$M_a =\{ f\in C[0,1] |\ f(a)=0 \}$ for $a$ $\in$ $[0,1]$. Is $M_a$ finitely generated in $C[0,1]$?

The following proof is found here. One can actually show that $M_a$ is not generated by countably many elements.

Assume the contrary that $M_a$ is generated by $(f_1, f_2, \dots, f_n, \dots)$. By normalizing we assume $|f_i(x)| \le 1$ for all $i, x$. Then define

$$f(x) = \sum_{i=1}^n \frac{\sqrt{|f_i(x)|}}{2^i}.$$

Then $f \in M_a$ and there exist $g_1, \dots, g_r \in C[0,1]$ so that

$$f = f_1 g_1 +\cdots +f_r g_r \Rightarrow |f(x)| \le M \sum_{i=1}^r |f_i(x)|,$$

where $M$ is an upper bound for $g_1, \dots, g_r$.

Now by continuity there is an open $U$ containing $a$, so that

$$\sqrt{|f_i(x) |} \le \frac{1}{2^{i+1}M} \ \ \ \ \ \forall x\in U,\ i=1, \dots, r.$$

This implies

$$|f(x) \vert \le M \sum_{i=1}^r |f_i(x)| \le M \sum_{i=1}^r \frac{\sqrt{|f_i(x)|}}{2^{i+1}M} \le \frac{1}{2} |f(x)| <|f(x)|$$

for all $x\in U$. This is impossible. Thus $M_a$ is not generated by countably many elements.

If we assume $M_a$ is finitely generated, say $M_a = \langle f_1,\dots, f_n \rangle$, such that $|f_k(x)| \le 1$ for $x \in [0,1]$.

Consider, $f(x) = \sum\limits_{k=1}^{n} \sqrt{|f_k(x)|} \in M_a$.

Then, there are functions $\{g_k\}_{1 \le k \le n} \in C[0,1]$ such that $f(x) = \sum\limits_{k=1}^{n} g_k(x)f_k(x)$, and set $M=\max\limits_{k=1}^{n} |g_k(x)|$.

Then, choose an open interval $U \subset [0,1]$ near $a$, such that $|f_k(x)| < \dfrac{1}{M^2}$, for each $k$ (which should exist by virtue of continuity of $f_k$'s and $f_k(a) = 0$).

So for $x \in U$, we have, $|f(x)| \le M\sum\limits_{k=1}^{n} |f_k(x)| < \sum\limits_{k=1}^{n} \sqrt{|f_k(x)|} = f(x)$ (contradiction!).