$A = \{1/n:n \in \mathbb{N}\}$ 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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    $\begingroup$ 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. $\endgroup$ – 211792 Feb 11 '17 at 1:25
  • $\begingroup$ RIght, that makes sense. Edited. $\endgroup$ – jackson5 Feb 11 '17 at 1:29
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    $\begingroup$ Do you know a theorem about a continuous bijection from a compact space to a Hausdorff space? $\endgroup$ – Ted Shifrin Feb 11 '17 at 1:31
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    $\begingroup$ 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? $\endgroup$ – symplectomorphic Feb 11 '17 at 2:09
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    $\begingroup$ 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 mathoverflow.net/questions/30661/… $\endgroup$ – symplectomorphic Feb 11 '17 at 2:19
  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.


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