Uniform continuity and boundedness in metric space

Question

For bounded subset $E$ of $\mathbb R$, uniform continuity of $f:E\to \mathbb R$ implies boundedness of $f(E)$.

the proof is pretty simple. take $\epsilon=1$. then there exist $\delta$ such that $$|x-y|<\delta\quad \to\quad |f(x)-f(y)|<1$$ Since E is bounded, we can cover E with finitely many $\delta$-balls.
so we have $|f(x)-f(y)|<$ (The number of $\delta$-balls) for any $x,y \in E$.

If the function is defined on metric space $(X,d)$ to metric space $(Y,\rho)$ instead of $E$ and $\mathbb R$, is the corresponding theorem still true?

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Edit:

Let $(X,d)$ and $(Y,\rho)$ be metric spaces and $(X,d)$ is bounded. if $f : X \to Y$ is uniform continuous, is $f(X)$ bounded in $Y$?

Answer 1

No.

What makes it hold in $\Bbb R$ is the precompactness (or more informatively, total boundedness) of bounded sets, which no longer holds in general metric spaces.

Let's consider an arbitrary infinite discrete metric space, then the space is bounded itself and any function defined on it is uniformly continuous, but of course needn't be bounded.