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Recall the following theorem:

Let $X$ be a compact space and $Y$ a Hausdorff space. Suppose that $f:X \rightarrow Y$ is a continuous bijection. Then f is homeomorphism.

Prove that the compactness assumption on $Y$ is necessary.

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  • $\begingroup$ $\mathbb R$ with discrete topology? $\endgroup$ – Hagen von Eitzen Aug 20 '14 at 20:57
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    $\begingroup$ You want two topologies on $X$, $\tau_1$ and $\tau_2$, with $\tau_2$ Hausdorff and $\tau_2\subsetneq \tau_1$. Then $\mathrm{id}_X:(X,\tau_1)\to(X,\tau_2)$ is a continuous bijection which is not a homeomorphism. $\endgroup$ – Thomas Andrews Aug 20 '14 at 21:00
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    $\begingroup$ Or you can take $[0,1)$ and glue it so it becomes $S^1$. $\endgroup$ – Amitai Yuval Aug 20 '14 at 21:02
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How can we find such an example? Since $f$ is a bijection, we may assume wlog that $X$ and $Y$ are the same set, just with different topologies (and then $f$ is just the identity map). The fact that $f$ is continuous then means that every $Y$-open set is also an $X$-open set. To prevent $f$ from being a homeomorphism, the converse should not hold, i.e. not all $X$-open sets are also $Y$-open. In other words: $X$ must have "more" open sets than $Y$. So why not take as many open sets as possible for $X$?

Thus let $Y$ be any Hausdorff space and $X$ the same set, but with the discrete topology. This works as an example (unless $Y$ is already discrete, of course - so how about $\mathbb Q$ with its usual topology for $Y$ and discrete topology for $X$?).

By the way, using the discrete topology is a bit of a sledge-hammer here. In principle. you really need just a single additional open set: Let $Y$ be $\mathbb Q$ with standard topology again, let $X$ be the same, but declare $\{0\}$ open. Or, what amounts to the same: Let $X=\mathbb Q\setminus([-1,0)\cup(0,1])$ and let $f(x)=x-\operatorname{sgn}(x)$. - Or, to make things "almost" compact, let $Y=[0,1]$ with standard topology and $X=\{-1\}\cup (0,1]$ with the obviois map (which is the same as letting $X=[0,1]$ but declaring $\{0\}$ open)

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  • $\begingroup$ +1, but: I'd say "the same set" rather than "the same space." Space implies a topology. You also need $Y$ not to have the discrete topology, of course. $\endgroup$ – Thomas Andrews Aug 20 '14 at 21:05

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