for a continuous function and open A such that f(A) is not open. I think $f:R \rightarrow R$, and $f(A)=0 $ for $A \in (0,1)$ works. since a singe point in $R$ is closed.

I have no idea to find a continuous function and closed A such that f(A) is not closed.

  • $\begingroup$ just an example $\endgroup$ – user378456 Oct 25 '16 at 3:51
  • 1
    $\begingroup$ If you are restricting yourself to functions $f:R\to R$, then $A$ must be a non-compact closed set. (Think about why this must be so.) $\endgroup$ – Braindead Oct 25 '16 at 3:55
  • $\begingroup$ Kind of confusing that the question ends with (closed), since [closed] has meaning on SE $\endgroup$ – QuantumFool Oct 25 '16 at 4:12

Let $f(x) = e^{-x^2}$ and $A= \mathbb{R}$.

Then $f(\mathbb{R}) = (0,1]$ which is neither open nor closed. $A$ is both open and closed.

  • $\begingroup$ why you put the -ve sign in the power of e? $\endgroup$ – user426277 Jun 3 '17 at 15:05
  • $\begingroup$ Now thanks I understood why. $\endgroup$ – user426277 Jun 3 '17 at 15:10

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