Tychonoff spaces and ideals

Question

Let $X$ be a tychonoff space and let $T$ be the set of all $f \in C(X)$ such that for any $g$ the equation $fg = 1$ has at most finitely many solutions. Under what conditions on $X$, the set $T$ is an ideal of $C(X)$?

Answer 1

Here is an easy direct proof that $T$ is an ideal. Clearly $T$ is closed under multiplication by elements of $C(X)$. Now suppose $f_0,f_1\in T$ but for some $g$, $gf_0+gf_1=1$ at infinitely many points. Without loss of generality, $\operatorname{Re}(gf_0)\geq 1/2$ at infinitely many points. Let $u:\mathbb{C}\to\mathbb{C}$ be continuous such that $u(z)=1/z$ if $\operatorname{Re}(z)\geq 1/2$, and let $h=u\circ(gf_0)$. Then $hgf_0=1$ at infinitely many points, contradicting the assumption that $f_0\in T$.

Answer 2

I think $T$ is always an ideal of $C(X)$.

First note that the condition $f\in T$ may be rephrased as: $f$ possesses a zero in any infinite closed subset $C\subset X$. Indeed, if $f\notin T$, there is $g\in C(X)$ for which the closed set $C:=\{fg=1\}$ is infinite and $f$ has no zero there. Conversely, if $C$ is a closed subset of $X$ where $f$ does not vanish, we can extend $1/f_{|C}$ to a $g\in C(X)$, therefore such that the set $\{fg=1\}$ is infinite.

As a consequence we have that any $f\in T$ (i) vanishes on any accumulation point $x$ of $X$ (because if $f(x)\neq 0$ then $f$ would be non-zero in a closed nbd ox $x$, an infinite set). Moerover (ii) on any closed discrete subset $C$ of $X$, $f$ is different from zero only on finitely many points (because any subset of $C$, in particular $\{f\neq0\}\cap C$ is closed in $C$ and also in $X$). Conversely, any $f$ in $C(X)$ verifying (i) and (ii) necessarily has a zero in any infinite closed set (indeed, any infinite closed set of $X$ either possesses an accumulation point, thus a zero of $f$ by (i), or is discrete, and has a zero by (ii)).

In conclusion, $f\in T$ if and only if $D(X)\subset\{f=0\}$ and for any discrete closed set $C$ the set $\{f\neq0\}\cap C$ is finite; from this characterization it follows plainly that $T$ is an ideal of $C(X)$.