isometry preserve dense property

Question

Let $A$ and $B$ are two metric spaces. let $f:A\to B$ be an isometry. If $C \subset A$ is dense in $A$, is it true that $f(C)$ is dense in $B$?

Thanks for your help in advance.

Answer 1

It depends how you define isometry. If you just define it as a distance-preserving map, then no. But if you define it as a bijection that preserves distance (like some textbooks do, like mine- I use Pugh's Real Analysis)- yes.

Proof

Suppose $A$ and $B$ are equipped with the metrics $m$ and $M$ respectively.

If $C$ is dense in $A$, then $\overline C = A$. To prove that $f(C)$ is dense in $B$, we need to show $\overline{f(C)} = B$.

It's clear that $\overline{f(C)} \subseteq B$, so it suffices to show $B \subseteq \overline{f(C)}$.

Suppose $y \in B$. If $f$ is surjective, $\exists$ $x \in A$ st $f(x) = y$.
Since $\overline C = A$, $x \in \overline C$. Then every open ball centered at $x$ contains a point in $C$, or, more formally, $\exists$ $x' \in C$ st $m(x, x') < r$, for every $r > 0$. Since $f$ is an isometry (a distance-preserving map), we have that $r > m(x, x') = M(y, f(x'))$. Then for each $y \in B$ and $r > 0$, $\exists$ $f(x') \in f(C)$ st $f(x') \in B_M(y, r)$, so $ \overline{f(C)} = B$.

Of course, if $f$ is not surjective, there are many counterexamples. drhab already provided an example in the comments, but another one is the map $f: \mathbb{R} \rightarrow \mathbb{R}^2, x \mapsto (x, 0)$. Take a dense subset of $\mathbb{R}$, say $\mathbb{Q}$, and note that $f(\mathbb{Q}$) is not dense in $\mathbb{R}^2$.