A space $X$ is regular if for all $x\in X$ and all closed $F\not\ni x$, there is $U,V$ open s.t. $x\in U$, $F\subset V$ and $U\cap V=\emptyset.$ In wikipedia, they talk about $T_3$ space as Regular and Hausdorff space (i.e. that are both). It looks strange for me... don't we have that Regular $\implies $ Hausdorff ? If yes, why such a separation axiom is considered ?

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    $\begingroup$ we don't. e.g. $X = \{a,b\}, \tau = \{\emptyset, X\}$. $\endgroup$ – mathworker21 Jul 14 at 14:07
  • $\begingroup$ @mathworker21: Yes. Corrected. Is your space really regular ? Actually there are no $x$ and no closed set $F$ s.t. $x\not\in F$... $\endgroup$ – user657324 Jul 14 at 14:07
  • $\begingroup$ you're wrong for two reasons. first, even if there were no $x$ and no closed $F$ such that $x \not \in F$, then the space would be regular (vacuously). second, there actually are $x$ and closed $F$ s.t. $x \not \in F$; take $x=a$ and $F = \emptyset$. $\endgroup$ – mathworker21 Jul 14 at 14:08
  • $\begingroup$ @mathworker21: ok, indeed... I would never think to such an example my self, but indeed it looks to work. Thank you. $\endgroup$ – user657324 Jul 14 at 14:09
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    $\begingroup$ the way to think of such an example is to note that regular does imply hausdorff if singletons are closed (since you can apply the definition of regular to $F$ being a singleton). so, to construct an example, you want singletons to not be closed. $\endgroup$ – mathworker21 Jul 14 at 14:11

As correctly noted in the comments and the answer of José Carlos Santos, any regular space in which singletons are closed is clearly Hausdorff. The condition "singletons are closed" is equivalent to the $T_1$ separation axiom. So if you want an example of a regular space which is not Hausdorff, you need to look at spaces which are not $T_1$.

But actually, it turns out that you need to go further and look at spaces which are not even $T_0$. This is why the examples suggested so far all have the trivial topology.

Recall that a space $X$ is $T_0$ if for any points $x,y\in X$ with $x\neq y$, there is some open set $U$ such that either $x\in U$ and $y\notin U$ or $x\notin U$ and $y\in U$.

Claim: If $X$ is a regular $T_0$ space, then $X$ is Hausdorff.

Proof: Let $x,y\in X$ with $x\neq y$. To show that $X$ is Hausdroff, we want to find disjoint open neighborhoods of $x$ and $y$. Since $X$ is $T_0$, we have (without loss of generality - swapping the names of $x$ and $y$ if necessary) some open set $U$ with $x\in U$ and $y\notin U$. Let $F = X\setminus U$. Then $F$ is closed and $x\notin F$, so by regularity we can find open sets $V$ and $W$ with $x\in V$, $F\subseteq W$, and $V\cap W = \emptyset$. But now note that $y\in F$, so $y\in W$, and $V$ and $W$ are disjoint open neighborhoods of $x$ and $y$, respectively.


In a space in which every singleton is a closed set (that is, in a $T_1$ space), yes, regular $\implies$ Hausdorff. But not in general. Take any set with more than one point, endowed with the trivial topology.


As Alex and José said, if there are such counterexamples, they aren't well behaved (i.e. they are not $T_0$). There is a general construction of a regular topological space which is not Hausdorff, the Partition Topology.

Its description is quite simple: Given a set $X$ and a partition $\mathcal{P}$ of $X$, let $\tau$ be the topology generated by the collection $\mathcal{P}$ (viz. the open sets are the arbitrary unions of these sets). Then this space is regular, since the closed sets and the open sets are the same, and if $x \notin F$ for some open set $F$, then let $U=F^c $ and $V=F$. It also clearly is not Hausdorff if the partition $\mathcal{P}$ is not constituted only of singletons.

For example, the following topological spaces fall inside the definition above:

  1. Any set $X$, endowed with the trivial topology $\{ X, \emptyset\}$.

  2. $\mathbb{N}$ endowed with the topology generated by $\{ \{2n-1, 2n \} \mid n \in \mathbb{N}\}$. (This is called the odd-even topology.)

  3. $\mathbb{R} - \mathbb{Z}$ endowed with the topology generated by $\{(n, n+1) \mid n \in \mathbb{Z} \}$. (This is called the deleted integer topology.)

  4. $\mathbb{R}^2 $ endowed with the topology generated by the vertical lines. (This example illustrates the partition does not need to be countable).

Furthermore, observe that, if $X$ is a regular non-Hausdorff space, and $Y$ is a regular, then $X \times Y$ is a regular non-Hausdorff space. In fact, if we can't separate $x_1$ and $x_2$ in $X$, we also can't separate $(x_1,y)$ and $(x_2, y)$ in $X \times Y$, and since the product of two regular spaces is regular, the result follows. Therefore, we can construct an even bigger family of regular non-Hausdorff space.

If you are looking for some strange spaces, you can always find some freaks here.


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