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Is the Cantor set homeomorphic to $B=\{\frac{1}{n}|n\in Z^+\}$? So far, I have concluded both sets are bounded and closed, therefore both are compact by the Heine-Borel Theorem. Neither are path connected or connected. Also, neither have the fixed-point property. So, I posit that they are homeomorphic. Now, assuming I am correct, I am supposed to construct an explicit homeomorphism between the two sets. How do I go about starting this part? And in general, how do I go about constructing explicit homeomorphisms from the Cantor set to another set? I have trouble grasping intuitively the Cantor set itself. Thanks in advance for the help!

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    $\begingroup$ There is not even a bijection, setting continuity aside. As for intuition about the Cantor set, it is closed, bounded, has empty interior, and is "perfect" i.e. all of its points are limit points. $\endgroup$
    – Ian
    Mar 20 '18 at 11:02
  • $\begingroup$ @Ian Which topological property do you suggest I use to show they are not homeomorphic sets? $\endgroup$ Mar 20 '18 at 11:03
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    $\begingroup$ Why do you have to do it topologically? Any homeomorphism is a bijection and their cardinalities are already different. $\endgroup$
    – Ian
    Mar 20 '18 at 11:05
  • $\begingroup$ Also, what makes you think $B$ is closed? $\endgroup$
    – quasi
    Mar 20 '18 at 11:05
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    $\begingroup$ @coreyman317: $B$ is closed in itself, sure. Every set is closed in itself. But $B$ is not closed in $\mathbb{R}$, hence $B$ is not compact. $\endgroup$
    – quasi
    Mar 20 '18 at 11:31
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$C$, the Cantor set, is uncountable, compact and has no isolated points (no $x \in C$ such that $\{x\}$ is open in $C$).

$B$ is countable, not closed (in $\mathbb{R}$) so not compact and all of its points are isolated points.

They hardly can be more different IMHO.

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