# Can somebody help me to understand this? (Baire Category Theorem)

Theorem $$\mathbf{6.11}$$ (Baire Category Theorem) Every residual subset of $$\Bbb R$$ is dense in $$\Bbb R$$.

$$\mathbf{6.4.5}$$ Suppose that $$\bigcup_{n=1}^\infty A_n$$ contains some interval $$(c,d)$$. Show that there is a set, say $$A_{n_0}$$, and a subinterval $$(c\,',d\,')\subset(c,d)$$ so that $$A_{n_0}$$ is dense in $$(c\,',d\,')$$. (Note: This follows, with correct interpretation, directly from the Baire category theorem.)

I'm trying to understand why there exists such subinetrval $$(c', d')$$ using Baire Category Theorem, but I don't how to apply it because I don't know which is the residual set in this case, i.e., a set whose complement can be represented as a countable union of nowhere dense sets.

$$\textbf{EDIT:}$$ I forgot to say that the sets $$A_n$$'s are closed.

• Consider the space $X = (c,d)$, and the sequence of sets $B_n = A_n \cap X$. Nov 4, 2013 at 17:08
• Hint: $\left(\bigcup A_n\right)^c$ is not dense, hence not residual... Nov 4, 2013 at 19:08
• @NateEldredge Do we need that the $A_n$'s are closed in this approach? Nov 5, 2013 at 19:17
• @Twink: No, I don't believe so. Nov 5, 2013 at 19:55

To apply the Baire category theorem (BCT), I assume that it is intended that the sets $\{A_n\}_{n=1}^{\infty}$ are closed subsets of $\mathbb{R}$. Now, let $c<a<b<d$ and consider $[a,b]$, which is a closed subset of a complete metric space, and is therefore a complete metric space itself. So, by the BCT it can not be the union of nowhere dense closed subsets. Since $\cup_{n=1}^{\infty} \big([a,b]\cap A_n\big) = [a,b]$ is the union of closed subsets, at least one of the sets $[a,b] \cap A_{n_0}$ must be somewhere dense. This means that there is a subset $(c',d') \subset [a,b]$ such that $[a,b]\cap A_{n_0}$ is dense in $(c',d')$. Now, since $[a,b] \cap A_{n_0}$ is closed, this will imply that $(c',d') \subset [a,b] \cap A_{n_0}$ (why!?).