I'm wondering if the metric spaces $S:=\{x+y\sqrt{2}:x,y\in\mathbb Q\}$ and $\mathbb Q^2$, with metrics induced by $\mathbb R$ and $\mathbb R^2$ respectively, are topologically equivalent. I know that the function $f:\mathbb Q^2\to S$ defined by $f(x,y):=x+y\sqrt{2}$ is bijective, but how can I show that $f$ is continuous / not continuous?
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3Are you okay with just saying it’s linear therefore continuous? – Randall Jul 30 '18 at 16:24
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1See also https://math.stackexchange.com/questions/355779/showing-mathbbq-is-homeomorphic-to-mathbbq2#1642728 and the surprising result in the answer. – gj255 Jul 30 '18 at 16:31
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1Maybe first show that $f(x,y):=x+y\sqrt{2}$ is continuous from $\mathbb R^2$ to $\mathbb R$. Then conclude it is continuous with the induced metrics. But that does not work for the inverse map, which is not continuous. The two spaces are homeomorphic, but not by that map. – GEdgar Jul 30 '18 at 16:31
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Can you name a map that works? @CEdgar – Ludvig Lindström Jul 30 '18 at 16:39
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1I think @GEdgar was who you meant to tag. – Cameron Buie Aug 05 '18 at 19:01
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Can I name a map? No. But I can cite: "Sierpinski proved that every countable metric space without isolated points is homeomorphic to the rationals" – GEdgar Aug 05 '18 at 23:22
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Ok, thanks! @GEdgar – Ludvig Lindström Aug 07 '18 at 17:54
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2Effectively a duplicate of Showing $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}^2$ – Cameron Buie May 12 '19 at 16:49