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$A = \{1/n:n \in \mathbb{N}\}$ and $B =A \cup \{0\}$.

$A$ and $B$ are subsets of $\mathbb{R}$, with Euclidean metric.

  1. Does a continuous bijection exist $f:A \rightarrow B$?

  2. Does a continuous bijection exist $f:B \rightarrow A$?

I can easily find bijections between $A$ and $B$ but none that are continuous. Intuitively, I feel like this is not possible, but how can I prove it? I also realize that $B$ is the closure of $A$, does that have any impact?

One approach I can think of is that if 1 AND 2 are true, then $A$ and $B$ are homeomorphic, but that is not possible, by the invariance theorem, since $B$ is closed while $A$ is not. But this does not preclude either 1 or 2 being true.

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    Though $A$ is not closed, it's also not open. Your point that $A$ and $B$ are not homeomorphic to each other still stands, though. – 2017-02-11
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    RIght, that makes sense. Edited. – 2017-02-11
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    Do you know a theorem about a continuous bijection from a compact space to a Hausdorff space? – 2017-02-11
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    Strictly speaking the question is not well-posed: you haven't told us what topologies you're using on the bare sets $A$ and $B$. If you're using the subspace topology induced by the usual topology on the reals, you should note that $A$ is a discrete space. What does this tell you about the answer to question 1? – 2017-02-11
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    @TedShifrin we haven't really learned about Hausdorff spaces, just metric spaces (which are Hausdorff, I have come to learn) – 2017-02-11
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    @symplectomorphic thank you for pointing that out, I have edited the question – 2017-02-11
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    Also, you have made a subtle error in your reasoning that "yes" answers to both 1 and 2 jointly imply that $A$ and $B$ are homeomorphic. Just because there is a continuous objection in both directions does not mean there is a *single* bijection that is continuous in both directions. Indeed, there are counterexamples: see http://mathoverflow.net/questions/30661/non-homeomorphic-spaces-that-have-continuous-bijections-between-them – 2017-02-11
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    @symplectomorphic of course, oversight on my part. Makes perfect sense, thanjs – 2017-02-11

1 Answers 1

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  1. Yes, because there is an obvious bijection $A\to B$ and $A$ is discrete. (Discreteness implies every function out of $A$ is continuous.)

  2. No, because $B$ is compact and $A$ is Hausdorff, so a positive answer would imply $A$ and $B$ are homeomorphic. This can't be right because $A$ is discrete but $B$ is not, and discreteness is a topological property.