# Why are there continuum many nowhere dense subsets of $\mathbb R$?

I am able to see why the closed nowhere dense subsets of $$\mathbb R$$ are equinumerous with $$\mathbb R$$: Every closed nowhere dense subset of $$\mathbb R$$ is the boundary of an open set (namely its compliment). $$\mathbb R$$ has $$2^{\aleph_0}$$ many open sets, so there are at most $$2^{\aleph_0}$$ many closed nowhere dense subsets of $$\mathbb R$$. Every Cantor set is a closed nowhere dense subset of $$\mathbb R$$, and there are $$2^{\aleph_0}$$ many different Cantor sets, so there are precisely $$2^{\aleph_0}$$ many closed nowhere dense subsets of $$\mathbb R$$.

My next thought was to take closures, but this is not injective. What am I missing?

• If $A$ is nowhere dense, surely any subset of $A$ is nowhere dense as well. Do you have any other requirements on the sets? Commented Nov 9, 2018 at 16:17
As pointed out in the comments, there are more than continuum many nowhere dense subsets of $$\mathbb R$$. In particular, the Cantor set $$C$$ has the same size as the reals, and any subset of $$C$$ is nowhere dense, so there are at least $$|2^{C}| = 2^{2^{\aleph_0}} > 2^{\aleph_0}$$ such sets.
• (Also, there are only $2^{2^{\aleph_0}}$ sets of reals, so the number you got is as large as possible.) Commented Nov 9, 2018 at 16:58