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Let $(X,d)$ be a complete metric space, $F\subset X$. Then $F$ is closed if and only if the relative metric space $(F, d_F)$ is complete.

Since $(X,d)$ is complete, Cauchy sequences in $X$ converge in $X$ with respect to $d$. Suppose $F$ is closed, then convergent sequences contained in $F$ converge in $F$. Since Cauchy sequences contained in $F$ are convergent, they must converge in $F$ under $(F, d_f)$.

Let $(x_k)_k\subset F$ be a Cauchy sequence converging to $x\in X$. Then for every $\epsilon > 0,\exists N\in\mathbb{N}$ such that $m,n\ge N\implies d_F(x_n,x_m)<\epsilon/3$. Since $(x_k)_k$ is convergent, $\forall\epsilon>0, \exists M\in\mathbb{N}$ such that $d_F(x_k, x)<\epsilon/3$. Fix $\epsilon>0$ and let $n\ge \max\{M,N\}$. Since $F$ is closed, $\exists y_\epsilon\in B_{\epsilon/3}(x)\cap F\ne \emptyset$.

Now, $$d_F(x_n, y)\le d_F(x_n, x_m)+d_F(x_m, x)+d_F(x,y)<\epsilon/3+\epsilon/3+\epsilon/3=\epsilon$$

Thus $(x_k)_k$ converges to some $y\in F$, but then $x=y$, so that $(F,d_F)$ is complete.

I'm wondering if I'm not being somewhat redundant in my attempt to make my proof more rigorous. Would appreciate some suggestions.

For the other direction, suppose $(F,d_F)$ is complete. Then every Cauchy sequence contained in $F$ converges in $F$ w.r.t. $d_F$. Let $(x_k)_k\subset F$ be Cauchy. Then $(x_k)_k$ converges to some point $x\in F$. Suppose there exists some limit point $y$ of $(F,d_F)$ not contained in $F$. Then there exists a sequence $(y_k)_k$ which converges to $y$. But then $(y_k)_k$ must be Cauchy, so that $(y_k)_k$ converges in $F$, a contradiction. Hence, $F$ is closed.

Please let me know if you find my proof satisfactory or if something better be fixed.

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If $(F,d_F)$ is closed, it is complete: suppose $(x_n)$ is a CAuchy sequence in $F$. It is still a Cauchy sequence in $(X,d)$ as $d$ and $d_F$ by definition agree on $F$. So it converges to some $x \in X$ as $X$ is complete. As $F$ is closed, a limit of a sequence from $F$ is in $F$, so $x \in F$ and $x_n \to x$ in $(F,d_F)$ as well.

If $(F,d_F)$ is complete: let $x_n \to x$ (in $(X,d)$) where all $x_n \in F$. We want to show $x \in F$. Well, any convergent sequence is Cauchy (in $(X,d)$ so also in $(F,d_F)$, so $(x_n)$ converges in $(F,d_F)$ to some $y \in F$, but as sequence limits are unique $x=y$ and $x \in F$. So $F$ is closed.

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  • $\begingroup$ Thanks. Can you please let me know if the technical part of my proof is satisfactory? $\endgroup$ – sequence Oct 17 '17 at 21:57
  • $\begingroup$ @sequence Your proof makes no sense to me, sorry. $\endgroup$ – Henno Brandsma Oct 18 '17 at 20:16
  • $\begingroup$ Actually I used the same argument (my wording may be a little less clear), but I also attempted to make it more formal. $\endgroup$ – sequence Oct 19 '17 at 22:01
  • $\begingroup$ What purpose does the $\frac{\varepsilon}{3}$ serve? What are you trying to show?@sequence $\endgroup$ – Henno Brandsma Oct 20 '17 at 8:18

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