Proof of Subsets of a Metric Space: the Restricted Metric, Sequences, Limits in Closed Subsets and Completeness of Closed Subsets

Each result cited is universally quantified over the data in its own statement.

Clause restriction. As $A\times A\subseteq X\times X$, $d_{A}$ is a map $A\times A\to\mathbb{R}$ (Domains and Restrictions of Maps §restriction-map). Let $a,b,c\in A$; then $a,b,c\in X$. The values of $d_{A}$ at the pairs of these points are those of $d$ (Domains and Restrictions of Maps §restriction-value). Hence the conditions Metric Space §nonnegative, Metric Space §zero, Metric Space §symmetric and Metric Space §triangle for $d$ at $a,b,c$ are the same conditions for $d_{A}$: $0\le d_{A}(a,b)$; $d_{A}(a,b)=0$ if and only if $a=b$; $d_{A}(a,b)=d_{A}(b,a)$; and $d_{A}(a,c)\le d_{A}(a,b)+d_{A}(b,c)$. So $d_{A}$ is a metric on $A$ by Metric Space §metric, and $(A,d_{A})$ is a metric space by Metric Space §metric-space.

Clause sequences. By Sequences §sequence, $(a_{k})$ is a map $\mathbb{N}\to A$. Its range lies in $A$, hence in $X$, so it is a map $\mathbb{N}\to X$ by Functions, Values of a Function, and Functions from One Class to Another §map, that is, a sequence in $X$. Let $a\in A$. For all $m,\ell\in\mathbb{N}$ we have $a_{m},a_{\ell}\in A$, so $d_{A}(a_{m},a)=d(a_{m},a)$ and $d_{A}(a_{m},a_{\ell})=d(a_{m},a_{\ell})$ (Domains and Restrictions of Maps §restriction-value). Therefore the following two equivalences hold for each pair $(\varepsilon,N)$ of a real number $\varepsilon>0$ and an $N\in\mathbb{N}$, so that no $N$ is chosen: $d_{A}(a_{m},a)<\varepsilon$ for every $m\in\mathbb{N}$ with $m\ge N$ if and only if $d(a_{m},a)<\varepsilon$ for every such $m$; and $d_{A}(a_{m},a_{\ell})<\varepsilon$ for all $m,\ell\in\mathbb{N}$ with $m\ge N$ and $\ell\ge N$ if and only if $d(a_{m},a_{\ell})<\varepsilon$ for all such $m,\ell$. By Convergent Sequence in a Metric Space §converges, applied in $(A,d_{A})$, a metric space by clause restriction, and in $(X,d)$, the first equivalence gives that $(a_{k})$ converges to $a$ in $(A,d_{A})$ if and only if it converges to $a$ in $(X,d)$. By Cauchy Sequence in a Metric Space §cauchy, applied likewise, the second gives that $(a_{k})$ is a Cauchy sequence in $(A,d_{A})$ if and only if it is one in $(X,d)$.

Clause closed-limits. Suppose, for a contradiction, that $x\notin A$, so $x\in X\setminus A$. Since $A$ is closed in $(X,\mathcal{T}_{d})$, $X\setminus A\in\mathcal{T}_{d}$ by Closed Subset of a Topological Space §closed, so $X\setminus A$ is open in $(X,d)$ by Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces §topology. The choices are made in this order: first $r$, then $N$. By Open Subset of a Metric Space §open there is a real number $r>0$ with $B_{d}(x,r)\subseteq X\setminus A$. By Convergent Sequence in a Metric Space §converges, applied in $(X,d)$ to $(a_{k})$ (a sequence in $X$ by the clause sequences) with $\varepsilon=r$, there is $N\in\mathbb{N}$ with $d(a_{m},x)<r$ for every $m\in\mathbb{N}$ with $m\ge N$. Since $N\ge N$ by Arithmetic and Order of the Natural Numbers §partial-order, $d(a_{N},x)<r$, and $d(x,a_{N})=d(a_{N},x)$ by Metric Space §symmetric. As $a_{N}\in A\subseteq X$ and $d(x,a_{N})<r$, Open Ball in a Metric Space §membership gives $a_{N}\in B_{d}(x,r)\subseteq X\setminus A$, so $a_{N}\notin A$, contradicting $a_{N}\in A$. Hence $x\in A$.

Clause closed-complete. By clause restriction, $(A,d_{A})$ is a metric space. Let $(a_{k})$ be a Cauchy sequence in $(A,d_{A})$. By clause sequences, $(a_{k})$ is a sequence in $X$ and a Cauchy sequence in $(X,d)$. Since $(X,d)$ is complete, $(a_{k})$ is convergent in $(X,d)$ by Complete Metric Space §complete, so by Convergent Sequence in a Metric Space §convergent it converges in $(X,d)$ to some $x\in X$. By clause closed-limits, $x\in A$. By clause sequences with $a=x$, $(a_{k})$ converges to $x$ in $(A,d_{A})$, so it is convergent in $(A,d_{A})$ by Convergent Sequence in a Metric Space §convergent. As $(a_{k})$ was an arbitrary Cauchy sequence in $(A,d_{A})$, $(A,d_{A})$ is complete by Complete Metric Space §complete.