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Todd Trimble
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It's classical (by a back-and-forth argument; see for instance A Shorter Model Theory by Hodges) that any two countable dense unbounded linear orders without endpoints are isomorphic. For example the linear order $\mathbb{Q}$ is isomorphic to $L = \mathbb{Q} \cap (0, 1)$, and they are homeomorphic under their order topologies.

So it will suffice to exhibit a homeomorphism $f: L \times L \to L$. One method might be to interleave two eventually periodic decimal expansions (forbidding infinite tails of 0's, say), using one expansion at the odd places and the other at the even places. The result is again eventually periodic, and is bicontinuous.

Todd Trimble
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