I was reading "Functional Analysis" by Rudin, and in the first chapter, the author introduces us to Topological Vector Spaces. Towards the end of this chapter, the author also gives some examples of how to get topological vector spaces from a family of seminorms, and one of the examples includes the space of smooth functions on an open set $\Omega \subseteq \mathbb{R}^n$. This space is denoted by $C^{\infty} \left( \Omega \right)$.
On this space, we first define seminorms as follows: First, we consider compact sets $\left\lbrace K_n | n \in \mathbb{N} \right\rbrace$ with the property that $K_n \subseteq \text{Int } K_{n + 1}$ and $\Omega = \bigcup\limits_{n \in \mathbb{N}} K_n$. Then, for each $n \in \mathbb{N}$, we define $p_n : C^{\infty} \left( \Omega \right) \rightarrow \left[ 0, \infty \right)$ as
$$p_n \left( f \right) = \sup \left\lbrace \left| D^{\alpha} f \left( x \right) \right| : x \in K_n \text{ and } \left| \alpha \right| \leq n \right\rbrace.$$
Here, we have used the notation that $\alpha = \left( \alpha_1, \alpha_2, \cdots, \alpha_n \right)$ is a multiindex (of length $n$) of non-negative integers, the operator $D^{\alpha} = \dfrac{\partial^{\left| \alpha \right|}}{\partial x_1^{\alpha_1} \partial x_2^{\alpha_2} \cdots \partial x_n^{\alpha_n}}$, with $\left| \alpha \right| = \alpha_1 + \cdots + \alpha_n$.
Then, the collection $\mathcal{P} = \left\lbrace p_n | n \in \mathbb{N} \right\rbrace$ is a separating family of seminorms which is closed under $\max$, and hence they induce a topology with the local base
$$V_n = \left\lbrace f \in C^{\infty} \left( \Omega \right) | p_n \left( f \right) \leq \dfrac{1}{n} \right\rbrace.$$
Moreover, since $\mathcal{P}$ is countable, the topology is metrizable, and a compatible metric is given by (from a result proved earlier in the book)
$$d \left( f, g \right) = \max\limits_{n \in \mathbb{N}} \left\lbrace \dfrac{p_n \left( f - g \right)}{2^n \left( 1 + p_n \left( f - g \right) \right)} \right\rbrace.$$
Some simple analysis also gives that with this metric, $C^{\infty} \left( \Omega \right)$ is a Frechet. Now the question is whether the space is normable. To prove this, the author proves that $C^{\infty} \left( \Omega \right)$ is not locally bounded and hence cannot be normable.
To this end, the author wants to use the following result:
Theorem: Every locally bounded topological vector space with the Heine-Borel property is finite-dimensional.
We start by proving that $C^{\infty} \left( \Omega \right)$ has the Heine-Borel property. Then, since it is not finite-dimensional, we must have that it is not locally bounded and hence not normable.
Let $E \subseteq C^{\infty} \left( \Omega \right)$ be a closed and bounded set. Then, by the characterizations of bounded sets in a topological vector space whose topology is induced by a family of seminorms, we have that for every $n \in \mathbb{N}$, there is some $M_n > 0$ such that for all $f \in E$, we have $p_n \left( f \right) \leq M_n$. As a consequence, the set $\left\lbrace D^{\beta} f | f \in E \right\rbrace$ is equicontinuous on $K_{n - 1}$, where $\left| \beta \right| \leq n - 1$.
Hence, by Arzela-Ascoli's Theorem, every sequence in these sets has a (uniformly) convergent subsequence. From here, the author writes that by Cantor's diagonalization trick, we can see that $E$ is compact (that is, every sequence in $E$ has a convergent subsequence). However, I am unable to see precisely how this works. Here is my attempt at this:
We start with a sequence $\left( f_j \right)_{j \in \mathbb{N}}$ in $E$. Then, this sequence has a uniformly convergent subsequence in $K_1$. Let us call this subsequence as $\left( f_j^{\left( 1 \right)} \right)_{j \in \mathbb{N}}$. We will now look at the subsequence $\left( f_j^{\left( 1 \right)} \right)_{j \in \mathbb{N}}$ alone. Again, it has a (uniformly) convergent subsequence, say $\left( \tilde{f}_j^{\left( 1, 0 \right)} \right)$ in $K_2$. Now, for a multiindex $\alpha$ with $\left| \alpha \right| = 1$, we have a subsequence of $\left( D^{\alpha} \tilde{f}_j^{\left( 1, 0 \right)} \right)_{j \in \mathbb{N}}$ which is uniformly convergent in $K_2$. Let us call this subsequence as $\left( D^{\alpha} \tilde{f}_{j}^{\left( 1, 1 \right)} \right)_{j \in \mathbb{N}}$ This, if the uniform limit of $\left( \tilde{f}_j^{\left( 1, 1 \right)} \right)$ is $\tilde{f}$, then $\tilde{f}$ is differentiable and $\lim\limits_{j \rightarrow \infty} D^{\alpha} \tilde{f}_j^{\left( 1, 1 \right)} = D^{\alpha} \tilde{f}$. So, now we work with the subsequence $\left( \tilde{f}_j^{\left( 1, 1 \right)} \right)_{j \in \mathbb{N}}$, and we call it as $\left( f_j^{\left( 2 \right)} \right)_{j \in \mathbb{N}}$.
Now, clearly, $\left( f_j^{\left( 2 \right)} \right)_{j \in \mathbb{N}}$ is a subsequence of $\left( f_{j}^{\left( 1 \right)} \right)_{j \in \mathbb{N}}$, and hence their limits agree on $K_1$. In this manner, for every $n \in \mathbb{N}$, we can construct a subsequence $\left( f_j^{\left( n \right)} \right)_{j \in \mathbb{N}}$ of $\left( f_j^{\left( n - 1 \right)} \right)_{j \in \mathbb{N}}$ that converges uniformly (along with its derivatives up to order $n$) on $K_n$ and the limits of the two sequences match on their common domain $K_{n - 1}$.
According to what is written in the book, I suppose that now we have to choose the diagonal sequence $\left( f_j^{\left( j \right)} \right)_{j \in \mathbb{N}}$, which is a subsequence of all the sequences that we have obtained. Suppose that $f_j^{\left( n \right)} \rightarrow f_n$, and $\lim\limits_{n \rightarrow \infty} f_n = f$. Then, it is clear that $f \in C^{\infty} \left( \Omega \right)$. However, we now have to prove that the sequence $\left( f_j^{\left( j \right)} \right)_{j \in \mathbb{N}}$ is convergent to $f$. And I am stuck with that. Somehow I need to convert the uniform convergence of all compact sets to the seminorms but that would also involve the derivatives of different order. So, the question is how to do it?
Any insights into this will be highly appreciated!