I'm trying to think of a counterexample to the claim if $f: X \to Y$ is continuous and $E \subseteq X$ is closed and bounded $f(E)$ is closed and bounded. Here, $X$ and $Y$ are metric spaces. Obviously, in Euclidean spaces this would be true, since continuity preserves compactness and compactness is equivalent to closed and bounded by Heine-Borel.

Help is appreciated.


Let $X=Y=\Bbb{R}$ and $f$ be the identity. Equip $Y$ with the standard metric and $X$ with the bounded metric $d(x,y)=\min(|x-y|,1)$. Then $f$ is continuous, and $X$ is closed and bounded (as a subset of itself) but its image $Y$ is not bounded.

More examples: $X=(0,1)$, $Y=\Bbb{R}$, $f(x)=1/x$, $E=X$. Here the image is neither closed nor bounded.

$X=(0,1)$, $Y=\Bbb{R}$, $f(x)=x$, $E=X$. Here the image is bounded but not closed.

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  • $\begingroup$ $f$ is even uniformly continuous in both directions, isn't it? $\endgroup$ – Stefan Hamcke Mar 11 '13 at 17:10
  • $\begingroup$ @StefanH.: yes. That truncated metric is the standard way to show that boundedness implies no uniform property, as I discuss here. $\endgroup$ – Chris Eagle Mar 11 '13 at 17:13

Actually you need a stronger condition on the metric space, namely total boundedness. This property says that for every $\epsilon>0$, there exists finitely many balls of radius $\epsilon$ such that their union covers the whole space. The proof is this:

Let $(M,d_1)$ total bounded metric space, and $f:M\rightarrow N$ a uniformly continuous function, and $E\subseteq M$ closed and bounded, choose a finite set of pairs $(x_i,\epsilon_i)$ with $x_i\in f(E)$ and $\epsilon_i>0$, in a way that the set of open sets $f^{-1}(B(x_i,\epsilon_i))$ cover $E$, it can be done for the condition of total boundedness and uniformly continuity. This says that $f(E)$ is bounded.

But as Chris Eagle showed, the set $f(E)$ need not be closed.

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  • 1
    $\begingroup$ What? $(0,1)$ is totally bounded, but can be continuously mapped onto $\Bbb{R}$. $\endgroup$ – Chris Eagle Mar 11 '13 at 17:44
  • $\begingroup$ @Chris Eagle I'm sorry i forget the conditon on N, but I already do the fix in the answer. $\endgroup$ – Daniel Mejia Mar 11 '13 at 17:49
  • $\begingroup$ But now your statement has no content. Obviously if $N$ is totally bounded then its subsets are bounded. $\endgroup$ – Chris Eagle Mar 11 '13 at 17:51
  • $\begingroup$ @Chris Eagle Yes you're right, the condition is on the function, this is an exercise of Rudin's book, that I did two years ago, I'm sorry. $\endgroup$ – Daniel Mejia Mar 11 '13 at 18:02

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