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Does there exist a continuous bijection $\mathbb{Q}\to \mathbb{Q}\times \mathbb{Q}$?

I am not able to find out how to proceed.

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related to… – SHOBHIT GAUTAM Aug 26 '13 at 8:24
up vote 11 down vote accepted

Every countable metric space without isolated points is homeomorphic to the rational numbers. In particular $\Bbb{Q\times Q}$.

This is a theorem of Sierpinski, and you can find the details here:

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Not only do such a maps exist, it is possible to construct explicit instances. Here is a sketch of how this can be done.

Let $\pi$ be an irrational real. For every $n \in \mathbb{Z}$ the set $\mathbb{Q} \cap (n+ \pi, n+\pi+1)$ is both open in $\mathbb{Q}$ and homeomorphic to $\mathbb{Q}$. A homeomorphism can be constructed by taking two rational sequences, one increasing and one decreasing to $\pi$ and constructing a piecewise linear, increasing function. Thus we obtain a homeomorpism $f: \mathbb{Q} \to \mathbb{Z} \times \mathbb{Q}$.

It is also possible to obtain an explicit bijection $g: \mathbb{Z} \to \mathbb{Q}$, which is always continuous, since $\mathbb{Z}$ is discrete. If we then put $h_1(x) = g(f_1(x))$ and $h_2(x) = f_2(x)$ then $h: \mathbb{Q} \to \mathbb{Q} \times \mathbb{Q}$ is the desired continous bijection.

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