Closed set of 1st category

Question

If $F$ is a closed subset of a complete metric space, is it possible for $F$ to be of the first category? This seems to lead to a contradiction. Since $F$ is a closed subset of a complete metric space, $F$ is itself a complete metric space. But the Baire category theorem implies that $F$ is of the second category; i.e. $F$ is not of the 1st category. What am I missing here?

The reason I ask is because my textbook says $F$ is of the first category if and only if the interior of $F$ is empty; yet it seems no such $F$ can exist.

Answer 1

Yes, it is possible; for an example in $\mathbb{R}$, consider $\{0\}$, or less trivially the Cantor set. Clearly $\{0\}$ is of first category in $\mathbb{R}$, and it's a good exercise to show that the Cantor set is nowhere dense (so also of first category).

So what about the contradiction you mention?

The crucial point is that when we say "$A$ is of first (or second) category," this is with respect to some larger space $B$. For a trivial example, consider the singleton $A=\{0\}$. Then clearly, as a subset of $\mathbb{R}$, $A$ is nowhere dense, so of first category; but also clearly, if we live within $A$, then every nonempty subset is dense - that is, every subset of $A$ is second category in $A$.

The Baire category theorem states that every complete metric space $X$ is a second category subset of itself; but this doesn't contradict $X$ being first category as a subset of some larger space $Y$. In fact, here's a good exercise:

Let $X$ be an arbitrary complete metric space. Show that there is a larger complete metric space $Y\supset X$ such that $X$ is first category in $Y$ (or, if you prefer, show that there is a metric space $Y$ and an isometric embedding $i: X\rightarrow Y$ such that $i(X)$ is a first-category subset of $Y$).

That is, no space is "absolutely second category".