Doubt in understanding 6.3 form Rudin functional analysis

Question

I was reading sec 6.3 form Rudin Functional analysis

I do not understand why highlighted set belong to $\beta$?

$D(\Omega)$ is test function space on open set

Please help me

Any Help will be appreciated

Answer 1

The set $S = \{\phi \in \mathscr{D}(\Omega): |\phi(x_n)| < c_n\; n \in \mathbb{N}\}$ is obviously convex and balanced.

Now take arbitrary compact $K \subset \Omega$ and consider $S \bigcap \mathscr{D}_K$. Since sequence $x_n$ does not have a limit point in $\Omega$ then $K \bigcap \{x_n:\; n \in \mathbb{N}\}$ is finite because $K$ is compact. Let $\{x_{n_1},\dots,x_{n_m}\} = K \bigcap \{x_n:\; n \in \mathbb{N}\}$. Then $S \bigcap \mathscr{D}_K = \{\phi \in \mathscr{D}_K: |\phi(x_{n_k})| < c_{n_k}\; k = 1,\dots,m\}$. This set is obviously open in $\mathscr{D}_K$ since it is a finite intersection of open sets $\bigcap\limits_{k = 1}^{m} \{\phi \in \mathscr{D}_K: |\phi(x_{n_k})| < c_{n_k}\} = S \bigcap \mathscr{D}_K \in \tau_K$.

By definition 6.3 (b) $S$ belongs to $\beta$.