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$X$ is a Tychonoff space (or a $T_{3\frac{1}{2}}$-space) iff it is $T_1$ and for every point $x\in X$ and a closed set $A$ not containing $x$ there is a continuous function $f\colon S\to[0,1]$ such that $f(x)=0$ and $f[A]=\{1\}$.

Is it true that if $X$ is Tychonoff space, then for a point $x$ such that $f(x)=0$ and closed $A\neq\emptyset$ without $x$ such that $f[A]=\{1\}$ and $f$ continuous, the family $f^{-1}\left[[0,r)\right]$ is a local basis for $x$? In the sense that for every $V$ open in $S$ and containg $x$ there is a real numer $r$ such that $f^{-1}[[0,r)]\subseteq V$.

If this fails in general for all Tychonoff spaces, is it true for connected Tychonoff spaces? If still not, what condition(s) does $X$ have to satisfy in order to guarantee this?

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  • $\begingroup$ You need to also let $f$ range over the family of continuous functions. Just one fixed $f$ (with various $r$) won't necessarily do. Your description of the family is not very clear. Do you mean $\{f^{-1}[[0,r)]:f\in C(X),0<r<1\}$, or, for a fixed $f$, the family $\{f^{-1}[[0,r)]:0<r<1\}$. The former family is indeed a base at $x$. $\endgroup$ – Mirko Feb 15 '17 at 0:04
  • $\begingroup$ But I am curious how the situation looks like when we have fixed $f$. Under what conditions pre-images under $f$ form a local basis? $\endgroup$ – Mad Hatter Feb 15 '17 at 0:07
  • $\begingroup$ well the constant function $f=0$ would always have $X=f^{-1}[[0,r)]$, for any $r>0$, this would rarely be a basis (only when $X$ is a singleton). $\endgroup$ – Mirko Feb 15 '17 at 0:09
  • $\begingroup$ @Mirko My question was ambiguous, I have improved it. $\endgroup$ – Mad Hatter Feb 15 '17 at 0:24
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    $\begingroup$ this follows directly from the definition of Tychonoff spaces. Say $x\in X$ and $U$ is a neighborhood of $x$. Let $G=X\setminus U$. There is a continuous $f$ with $f(x)=0$ and $f[G]=\{1\}$. Then $x\in f^{-1}[[0,\frac12)]\subseteq U$. $\endgroup$ – Mirko Feb 16 '17 at 18:13
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For the first version (without $A$) I made a comment that for the constant function $f=0$, we have $X=f^{-1}[[0,r)]$ for every $r>0$, hence $\{f^{-1}[[0,r)]:r>0\}$ is not a basis at $x$ (unless $X=\{x\}$, a singleton).

For the revised version, the case is not much different. You could find a closed $B$ disjoint from $A$ (you mean $x\not\in A\not=\emptyset$, I assume) such that $x$ is contained in the interior of $B$. If $X$ is also normal, then we could find continuous $f$ with $f[B]=\{0\}$ (and $f[A]=\{1\}$), and again $\{f^{-1}[[0,r)]:r>0\}$ cannot be a basis at $x$ since $B\subseteq f^{-1}[[0,r)]$ for each $r>0$. (We need not assume that $X$ is normal. Say, $f$ is any continuous function with $f(x)=0$ and $f[A]=\{1\}$, where $x\not\in A$ and $A$ is closed. As long as there is $p\in X\setminus A$, with $p\not=x$, then we may find continuous $g$ with $g(p)=0$ and $g[A]=\{1\}$ (also assuming, if we wish, that $0\le f(y)\le 1$ and $0\le g(y)\le 1$ for all $y\in X$). Then $h(y)=f(y)\cdot g(y)$ would, just like $f$, have the properties that $h(x)=0$ and $h[A]=\{1\}$, but, nevertheless, $\{h^{-1}[[0,r)]:r>0\}$ cannot be a basis at $x$, as $\{x,p\}\subseteq h^{-1}[[0,r)]$ for all $r>0$.)

I think your question might be better stated in the following form. For a Tychonoff space $X$ and a point $x$, does there exists a suitable continuous $f$ such that $\{f^{-1}[[0,r)]:r>0\}$ is a basis at $x$?

In this form, a necessary condition is that $X$ is first countable at $x$, by which I mean that there is a countable local basis (of open sets) at $x$. Indeed, if $\{f^{-1}[[0,r)]:r>0\}$ is a basis at $x$ then so is $\{f^{-1}[[0,\frac1n)]:n=1,2,3,...\}$, and the latter family is countable.

It seems that first countable is also a sufficient condition. Indeed, let $\{U_n:n=1,2,3,...\}$ be a countable local basis at $x$, then $C_n=X\setminus U_n$ is a closed set missing $x$ and there is a continuous $f_n$ such that $f_n(x)=0$ and $f_n[C_n]=\{1\}$ (and $0\le f(y)\le 1$ for all $y\in X$). You could define $f(y)=\sum_{n=1}^\infty \frac{f_n(y)}{2^n}$ and verify that it works (i.e., that $\{f^{-1}[[0,r)]:r>0\}$ is a basis at $x$, use that $f^{-1}[[0,\frac1{2^n})]\cap C_n=\emptyset$).

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  • $\begingroup$ Thanks a lot! Very helpful answer. $\endgroup$ – Mad Hatter Feb 15 '17 at 1:03
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    $\begingroup$ You are welcome! I just added a comment that I did not need to assume that $X$ was normal, in the second paragraph (and added some more details.) $\endgroup$ – Mirko Feb 15 '17 at 1:20

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