This is a quick question that I have not managed to answer myself: let $X$ be a metrizable topological space, let $A\subset X$ be a closed, bounded subset. $f:X\to Y$ is a homeomorphism, must $f(A)$ be closed and bounded?

I know that if $X=\mathbb{R}^n$ the answer is yes, because compactness is a topological property, and also that closedness and boundedness alone are not enough to guarantee that $f(A)$ is either closed or bounded (consider $[0,\infty)$ and $[0,1)$), but what happens if we have both?


For metric spaces: Take $X = (0..1)$ and take the inclusion $ι\colon X → ℝ$. Now take $A = X$. Or even take $f\colon X → ℝ,~x↦ \frac{x}{1-x}$ and $A = X$ if you want $f(A)$ to be neither bounded nor closed in $ℝ$.

  • $\begingroup$ Thank you, so I can conclude from this that "closedness and boundedness" is not a topological property? $\endgroup$ – Dr. Nov 24 '14 at 21:38
  • $\begingroup$ It’s not, no. There’s no reasonable way of interpreting “boundedness” in arbitrary topological spaces. $\endgroup$ – k.stm Nov 24 '14 at 21:48
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    $\begingroup$ @Dr.: It's worth noting that, even in metrizable topological spaces, there needn't be a natural way to interpret boundedness. For example, $X=(0,1)$ is certainly bounded as a subspace of $\Bbb R$ under the usual metric. However, if we instead consider the function $\rho:X\times X\to\Bbb R$ given by $$\rho(x,y)=\left|\frac1x-\frac1y\right|,$$ which turns out to be a metric on $X$ topologically equivalent to the usual one, we find that $X$ is not bounded with respect to $\rho.$ $\endgroup$ – Cameron Buie Nov 24 '14 at 23:21

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