In our class, our professor said cantor set is clopen (both closed and open). One argument is that the interior of a cantor set intersecting with a closed interval, say $A = \Delta \cap [0,1/3]$ whose interior is $A$ itself. From my point of view, the interior should be empty. Is there any difference if we considered $A$ as a subset of $\Delta$(cantor set itself) rather than $\mathbb{R}$?

  • $\begingroup$ let me clarify your question: we know that the cantor set is closed in $[0,1]$ as well as in $\mathbb{R}$. But, in which set you want to show that it is open also? $\endgroup$ – Groups Sep 29 '15 at 6:23
  • $\begingroup$ I want to show that cantor set intersecting [0,1/3] is open in cantor set. $\endgroup$ – metricspace Sep 29 '15 at 6:31

Let $X$ be a metric space; let $A \subset X$; let $B \subset A$; then $B$ is called open (resp. closed) in $A$ if and only if there is some $C$ open (resp. closed) in $X$ such that $B = C \cap A$. From this definition it follows that the Cantor set is clopen in itself.

However, the Cantor set is not clopen in $\mathbb{R}$, as you have remarked.

  • $\begingroup$ So C is this case could be an open interval like say (-1/2, 1/2), and A would just be the cantor set. B is the intersection in question, right? $\endgroup$ – metricspace Sep 29 '15 at 6:29
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    $\begingroup$ Yes; moreover, we know that every metric subspace is clopen in itself. :) $\endgroup$ – Megadeth Sep 29 '15 at 6:34

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