# Slight generalisation of the Baire category theorem?

By the Baire category theorem, one cannot write a complete metric space $X$ as a countable union of closed nowhere dense subsets of $X$. Can this be generalised to say that there is no injection $f: X \hookrightarrow X$, $f(X) \subseteq \cup_{n \ge 1} X_n$, where the $X_n$ are closed and nowhere dense? It seems like not much of a jump, and that it should be true, but I can't quite get it out. Thanks for any insight.

• Presumably you meant to say that you can’t write $X$ as a countable union of closed, nowhere dense sets. For the question itself, are you requiring $f$ to be continuous? – Brian M. Scott May 28 '13 at 7:03
• Thanks for the pick up. No I'm not. – Mr. Chip May 28 '13 at 7:05

Since you don’t require continuity, there is an injection of $\Bbb R$ into the middle-thirds Cantor set $C$, which is closed and nowhere dense, because $|\Bbb R|=|C|=2^\omega=\mathfrak c$. If you insist on a union of countably infinitely many closed, nowhere dense sets, just inject each $[n,n+1)$ into a Cantor set in $\left[n,n+\frac12\right]$.
• Thank you for your response. I find it kind of perturbing though, because it seems to contradict an assignment question I'm working on. It goes as follows: "If $X = A \cup B$ is complete and $B$ is a countable union of closed nowhere dense subsets of $X$, show $A$ cannot be embedded in any countable union of closed nowhere dense subsets in $X$." But if we take $X = A = \mathbb{R}$ and your injection, we have a contradiction. Is it possible I'm misinterpreting 'embedding'? – Mr. Chip May 28 '13 at 7:21