Rudin's "Functional Analysis" theorem 6.5

Question

In the proof of part (a) of theorem 6.5 (pg. 139 of the first edition) he states that:

Since $\mathcal{D}_k\cap W$ is open in $\mathcal{D}_k$, we have proved that $\mathcal{D}_k\cap V \in \tau_k$

Why is that true? Thanks.

Answer 1

To provide some context, $\mathscr{D}_K$ is the $C^\infty$ functions with support in $K\subset \Omega$, $K$ being compact. They have a topology $\tau_K$ given by seminorms. There is also a topology $\tau$ for $\mathscr{D}$, and we have an arbitrarily chosen $V\in \tau$ by hypothesis. We have also by hypothesis an arbitrarily chosen $\phi\in \mathscr{D}_K\cap V$. We are told in the proof that by Theorem 6.4 we can find $W$ which satisfies $\phi+W\subset V$ and is convex, balanced, and satisfies $\mathscr{D}_K \cap W\in\tau_K$.

We want to infer that $\mathscr{D}_K \cap V\in\tau_K$, i.e., that $\mathscr{D}_K\cap V$ is open in the topology $\tau_K$. That inference can be made because $\phi$ was an arbitrary element in $\mathscr{D}_K\cap V$, and we have a $\tau_K$-neighborhood (i.e., neighborhood in the topology $\tau_K$) surrounding $\phi$ and contained in $\mathscr{D}_K\cap V$, that $\tau_K$-neighborhood being $\phi+(\mathscr{D}_K \cap W)$. The fact that the latter is contained in $\mathscr{D}_K\cap V$ follows from $\phi+W\subset V$. The general principle being applied is that, if every point of a set contains a neighborhood which lies within that set, then the set is open.

I think I've just paraphrased Rudin.