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I can't figure out this out.

Cantor set is closed in $\mathbb{R}$.

$\mathbb{R}$ is a complete metric space.

Every closed subset of a complete space is also complete; thus, so is the Cantor set.

Complete space can't be written as a countable union of nowhere dense sets; thus, the Cantor set has no such representation. (1)

Cantor set is nowhere dense (its interior is empty, and no intervals are contained in it)

Thus, the Cantor set is a finite (then countable) intersection of nowhere dense sets.

Contradiction with statement (1).

Where is a mistake?

Does representation in (1) refer only to infinite countable intersections?

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    $\begingroup$ A complete metric space $X$ cannot be written as a countable union of sets that are nowhere dense in $X$. The Cantor set is not a countable union of sets that are nowhere dense in the Cantor set, even though it is itself nowhere dense in $\Bbb R$. $\endgroup$ – Brian M. Scott Jan 22 '15 at 22:00
  • $\begingroup$ A complete metric space $X$ is a Baire space, so no somewhere dense set can be written as a countable union of nowhere dense sets in $X$. In particular $X$ cannot be written as such a union. $\endgroup$ – Stefan Hamcke Jan 22 '15 at 22:00
  • $\begingroup$ You could have made the same mistake with a one-point subspace of $\mathbb{R}$, which is also complete. $\endgroup$ – user208259 Jan 22 '15 at 22:02
  • $\begingroup$ I understand. Tnx $\endgroup$ – zariski Jan 22 '15 at 22:06
  • $\begingroup$ Another question,related to this. Can we conclude by the previous statements,that Cantor set is uncountable? If Cantor set C would be countable,it could be written as a countable union of singletons (which are nowhere dense in C) and that would be a contradiction with the statement (1)? $\endgroup$ – zariski Jan 22 '15 at 22:13
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the assumption that since its finite thus countable is false. Countable means a bijection with N. It can be infinite (countably infinite, or simply just stated countable)

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