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I'm attempting to find a counterexample for the following question:

If $f: X \rightarrow Y$ is continuous and open, then it is also closed.

Would an example involving the discrete topology on $X$ (where every set is open) work to show that $f$ is not necessarily closed, because it maps an open set in $X$ to a closed set in $Y$?

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Mapping an open set to a closed set does not imply $f$ is not closed; instead, you need to exhibit a function that maps a closed set to a non-closed set. –  Neal Mar 12 '13 at 19:53

3 Answers 3

up vote 3 down vote accepted

Take $f:\mathbb{R}\rightarrow \mathbb{R}$ (both with standard topology) and $$f(x)=\frac{x}{1+|x|}$$ then $f(\mathbb{R})=(-1,1)$ is not closed.

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HINT: Try the map $\pi:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x$, and consider the graph of $f=\frac1x$.

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The projection map $p_1:\mathbb R ^2 \rightarrow \mathbb R$ defined by $p_1(x,y)=x$. Take the standard topologies on $\mathbb R^2, \mathbb R$. Then $p_1$ is continuous, open and not closed. To see that it's not closed take the image of one of the branches of $\tan(x)$ (it will be an open interval).

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