Suppose $\mathcal{F}$ is a family of real-valued continuous functions on a compact topological space $X$.

Question

Suppose $\mathcal{F}$ is a family of real-valued continuous functions on a compact topological space $X$ such that

• $f \geq 0$ everywhere for all $f \in \mathcal{F}$;
• if $f,g \in \mathcal{F}$ then $f+g \in \mathcal{F}$;
• for every $f \in \mathcal{F}$ there is an $x \in X$ s.t. $f(x)=0$.

Prove that there is some point $a \in X$ such that $f(a)=0$ for every $f \in \mathcal{F}$.

Attempt. Take an $f_1 \in \mathcal{F}$, then there exists $x_1 \in X$ such that $f_1(x_1)=0$. If every $f \in \mathcal{F}$ annihilates $x_1$ then we're done. Otherwise, there must be an $f_2 \in \mathcal{F}$ that doesn't annihilate $x_1$, then by the properties above (specifically apply the 3rd and 1st axioms to $f_1 + f_2$) we know that there is an $x_2 \in X$ such that $f_1(x_2)=f_2(x_2)=0.$ If every $f \in \mathcal{F}$ annihilates $x_2$ then we're done. Otherwise continue the process.

We have a sequence $(x_n)_{n \geq 1}$ in a compact space, thus there is a convergent subsequence with limit $x\in X$. Clearly all the $f_i$ that we constructed must annihilate $x$.

But how do we prove that every $f \in \mathcal{F}$ annihilates $x$? The problem comes from the fact that $\mathcal{F}$ may not be countable.

Answer 1

Assume for contradiction, that for every $x \in X$, there is a function $f_x \in \mathcal{F}$ such that $f_x(x) > 0$. Then, using continuity, there exists an open neighbourhood $N_x$ of $x$ on which $f_x > 0$.

$N_x$ form an open cover of $X$, so take a finite subcover $N_{x_1}, \dotsc, N_{x_n}$. So, for every $x \in X$, one of the $f_i(x) > 0$ for some $i = 1,\dotsc, n$. Thus, $f_1 + \dotsc + f_n > 0$ at all points in $X$, a contradiction.

Answer 2

A family $F$ of sets has the FIP (Finite Intersection Property) iff $\cap G\ne \emptyset$ for every finite non-empty $G\subset F.$

A space $X$ is compact iff any non-empty family $F$ of closed subsets of $X,$ such that F has the FIP, satisfies $\cap F \ne \emptyset.$

If $f\in \mathcal F$ then $f^{-1}\{0\}$ is closed in $X$ and not empty.

For $n\in \Bbb N,$ if $f_1,..,f_n\in \mathcal F$ then $f_1+...+f_n\in \mathcal F. $So for some $x\in X$ we have $0=f_1(x)+...+f_n(x)$ for some $x\in X,$ which implies $0=f_1(x)=...=f_n(x).$ That is, $$\cap_{j=1}^n f_j^{-1}\{0\} \ne \emptyset.$$

So $F=\{f^{-1}\{0\}: f\in \mathcal F\}$ is a closed family with the FIP. Since $X$ is compact, therefore $\cap F \ne \emptyset.$

(Assuming that $\mathcal F\ne \emptyset$ .)

APPENDIX: "Closed family" means "family of closed sets"..... Suppose $F$ is a non-empty family of closed subsets of a space $X$ and $F$ has the FIP but $\cap F= \emptyset.$ Then $\{X\setminus f:f\in F\}$ is an open cover of $X$ with no finite sub-cover so $X$ is not compact...... Suppose $X$ is not compact. Let $C$ be an open cover of $X$ with no finite sub-cover. Consider $F=\{X\setminus \cup D: D$ finite and $D\subset C\}.$.... Every statement about open sets has a "dual" statement about their complements, the closed sets.

Answer 3

a space is compact iff it has finite intersection property. the family $\{f^{-1}(0):f\in\mathcal{F}\}$ has finite intersection by assumption