Why is the topology of compactly supported smooth function in $\mathbb R^d$ not first countable?

Question

In other words, given a countable sequence of neighborhoods of $f(x)=0$, how to construct another open neighborhood that doesn't contain any of these neighborhoods? Thanks.

Answer 1

Denote by $B_m$ the open ball in $R^d$ of radius $m$, and let $\bar B_m$ denote the closed ball of radius $m$. Given a countable collection of non-empty open neighborhoods of $f=0$, say $U_n$, fix, for each $n$, a non-zero function $f_n\in U_n$ such that $f_n$ has support in $\bar B_{m_{n+1}}\setminus B_{m_n}$, where $m_n$ is a strictly increasing sequence, picked inductively large enough so that such a function $f_n$ exists. Pick also $x_n\in \bar B_{m_{n+1}}\setminus B_{m_n}$ with $f_n(x_n)\neq 0$. The following set is open in the space of smooth compactly supported functions (see page 152 of the book 'functional analysis' by rudin) $$ U=\{\phi\in C_c^\infty \,| \; |\phi(x_n)|<|f_n(x_n)|\}. $$ Since $f_n\notin U$, no $U_n$ is contained in $U$.