Unit semiball of the supremum of seminorms is closed

Question

I don't know how to prove the second part of this exercise (ex. 7.2 in F. Treves, "Topological vector spaces, distributions and kernels"). Let $\mathcal{P}$ be a family of continuous seminorms on a locally convex topological vector space $X$. Suppose that for every $x \in X$, $$ p_0(x) = \sup_{p \in \mathcal{P}} p(x) $$ is finite. Prove that $p_0$ is a seminorm and that its closed unit semiball $B= \{x \in X \; : \; p_0(x) \leq 1\}$ is a closed subset of $X$.

Verifying that $p_0$ is a seminorm is obviously trivial. I have some questions for the second part. Specifically, should I assume that the family $\mathcal{P}$ is the family of seminorms that induces the topology on $X$? From the statement I'd say that $\mathcal{P}$ is an arbitrary family of seminorms, such that each seminorm $P \in \mathcal{P}$ is continuous with respect to the locally convex topology on $X$, which is induced by some other family of seminorms $\mathcal{Q}$. We don't necessarily have $\mathcal{P} = \mathcal{Q}$. But then how to show that the unit semiball is closed, since the topology is characterized by $\mathcal{Q}$, and not by $\mathcal{P}$?

Also, if I was able to show that $p_0$ is continuous, it would immediately follow that $B$ is closed. Is it possible for $p_0$ not to be continuous? Can you give me some example for this situation?

Answer 1

To prove closedness, observe that $$ \{ x: \ p_0(x) \le 1 \} = \bigcap_{p\in \mathcal P} \{x: \ p(x)\le 1\}. $$

Answer 2

I don't think you should suppose that the given locally convex topology on $X$ is necessarily induced by the family $\mathcal{P}$.

As for the problem itself, if $x\notin B$ then $p(x)>1$ for some $p\in \mathcal{P}$ and by continuity of $p$ there is an open set containing $x$ where $p(x)>1$, and in particular $p_0(x)>1$, hence the complement of $B$ is open.

I don't know if $p_0(x)$ must be continuous at zero, which is equivalent to it being continuous everywhere.