# Cantor set--nowhere dense, complete

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.

Where is a mistake?

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

• 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 $\Bbb R$. – Brian M. Scott Jan 22 '15 at 22:00
• 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. – Stefan Hamcke Jan 22 '15 at 22:00
• You could have made the same mistake with a one-point subspace of $\mathbb{R}$, which is also complete. – user208259 Jan 22 '15 at 22:02
• I understand. Tnx – zariski Jan 22 '15 at 22:06
• 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)? – zariski Jan 22 '15 at 22:13