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?
Answer
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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