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$\mathbb{R}^2$ with different topologies on it are homeomorphic as a topological space? for example with discrete topology and usual topology, what I need is a continous bijection with inverse is continous, from usual to discrete any continous map is finally constant map,so I think they are not homeomorphic. thank you for help and discussion

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An even easier example is defining the topology with only $\left\lbrace \mathbb{R}^2,\emptyset\right\rbrace$ as open sets, that obviously cannot be homeomorphic to the usual topology in $\mathbb{R}^2$. –  MyUserIsThis Jun 16 '13 at 18:05
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No, they are not homeomorphic. $\Bbb R^2$ with the discrete topology is not connected, and $\Bbb R^2$ with the usual topology is connected, so there isn’t even a continuous map from $\Bbb R^2$ with the usual topology onto $\Bbb R^2$ with the discrete topology: continuous maps preserve connectedness. There are many other ways to see that they aren’t homeomorphic. For instance, the usual topology is separable, and the discrete topology isn’t. The usual topology has a countable base; the discrete topology does not.

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Suppose there exists a homeomorphism $f$ between $\mathbb R^{2}$ with the usual topology, and $\mathbb R^{2}$ with the discrete topology. Then $f(S^{1})$ is compact, as continuous functions preserve compactness, and it is also infinite, as $S^{1}$ has infinitely many points. However in a discrete space, only finite sets are compact. So there cannot exist a homeomorphism. In fact, there cannot even be an injective continuous such $f$, as that is all we used in the above.

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