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Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and let $f: X \to Y$ be a uniformly continuous function. If $A \subset X$ is bounded, must $f(A) \subset Y$ be bounded?

It is clear to me that in metric spaces that satisfy the Heine-Borel property, such as $\mathbb{R}^{n}$, the answer to this question is yes. However, I can see no reason why this should hold for arbitrary metric spaces. Any ideas? Thanks!

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A very silly counterexample: Let $(Y,\rho)$ be an unbounded metric space and let $X = Y$ be equipped with the metric $d(x,y) = 1$ for $x \neq y$. Every subset of $X$ is bounded. The function $f(x) = x$ is uniformly continuous $(X,d) \to (Y,\rho)$ since we can take $\delta \lt 1$ for every $\varepsilon \gt 0$. However, $f(X) = Y$ is unbounded as well as $f(A)$ for every unbounded $A \subset Y$. –  t.b. Mar 28 '12 at 2:38

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Consider the space $X\subset \ell^\infty$ consisting of elements of the form $(1,\ldots,1,t,0,\ldots)$ such that $t\in [0,1]$. Note that $X$ is bounded under the $\sup$ metric. Define the function $f:X\to \mathbb R$ by $f((x_n))=\sum\limits_{n=1}^\infty x_n$. The function $f$ is uniformly continuous, as given any $\epsilon>0$ if we let $\delta=\max\{\epsilon,1\}$ we have that $\|x-y\|<\delta\implies |f(x)-f(y)|<\epsilon$ for $x,y\in X$. Yet $f(X)$ is clearly unbounded.

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Ah, excellent. I suspected that I would have to look to some infinite dimensional Banach space to escape Heine-Borel. Thanks! –  Isaac Solomon Mar 28 '12 at 3:07

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