# Doubt in understanding 6.3 form Rudin functional analysis

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

Any Help will be appreciated

• Your question is unclear. Are you tried to say "I do not understand highlighted set , which belongs to $\beta$", or "I do not understand why the highlighted set belongs to $\beta$". In either case, it would help if you wrote out the definition of $\mathscr D(\Omega)$ for us. – Omnomnomnom Jan 16 at 13:51
• By your previous question, it seems that $\mathscr D(\Omega)$ is the set of all $C^\infty$ functions over $\Omega$ with compact support. – Omnomnomnom Jan 16 at 13:54
• Yes Sir.That is defination Thanks a lot – MathLover Jan 16 at 13:54

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 $$\beta$$.