Baire Category Theorem?

TheEmptySet · #1 $Old$ November 21st, 2008, 04:25 PM

Alright this question is from Carothers Real Analysis.

Let $(r_n)$ be an enumeration of $\mathbb{Q}$ . For each each n, let $I_n$ be the open interval centered at $r_n$ of radius $2^{-n}$ , and let $U= \cup_{n=1}^{\infty}I_n$ . Prove that U is a proper, open subset, dense subset of $\mathbb{R}$ and that $U^{c}$ is nowhere dense in $\mathbb{R}$ .

Where I am stuck is try to show that $U$ is a proper subset of $\mathbb{R}$ . My first thought was a proof by contradiction, by assuming that $\cup_{n=1}^{\infty}I_n=\mathbb{R}$ . Then by the Baire Categroy theorem one of the $int\{ \overline{I}_n\} \ne \emptyset$ but this didn't seem to go anywhere.

Thanks for any input

Moo · #2 $Old$ November 22nd, 2008, 01:24 AM

Hello,

I don't know if it leads anywhere, but you could see if $U$ is a closed set or not.
If it's not, then it cannot be $\mathbb{R}$ nor $\emptyset$ , which are both open and closed.

Opalg · #3 $Old$ November 22nd, 2008, 02:47 AM

The total length of the intervals in U is 2, so it's a very long way from being the whole of $\mathbb{R}$ .

More precisely, the Lebesgue measure of U is at most 2, while the measure of $\mathbb{R}$ is infinite.