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The Cantor Middle Third set in $[0,1]$ is formed by removing middle third open interval at each stage. Then Cantor set is the intersection of those remaining intervals.

More precisely, given $[0,1],$ let $F_1=[0,\frac{1}{3}]\cup[\frac{2}{3},1],$ where we remove middle third open interval $(\frac{1}{3},\frac{2}{3})$ from $[0,1].$ Proceed in similar fashion, remove middle third open interval from $[0,\frac{1}{3}]$ and $[\frac{2}{3},1]$ and $F_2$ be the remaining sets. The Cantor set is $C:=\bigcap_{n=1}F_n.$

Question: Instead of removing open middle third interval, what happens if we remove closed middle third interval at each stage and take intersection like Cantor set? How many elements are left?

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The answer is uncountably many. It is well-known that the original Cantor set consists of all numbers in the unit interval that can be written in ternary using no $1$'s (including the ones that end in infinitely many $2$'s, if that lets you get away from having a $1$ in there).

If you remove closed intervals instead of the open ones, that means that you're also taking away all the numbers that can be written with a terminating ternary expansion, of which there are only countably many (since it's a subset of the rational numbers)

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The new set has the same cardinality as that of Cantor set in that the end points of all the removed intervals form a countable set.

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