3
$\begingroup$

The question pretty much says it all. I need a completely regular (the definition not requiring $T_1$) topological space which is $T_0$ but not $T_1$. I've sifted through Counterexamples in Topology but I just can't find one. I've also tried google and searching this site but I'm not coming up with anything.

It seems the counterexample should be simple but everything I try doesn't work (like Sierpinski topology, open extension topology, topology on $\mathbb{R}$ whose open sets are $(a,+\infty)$, etc.).

EDIT: By a completely regular topological space I mean a topological space $(X,\tau)$ so that for every $x\in X$ and closed $\mathscr{C}\subseteq X$ such that $x\not\in\mathscr{C}$ there exists a continuous function $f:X\to[0,1]$ so that $f(x)=0$ and $f(\mathscr{C})=\{1\}$.

By a $T_0$ space I mean a topological space in which any two distinct points are topologically distinguishable.

By a $T_1$ space I mean a space in which for any two distinct points both of them are contained in neighborhoods which do not contain the other.

I was thinking completely regular + $T_0$ did not imply $T_1$ since the definitions of Tychonoff/$T_{3\frac{1}{2}}$ spaces I've seen require completely regular and $T_1$ rather than completely regular and $T_0$.

EDIT EDIT: Turns out no such counterexample exists.

$\endgroup$
1
  • 2
    $\begingroup$ What exactly do you mean by "completely regular" if it doesn't imply $T_1$? $\endgroup$
    – MJD
    May 13, 2013 at 16:06

1 Answer 1

5
$\begingroup$

If $X$ is completely regular (closed set and point can be separated by a continuous real-valued function), then it is also regular. Let $x\in X$ be a point and $A\subset X$ be a closed set not containing $x$, then there is a continuous function $f:X\to\Bbb R$ with $f(x)=0$, $f(A)=1$, so $f^{-1}([0,\frac12))$ and $f^{-1}((\frac12,1])$ are disjoint open neighborhoods of $x$ and $A$, respectively.

To show that a regular $T_0$-space $X$ is $T_2$, let $x,y\in X$. There is a closed set $A$ containing one point, say $y$, but not the other. By regularity, there are disjoint open sets $U\ni x$ and $V\supseteq A\ni y$. So $X$ is Hausdorff.

$\endgroup$
10
  • $\begingroup$ So if we take the indiscrete topology, it's completely regular? $\endgroup$
    – Asaf Karagila
    May 13, 2013 at 16:25
  • $\begingroup$ @AsafKaragila Yes it is $\endgroup$ May 13, 2013 at 16:32
  • 1
    $\begingroup$ @Asaf: Under some definitions, yes. It is more common to include T$_1$-ness in the definition of regularity, complete-regularity, normality, etc., but this isn't always the case. $\endgroup$
    – user642796
    May 13, 2013 at 16:32
  • 9
    $\begingroup$ @Arthur: That’s exactly the reverse of my training and experience. For me the $T$ properties are the hierarchy, and the named properties are the ‘naked’ properties. A regular space need not be $T_0$. $\endgroup$ May 13, 2013 at 17:27
  • 1
    $\begingroup$ Kelley and Willard also do the $T_3$ = regular + $T_0$ way! $\endgroup$
    – danzibr
    May 14, 2013 at 0:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .