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I read that the Cantor set is an uncountable perfect set that does not contain isolated points. At the same time, it is totally disconnected and does not contain intervals of any kind.

I could not figure out where the irrational numbers can be located inside the Cantor set. We know that the endpoints of the Cantor set are countable since they are rational numbers. Now, the irrationals should be 'inside' the interval between the rational endpoints.... But if there are no intervals, where would the irrationals lie? s.t. the Cantor ser is classified as uncountable.

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    $\begingroup$ The numbers in the Cantor set are those reals between $0$ and $1$ (inclusively) that can be expressed in ternary expansion without using any $1$s; as in the case of decimal expansion, the irrationals among those are precisely the ones whose ternary expansion is non-periodic, while the rationals are those whose ternary expansion is eventually periodic. $\endgroup$ Commented Jul 2, 2019 at 4:25
  • $\begingroup$ @ArturoMagidin Why are you answering in a comment? $\endgroup$
    – Arthur
    Commented Jul 2, 2019 at 4:35
  • $\begingroup$ Okay, but from another perspective, if the endpoints of $C_n$ for any $n$ are countable then can irratuinals can be mapped to those in between endpoints ? $\endgroup$
    – Link L
    Commented Jul 2, 2019 at 5:12

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The Cantor set has lots of points that are not end points of the intervals that are removed in its construction. The irrational numbers within the Cantor sets are not the end points of those intervals. Your confusion is caused by thinking that every points between $0$ and $1$ is either in one of the intervals removed or an end point of one of those intervals.

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  • $\begingroup$ Okay thanks... just a question, if the irrational numbers are not endpoints of intervals, how is it that at infinity, we have that the Cantor set does not contain any interval? -i.e. would it not remove the interior of the intervals? $\endgroup$
    – Link L
    Commented Jul 2, 2019 at 6:10
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    $\begingroup$ @LinkL it is not easy to imagine what points are left after all the intervals are removed. Using expansion to base $3$ we can show that Cantor set is uncountable thereby proving that there must be points in it other than the end points. $\endgroup$ Commented Jul 2, 2019 at 6:41
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"the elements of the Cantor set are precisely those elements in [0,1] with a ternary expansion consisting of 0's and 2's.'' and I add, uncountabley many irrationals are found therein.

Please see:

Is the Cantor set a subset of rational numbers, and is it countable or uncountable?

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