note that: {{begin-eqn}} {{eqn | o =        | r = \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }       | c =  }} {{eqn | r = \size {\paren {x_1 - y_1} + \paren {x_2 - y_2} }       | c =  }} {{eqn | o = \le       | r = \size {x_1 - y_1} + \size {x_2 - y_2}       | c = Triangle Inequality for Real Numbers }} {{eqn | n = 1       | o = \le       | r = \sqrt 2 \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}       | c = $p$-Product Metrics on Real Vector Space are Topologically Equivalent }} {{end-eqn}} Let $\epsilon \in \R_{>0}$. Let $x = \tuple {x_1, x_2} \in \R^2$. Let $\delta = \dfrac \epsilon {\sqrt 2}$. Then: {{begin-eqn}} {{eqn | lo= \forall y = \tuple {y_1, y_2} \in \R^2:       | l = \map {d_2} {x, y}       | o = <       | r = \delta       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}       | o = <       | r = \delta       | c = {{Defof|Euclidean Metric on Real Number Plane}} }} {{eqn | ll= \leadsto       | l = \sqrt 2 \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}       | o = <       | r = \delta \sqrt 2       | c =  }} {{eqn | ll= \leadsto       | l = \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }t       | o = <       | r = \epsilon       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = \size {\map f x - \map f y}       | o = <       | r = \epsilon       | c = Definition of $f$ }} {{eqn | ll= \leadsto       | l = \map d {\map f x, \map f y}       | o = <       | r = \epsilon       | c = {{Defof|Euclidean Metric on Real Number Line}} }} {{end-eqn}} Thus it has been demonstrated that: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in \R^2: \map {d_2} {x, y} < \delta \implies \map d {\map f x, \map f y} < \epsilon$ Hence by definition of continuity at a point, $f$ is continuous at $x$. As $x$ is chosen arbitrarily, it follows that $f$ is continuous for all $x \in \R^2$. The result follows by definition of continuous mapping. {{qed}}
By definition, $f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ iff: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \left[{B_\delta \left({a; d_1}\right)}\right] \subseteq B_\epsilon \left({f \left({a}\right); d_2}\right)$ For a mapping $f: X \to Y$ we have: :$f \left[{U}\right] \subseteq V \iff U \subseteq f^{-1} \left[{V}\right]$ where $U \subseteq X$ and $V \subseteq Y$. Hence the result. {{qed}}
From Open Ball of Point Inside Open Ball: : $\exists \delta \in \R: B_\delta \left({x}\right) \subseteq B_\epsilon \left({a}\right)$ Thus by definition $B_\delta \left({x}\right)$ is a neighborhoods of $x$ in $M$. {{qed}}
=== $\epsilon$-Ball Definition implies Definition by Neighborhood === Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$ where $\map {B_\epsilon} {\map f a; d_2}$ denotes the open $\epsilon$-ball of $\map f a$ with respect to the metric $d_2$, and similarly for $\map {B_\delta} {a; d_1}$. Let $N'$ be a neighborhood of $\map f a$ in $M_2$. Then : {{begin-eqn}} {{eqn | lo= \exists \epsilon \in \R_{>0}:       | l = \map {B_\epsilon} {\map f a; d_2}       | o = \subseteq       | r = N'       | c = {{Defof|Neighborhood (Metric Space)}} }} {{eqn | ll= \leadsto       | lo= \exists \delta \in \R_{>0}:       | l = f \sqbrk {\map {B_\delta} {a; d_1} }       | o = \subseteq       | r = \map {B_\epsilon} {\map f a; d_2}       | c = as $f$ is $\tuple {d_1, d_2}$-continuous at $a$ }} {{end-eqn}} But by Open Ball is Neighborhood of all Points Inside, $\map {B_\delta} {a; d_1}$ is a neighborhood of $a$ in $M_1$. Thus letting $N := \map {B_\delta} {a; d_1}$: :$f \sqbrk N \subseteq \map {B_\epsilon} {\map f a; d_2} \subseteq N'$ and so the condition for $f$ to be $\tuple {d_1, d_2}$-continuous at $a$ by Neighborhood is fulfilled. {{qed|lemma}} === Definition by Neighborhood implies $\epsilon$-Ball Definition === Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that: :for each neighborhood $N'$ of $\map f a$ in $M_2$ there exists a corresponding neighborhood $N$ of $a$ in $M_1$ such that $f \sqbrk N \subseteq N'$. Let $\epsilon \in \R_{>0}$. Consider the open $\epsilon$-ball $\map {B_\epsilon} {\map f a; d_2}$ of $\map f a$. By Open Ball is Neighborhood of all Points Inside, $\map {B_\epsilon} {\map f a; d_2}$ is a neighborhood of $\map f a$ in $M_2$. Let $N' := \map {B_\epsilon} {\map f a; d_2}$. By definition of $\tuple {d_1, d_2}$-continuity at $a$, there exists a neighborhood $N$ of $a$ in $M_1$ such that: :$f \sqbrk N \subseteq N'$ By definition of neighborhood: :$\exists \delta \in \R_{>0}: \map {B_\delta} {a; d_1} \subseteq N$. Therefore: :$\map f {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$ and so the condition for $f$ to be $\tuple {d_1, d_2}$-continuous at $a$ by open $\epsilon$-ball is fulfilled. {{qed}}
=== $(1)$ implies $(2)$ === Let $G \subseteq X$ be open in $\left({X, d}\right)$. Let $x \in G$. Let $\left\langle{x_n}\right\rangle$ be a sequence in $X$ such that $x_n \to x$. By definition of open set, there exists $\epsilon > 0$ such that: : $B_\epsilon \left({x}\right) \subseteq G$ where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$. Since $x_n \to x$, there exists $n_0 \in \N$ such that: : $\forall n \ge n_0: d \left({x_n, x}\right) < \epsilon$ Thus: : $\forall n \ge n_0: x_n \in B_\epsilon \left({x}\right) \subseteq G$ {{qed|lemma}} === $(2)$ implies $(1)$ === Let $G \subseteq X$. Let: :$\forall x \in G: \forall \left\langle{x_n}\right\rangle \in X: x_n \to x: \exists n_0 \in \N: \forall n \ge n_0: \left\langle{x_n}\right\rangle \in G$ Aiming for a contradiction, suppose $G$ is ''not'' open in $\left({X, d}\right)$. Then: : $\exists x \in G: \forall \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \nsubseteq G$ Therefore, for $n = 1, 2, \ldots$ we can inductively find a sequence: : $x_n \in B_{1/n} \left({x}\right) \cap \left({X \setminus G}\right) \implies x_n \notin G$ and: : $\forall n \in \N: d \left({x_n, x}\right) < \dfrac 1 n \implies x_n \to x \in G$ and $\forall n \in \N: x_n \notin G$ This contradicts hypothesis $(2)$. Thus, by Proof by Contradiction, $G$ is open in $\left({X, d}\right)$. {{qed}}
By definition, $f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ {{iff}}: :for each neighborhood $N$ of $\map f a$ in $M_2$ there exists a corresponding neighborhood $N'$ of $a$ in $M_1$ such that $f \sqbrk {N'} \subseteq N$. For a mapping $f: X \to Y$ we have: :$f \sqbrk U \subseteq V \iff U \subseteq f^{-1} \sqbrk V$ where $U \subseteq X$ and $V \subseteq Y$. Hence the result. {{qed}}
Let $a \in A$. Then $A$ is a neighborhood of $a$ in $M$. {{qed}}
By definition of neighborhood: : $\exists \epsilon \in \R_{>0}: B_\epsilon \left({a}\right) \subseteq N$ where $B_\epsilon \left({a}\right)$ is the open $\epsilon$-ball of $a$ in $M$. By Subset Relation is Transitive: :$B_\epsilon \left({a}\right) \subseteq N'$ The result follows by definition of neighborhood of $a$. {{qed}}
By definition of neighborhood: :$\exists \epsilon_1 \in \R_{>0}: \map {B_{\epsilon_1} } a \subseteq N$ where $\map {B_{\epsilon_1} } a$ is the open $\epsilon_1$-ball of $a$ in $M$. :$\exists \epsilon_2 \in \R_{>0}: \map {B_{\epsilon_2} } a \subseteq N'$ where $\map {B_{\epsilon_2} } a$ is the open $\epsilon_2$-ball of $a$ in $M$. Thus by definition of intersection: :$\map {B_\epsilon} a \subseteq N \cap N'$ where $\epsilon = \min \set {\epsilon_1, \epsilon_2}$ The result follows by definition of neighborhood of $a$. {{qed}}
By definition of neighborhood: : $\exists \epsilon \in \R_{>0}: B_\epsilon \left({a}\right) \subseteq N$ By Open Ball is Neighborhood of all Points Inside, $N' = B_\epsilon \left({a}\right)$ fulfils the conditions of the statement. {{qed}}
Let $N$ be a neighborhood of $a$ in $M$. Then by definition: :$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subseteq N$ where $\map {B_\epsilon} a$ is the open $\epsilon$-ball at $a$. The result follows from Open Ball in Real Number Line is Open Interval. {{qed}}
Let $S = \map {B_\epsilon} x$ be an open $\epsilon$-ball at $x$. Let $y \in \map {B_\epsilon} x$. Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | m = \map {B_\epsilon} x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \map d {y, x}       | o = <       | m = \epsilon       | c = {{Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}} }} {{eqn | ll= \leadstoandfrom       | l = \size {y - x}       | o = <       | m = \epsilon       | c = {{Defof|Euclidean Metric on Real Number Line}} }} {{eqn | ll= \leadstoandfrom       | l = -\epsilon       | o = <       | m = y - x       | mo= <       | r = \epsilon       | c = {{Defof|Absolute Value}} }} {{eqn | ll= \leadstoandfrom       | l = x - \epsilon       | o = <       | m = y       | mo= <       | r = x + \epsilon       | c =  }} {{eqn | ll= \leadstoandfrom       | l = y       | o = \in       | m = \openint {x - \epsilon} {x + \epsilon}       | c = {{Defof|Open Real Interval}} }} {{end-eqn}} As the implications go both ways: :$\map {B_\epsilon} x \subseteq \openint {x - \epsilon} {x + \epsilon}$ and :$\map {B_\epsilon} x \supseteq \openint {x - \epsilon} {x + \epsilon}$ By definition of set equality: :$\map {B_\epsilon} x = \openint {x - \epsilon} {x + \epsilon}$ Hence the result. {{qed}}
Let: {{begin-eqn}} {{eqn | l = \alpha       | r = \frac {a + b} 2 }} {{eqn | l = \epsilon       | r = \frac {b - a} 2 }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = a       | r = \alpha - \epsilon }} {{eqn | l = b       | r = \alpha + \epsilon }} {{end-eqn}} Thus: :$\openint a b = \openint {\alpha - \epsilon} {\alpha + \epsilon}$ From Open Ball in Real Number Line is Open Interval: :$\openint {\alpha - \epsilon} {\alpha + \epsilon} = \map {B_\epsilon} \alpha$ where $\map {B_\epsilon} \alpha$ is the open $\epsilon$-ball of $\alpha$ in $\R$. Hence the result {{qed}}