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I know the Cantor set is uncountable, but I just came with an argument that shows it is countable. Obviously my argument is wrong, but I just don't know where is the mistake. Here it is.

Let $C$ be the Cantor set, we know $C$ is closed, so $A = [0,1]\setminus C$ is open. As $A$ is open, we know there is a collection of open disjoint intervals $(a_n,b_n)_{n=1}^\infty$ such that $A=\bigcup_{n=1}^\infty(a_n,b_n)$.

Given any intervals $(a_n,b_n)$ and $(a_m,b_m)$ (suppose $b_n\leq a_m$), we may have or not points of $C$ in $[b_n, a_m]$.

If $b_n<a_m$, is not possible that $[b_n, a_m]\subset C$, for $C$ is nowhere dense. In this case, there is more open intervals inside $[b_n, a_m]$. So we get more intervals and to the same reasoning.

If $b_n=a_m$, it is possible that this point is in $C$.

My problem is how to manage the first case. Looks like doesn't matter if there is more open intervals inside this 'gaps', in the final I will have only the cases where $b_n=a_m$. In other words, these are the only possible points in $C$, therefore $C$ is countable.

Michael Hardy
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Integral
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    All but countably many of the points in the Cantor set are not endpoints of the intervals which were deleted. Indeed, all but countably many of these are irrational. None of the endpoints of deleted intervals are irrational. – Ian Jan 30 '15 at 22:35
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    For example, $1/4 \in C$, but is not an endpoint of an interval. – GEdgar Jan 30 '15 at 22:36
  • You can find similar questions both on the Related links here, and in the Linked questions of my suggested dup. – Asaf Karagila Jan 30 '15 at 22:38
  • I see, thanks for the links. – Integral Jan 30 '15 at 22:39

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