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Let $X$ be a topological space, $C\subseteq X$ be closed, $X\ni x\notin C$, and $f\in C(X,\mathbb R)$. Suppose further that $f(x)$ and $f(C)$ are contained in disjoint open neighborhoods of $\mathbb R$. Is this enough to enough to ensure that $x$ and $C$ are separated by a function, i.e., that there exists some $g\in C(X,\mathbb R)$ such that $g(x)=\{0\}$ and $g(C)=\{1\}$? Please give a proof or counterexample.

I'm asking because I recently looked up the definition of Tychonoff spaces and wondered if we could give an equivalent definition along the above lines.

It seems intuitively plausible to me that this is the case, but I'm guessing it's false because none of the definitions of "separation by a function" that I can find mention this.

This is a follow-up to this post, where I tried to generalize the question I'm asking here, but ended up asking the wrong thing.

WillG
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    Assume WLOG that $f(x)=0$. Now it is easy to see that we can make a continuous function $h$ which takes $h(0)=0$ and $h(y)=1$ for all $y$ at least $\epsilon$ from $0$. As $\Bbb{R}$ is a metric space we can choose $\epsilon$ small enough so that $h\circ f$ seperates $x$ and $C$. (Assuming I haven't misunderstood anything.) – Fishbane Sep 16 '22 at 16:16
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    You don't even need $C$ to be closed, and the open neighborhood of $f(C)$ is useless. You just need an open subset of $\mathbb R$ which contains $f(x)$ and not $f(C).$ – Anne Bauval Sep 16 '22 at 16:59
  • @Fishbane Why not an official answer? – Paul Frost Sep 17 '22 at 22:05
  • @PaulFrost Feel free to write it up yourself. I generally prefer writing longer answers and I feel like I don't have much else to say on this. – Fishbane Sep 18 '22 at 00:33

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