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

Also, does the answer change if we assume $A$ and $B$ are closed? What if $A$ is closed and $B$ is a singleton? 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.

WillG
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  • I would look into Urysohn's Lemma https://en.wikipedia.org/wiki/Urysohn's_lemma and the Separation Axioms https://en.wikipedia.org/wiki/Separation_axiom if you haven't already as I think that the different separation axioms might answer your question. – Steven Creech Sep 16 '22 at 15:36
  • @StevenCreech I should have mentioned, I was looking into both of those and couldn't find an answer. – WillG Sep 16 '22 at 15:45

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I don't know if this works as a counter-example, but think of the line with two origins. Then just take the projection to $\mathbb{R}$ as the function $f$, and $A=(-\infty,0)$ and $B=(0,\infty)$. These should not be separable by a function $g$. Does this solve your problem?

Steven Creech
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  • It answers my question, but now I realize I asked the wrong question! See here for the question I should have asked. – WillG Sep 16 '22 at 16:01
  • Are the two origins doing anything here, or does this work just as well with $X=\Bbb R$? – Karl Sep 16 '22 at 16:23