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Prove the following homeomorphism: $ R_{ \geqslant 0} \times R_{ \geqslant 0} \, \cong \,R_{ \geqslant 0} \times R $

where R are the real numbers and $ R_{ \geqslant 0} = \left\{ {x \in R:x \geqslant 0} \right\} $

If $ \,R_{ \geqslant 0} \, \cong \,\,R $ it´s done , but I think that this is not true, since it´s obvious that a continuous bijective function $ f:A \subset R \to R $ must be strictly increasing or strictly decreasing, so considering $ f(0) $ it´s easy to see that a continuous injective function between this two sets cannot be surjective, so there are not homeomorphic, but does not imply necesarly that the product it´s not. What can I do?

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Here's a very similar post:… You'll need to modify the solutions offered in the comments slightly, but most of the maps suggested can be extended to the boundary. – Brett Frankel Mar 3 '12 at 21:55

Here is a hint: Think of the two given subsets of the plane in polar coordinates.

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Good hint, it´s only sending $ r\left( {\cos u,sen\,u} \right) \to r\left( {\cos 2u,sen\,2u} \right) $ and it´s easy to check that it´s bijective, the continuity of the map and the inverse map, it´s only because they are restrictions of continuous maps – August Mar 3 '12 at 21:58

An alternative to Harald Hanche-Olsen's answer is that the map $z\mapsto z^2$ is a homeomorphism from the closed "first quadrant" in the complex plane to the closed upper half-plane.

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