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If the space $X$ is completely regular, we Know that The collection {$intZ(f)$:$f$ is a continuous function from $X$ to the real numbers} is an open base for open subsets of the space $X$ (i.e. If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f:X→\mathbb{R}$ such that $x∈intZ(f)⊆Z(f)⊆U_x)$. I have two questions about converse of this theorem. these questions are almost the same, but I think these are different.

1.If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f:X→\mathbb{R}$ such that $x∈intZ(f)⊆U_x$, then $X$ is completely regular.

2.If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f:X→\mathbb{R}$ such that $x∈intZ(f)⊆Z(f)⊆U_x$, then $X$ is completely regular.

I think these two claims have counterexamples and these conditions don't emply the complete regularity of $X$.

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When you write $U$ above, that always means a neighborhood of $x$, right? –  David White Apr 29 '12 at 20:49
    
What is your definition of completely regular? The definition in section 33 of Munkres (page 211) seems equivalent to your condition 2 (well, you have to assume $X$ is $T_1$ also). Munkres requires that for every $x$ and every closed $A$ not containing $x$, there is a cont. function with $f(x)=0$ and $f(A)=1$. So it's equivalent to your (2) with $U$ being the complement of $A$. We just need to know that the zero-set of $f$ doesn't have empty interior, which shouldn't be hard to prove. –  David White Apr 29 '12 at 20:51
    
Dear david My notation U means a neighborhood of x. My definition of compeletly regular space is the same as yours. As you know when A is closed and x is in the complement of A. by my condition, there is a continuous function f which x∈intZ(f)⊆Z(f)⊆X-A. But we only know that f is nonzero in A. How do we conduct f to the value 1(i.e. how can we define f such that f(A)={0}) At the End I have another question of you.do you have the same VIEWPOINT about the first of my claim? –  Ali Reza Apr 30 '12 at 6:45
    
In my comment and in Munkres $f(A)=1$, so we don't have to "conduct $f$ to the value 1." If you're curious about taking a function and tweaking it in a clever way so that it's 0 on some set and 1 on some other set then I recommend reading the proof of Urysohn's Lemma in Munkres. Also, Exercise 8 in Section 33 might interest you. Anyway, I'm now convinced that (2) is exactly the definition of completely regular. Also, (1) is not enough. When $U=int(Z(f))$ you can have points in $\overline{U}$ which $f$ takes to $0$, and if $A = U^c$ then those points would be in $A$, i.e. $f(A)=1$ would fail. –  David White Apr 30 '12 at 12:20
    
try: Gillman Jerison "Rings of continuous funtions", see index p. 297 looking for property of $O_p$ and $O^p$. –  Buschi Sergio May 1 '12 at 21:30
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1 Answer 1

Here's a counterexample to 1.

Let $T$ be the Tychonoff plank, i.e., the product $(\omega_1+1)\times(\omega+1)$ with the point $\langle\omega_1,\omega\rangle$ removed.

Consider the set $\omega\times T\cup\lbrace \infty\rbrace $ topologized so that $\omega\times T$ has the product topology and is an open subset itself, and the basic neighbourhoods of $\infty$ are of the form $U_n(\infty) = (\omega\setminus n)\times T\cup\lbrace \infty\rbrace $.

We construct a quotient space by identifying $\langle n,\alpha,\omega\rangle$ and $\langle n+1,\alpha,\omega\rangle$ whenever $n$ is odd and $\alpha\in\omega_1$, and identifying $\langle n,\omega_1,i\rangle$ and $\langle n+1,\omega_1,i\rangle$ whenever $n$ is even and $i\in\omega_1$.

The resulting space $C$, the Tychonoff corkscreww, is regular but not completely regular (the copy of $\lbrace 0\rbrace \times T$ and $\infty$ cannot be separated by continuous functions).

For each odd $n$ define $f_n:C\to[0,1]$ by $f_n(\infty)=0$ and $$ f(m,\alpha,i)= \begin{cases} 2^{-i} &\text{ if } m\le n \text{ and }i<\omega\cr 0 &\text{ if } m> n \text{ and }i<\omega \end{cases} $$ and $f_n(m,\alpha,\omega)=0$ for all $m$ and $\alpha$.

Then the interiors $\operatorname{int}Z(f_n)$ form a local base at $\infty$.

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