# A strong Hausdorff condition

Is the following strong form of Hausdorff equivalent to usual Hausdorff?

$X$ is strong Hausdorff if given distinct elements $x,y$ in $X$ there are open sets $U,V \subseteq X$ with $x \subseteq U$, $y \subseteq V$ and $\overline{U} \cap \overline{V} = \emptyset$.

I think that it is not equivalent, but haven't been able to prove this.

No, this property is called being Urysohn (or sometimes completely Hausdorff), see Wikipedia. There we also find an example (from Steen and Seebach's book) of a Hausdorff but not Urysohn space: the relatively prime integer topology: online explanation.

As Henno Brandsma's answer shows, this property is not equivalent to Hausdorffness. Another example (of my own construction, though I highly doubt it is original) is as follows:

Let $X = ( \mathbb{N} \times \mathbb{Z} ) \cup \{ -\infty , +\infty \}$. For notational convenience, by $\mathbb{Z}^{>0}$ I will denote the positive integers, and by $\mathbb{Z}^{<0}$ I will denote the negative integers.

We topologise $X$ as follows:

• each $\langle i,n \rangle$ with $n \neq 0$ is isolated.
• the basic open neighbourhoods of $\langle i,0 \rangle$ are of the form $\{ \langle i,0 \rangle \} \cup \{ \langle i,n \rangle : |n| \geq k \}$ for $k > 0$.
• the basic open neighbourhoods of $-\infty$ are of the form $\{ -\infty \} \cup A$ where $A \subseteq \mathbb{N} \times \mathbb{Z}^{<0}$ is such that $\{ i \in \mathbb{N} : \{ n \in \mathbb{Z}^{<0} : \langle i,n \rangle \notin A\text{ is infinite} \} \}$ is finite.
• the basic open neighbourhoods of $+\infty$ are of the form $\{ -\infty \} \cup A$ where $A \subseteq \mathbb{N} \times \mathbb{Z}^{>0}$ is such that $\{ i \in \mathbb{N} : \{ n \in \mathbb{Z}^{>0} : \langle i,n \rangle \notin A\text{ is infinite} \} \}$ is finite.

(The basic idea is that the subspaces $( \mathbb{N} \times \mathbb{Z}^{<0} ) \cup \{ - \infty \}$ and $( \mathbb{N} \times \mathbb{Z}^{>0} ) \cup \{ +\infty \}$ are copies of the Arens-Fort space, and each $\langle i,0\rangle$ is a limit point of the sections $\{ i \} \times \mathbb{Z}^{<0}$ and $\{ i \} \times \mathbb{Z}^{>0}$.)

It is fairly easy to show that $X$ is Hausdorff.

However, if $U$ and $V$ are (basic) open neighbourhoods of $-\infty$, $+\infty$, respectively, then there must be an $i \in \mathbb{N}$ such that $\{ n \in \mathbb{Z}^{<0} : \langle i,n \rangle \in U \}$ and $\{ n \in \mathbb{Z}^{>0} : \langle i,n \rangle \in V \}$ are both infinite, and so $\langle i,0 \rangle \in \overline{U} \cap \overline{V}$.

Here's yet another example based on the examples here. For each integer $n$, let $I_n$ denote the open interval $(n,n+1)$.

Let $X = \mathbb{R} \cup \{p_0,p_1\}$, where $p_0$ and $p_1$ are distinct points not belonging to $\mathbb{R}$. Topologize $X$ by declaring $U \subseteq X$ to be open if and only if the following hold:

• $U \cap \mathbb{R}$ is open with respect to the standard topology on $\mathbb{R}$
• If $p_0 \in U$, then $U$ contains all but finitely many of $I_0,I_2,I_4,\ldots$
• If $p_1 \in U$, then $U$ contains all but finitely many of $I_1,I_3,I_5,\ldots$

This space is Hausdorff. However, if $U$ is an open neighbourhood of either $p_0$ or $q_0$, then $\overline U$ contains all but finitely many of the positive integers. Thus, neighbourhoods of $p$ and $q$ can never have disjoint closures.

• You can also replace $\mathbb{R}$ with $\mathbb{Q}$ and the $I_n$ by their rational counterparts to get an example $X$ that is also countable and totally disconnected (having only singleton connnected subsets). This is what is done under the link. Aug 11, 2014 at 18:00