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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.

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3 Answers 3

up vote 6 down vote accepted

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.

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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 $\langle i,0 \rangle \in \overline{U} \cap \overline{V}$.

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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.

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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. –  Mike F Aug 11 at 18:00

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