Proof of Uniform boundedness principle (why are the sets closed)

Question

I have a question on the proof of the UBP:

Let $(X,d)$ be a complete metric space and $\mathcal{F}$ be a familly of continuous functions such that for each $x \in X$ $\sup_{f \in \mathcal{F}}|f(x)| < \infty$. Then there is an $M \in \mathbb{R}$ and open ball $B \subset X$ such that $\sup_{f \in \mathcal{F}, x \in B }|f(x)| \leq M$.

Proof: The sets $C_n= \{ x \in X: \sup_{f \in \mathcal{F}}|f(x)| \leq n\} $ are closed and satisfy $X = \cup_{n \geq 0} C_n$, so by the Baire Category theorem at least one of them must have an interior point, i.e there is an open ball and an integer $M$ such that the statement holds.

My question is why you require that the sets $C_n$ are closed? I mean the BTC says that a complete metric space can't be written as the countable union of nowhere dense sets, not necessarily closed..? What hapends if you don't make this requirement?

Answer 1

If a complete metric space $X$ is written as a union of a sequence of sets $\{C_n\}$ then BCT says some $C_n$ is not nowhere dense, which means the closure of $C_n$ has a an interior point for some $n$. If you do not know what the closure is is, then this property becomes useless. When $C_n$ 's are closed one of them has an interior point and this is usually helpful in completing the argument.

Answer 2

Nowhere dense: $\left(\overline{C_{n}}\right)^{\circ}=\emptyset$, but now $\overline{C_{n}}=C_{n}$, so $\left(C_{n}\right)^{\circ}=\emptyset$ in this case, so there must be some $n$ such that $\left(C_{n}\right)^{\circ}\ne\emptyset$.