3
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $f$ 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
-1
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.