# A set intersecting the graph of any continuous function in a finite set

New version of the problem I am looking for a characterization of those completely regular and hausdorff spaces $X$ such that the follwing is true:

If $f :X\longrightarrow \Bbb{R}$ is continuous such that for any continuous function $g: X \longrightarrow \Bbb{R}$, $\mbox{Graph}(g)\cap A_f$ is finite, then $A_f$ is finite. (where $A_f := \{ (x, \frac 1{f(x)}) \; | \; f(x)\not = 0 \}$).

As @NikWeaver has pointed out below this is now true in general.

Solution to the new version: $X$ has this property if and only if it does not contain a countably infinite set of isolated points every infinite subset of which has a cluster point.
In one direction, if $X$ does contain such a configuration, then the counterexample given in my other answer generalizes straightforwardly. Conversely, suppose it contains no such configuration and suppose $f: X \to \mathbb{R}$ is continuous and $A_f$ is infinite. There are two possibilities. First, suppose that for some $\epsilon > 0$ the set $\{x \in X: |f(x)| \geq \epsilon\}$ is infinite. Then find a continuous function $\phi: \mathbb{R} \to \mathbb{R}$ such that $\phi(t) = \frac{1}{t}$ for $|t| \geq \epsilon$. The function $g = \phi \circ f$ then has $\Gamma(g) \cap A_f$ infinite, so $f$ is not a counterexample. Otherwise, suppose that the set $\{x \in X: |f(x)| \geq \epsilon\}$ is finite for every $\epsilon > 0$. By the assumption on $X$ we can find an infinite subset $S$ of $\{x: f(x) \neq 0\}$ that has no cluster point. Then define $g(x) = \frac{1}{f(x)}$ when $x \in S$ and $g(x) = 0$ when $x \not\in S$. Then $g$ is continuous and $\Gamma(g) \cap A_f$ is infinite, so again $f$ is not a counterexample.
• Incidentally, this condition is weaker than "does not contain a countably infinite set of isolated points that has a limit point". For instance, let $X$ be $\mathbb{N}$ together with one point of $\beta \mathbb{N}$. This $X$ does contain a copy of $\mathbb{N}$ that has a cluster point, but every infinite subset of $\mathbb{N}$ can be split into two infinite pieces, only one of which has a cluster point. – Nik Weaver Sep 30 '13 at 22:59
• @LisaK: the generalization is this. If $X$ contains such a configuration, then identify it with ${\bf N}$ and define $f$ as in my other answer. Let $f$ be zero off the configuration. A cluster point of a set $A$ is a point $x$ every neighborhood of which has infinite intersection with $A$. – Nik Weaver Jan 10 '15 at 14:27
No. Let $X$ be the one-point compactification of $\mathbb{N}$ and define $f(n) = \frac{1}{n}$, $f(\infty) = 0$. Then $A_f$ is infinite but $\Gamma(g) \cap A_f$ is finite for any continuous $g$ since continuous functions are bounded.