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.

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

1 Answer 1


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!?).


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