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Suppose that $(X, d_X)$ and $(Y, d_Y )$ are complete metric spaces. Let $i \colon X → Y$ be an isometry. Suppose that $F ⊂ X$ is closed. How can I prove that $i(F)$ is a closed subset of $Y$?

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4 Answers 4

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Take a sequence $(i(x_n))_{n\in\mathbb{N}}$ in $i(F)$ which converge to $l\in Y.$ You have that the sequence is convergent so Cauchy, and $$d_Y(i(x_n),i(x_m))=d_X(x_n,x_m)\underset{m,n\to+\infty}{\longrightarrow} 0$$ which implies that $(x_n)_{n\in\mathbb{N}}$ is Cauchy so converge to $x_*\in F$ because $F$ is complete. By continuity of $i,$ you finally have $l=\lim\limits_{n\to+\infty}i(x_n)=i(x_*)\in i(F).$

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  • $\begingroup$ Does this mean that $i(F)$ is closed? $\endgroup$
    – user189013
    Nov 8, 2015 at 21:41
  • $\begingroup$ Yes : in a metric space, a set $F$ is closed if and only if each convergent sequence of elements of $F$ has its limit in $F.$ $\endgroup$
    – Balloon
    Nov 8, 2015 at 21:42
  • $\begingroup$ a yes that's true thanks! Also, does the $(x_*)$ denote something specific? $\endgroup$
    – user189013
    Nov 8, 2015 at 21:44
  • $\begingroup$ No, $i(x_*)$ is just your isometry $i$ evaluated on the limit $x_*$ found before. $\endgroup$
    – Balloon
    Nov 8, 2015 at 21:44
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    $\begingroup$ @CloudJRK: Indeed, only the one of $(X,d_X)$ is needed here! $\endgroup$
    – Balloon
    Nov 28, 2020 at 19:55
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Hint: Let $y_n\in i(F)$ (Hence there exists $x_n\in F$ such that $y_n=i(x_n)$), and suppose that $y_n\to y$. Use that $y_n$ is a Cauchy sequence and deduce that $x_n$ is also a Cauchy sequence..

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  • $\begingroup$ Does the completeness of Y necessary? $\endgroup$
    – Cloud JR K
    Nov 28, 2020 at 14:08
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The image of a complete metric space under an isometry is itself complete (just think about Cauchy sequences), so you're really asking if all complete metric subspaces $X\subset Y$ must be closed. This certainly must be the case. Suppose $X$ is a complete metric subspace of $Y$ which is not closed, then there exists some $y\in Y$, $y\notin X$, such that if every open ball of radius $1/n$ centered at $y$ intersects $X$. Then, picking elements from the intersection of $X$ with each open ball of radius $1/n$ centered at $y$, you get a Cauchy sequence in $X$ converging to $y$, but $y\notin X$, so this would contradict completeness of $X$.

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Let $a\notin i(F)$. Assume for all $r>0$ there exists $y\in i(F)$ with $d_Y(y,a)<r$. Then we find a sequence $(x_n)$ in $X$ such that $y_n\to a$, so that $(y_n)$ is Cauchy, so that $(x_n)$ is Cauchy, so that $x_n\to x$ for some $x\in X$, and as $F$ is closed, $x\in F$. For $n$ large enough, we have both $|x_n-x|<\frac12|f(x)-a|$ and $|f(x_n)-a|<\frac12|f(x)-a|$, hence $|f(x)-a|\le |f(x_n)-f(x)|+|f(x_n)-a|<|f(x)-a|$, contradiction. Hence for some $r>0$, the $r$-ball around $a$ is disjoint from $i(F)$. In other words, the complement of $i(F)$ is open, $i(F)$ is closed.

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