Does "regular" implies collectionwise hausdorff?

A topological space is said to be collectionwise Hausdorff if given any closed discrete collection of points in the topological space, there are pairwise disjoint open sets containing the points.

I believe it and try to proof it, however I'm not sure. Thanks in advance:)


1 Answer 1


No, it doesn't. The Sorgenfrey plane is regular, as the Sorgenfrey line is normal, but it isn't collectionwise Hausdorff as $\{(-x,x)\mid x \in \mathbb R\}$ is discrete and closed, but there aren't pairwise disjoint open neighbourhoods.

To prove this, let $U_x = [-x,-x+\epsilon_x) \times [x,x+\epsilon_x)$ be a neighbourhood of $(-x,x)$ in the Sorgenfrey plane, then there is some $\epsilon > 0$ such that $E_\epsilon= \{x \in \mathbb R\mid \epsilon_x > \epsilon \}$ is uncountable (as $\mathbb R = \bigcup_n E_{1/n}$). Then $E_\epsilon$ has an accumulation point in $\mathbb R$, choose $x_1, x_2 \in E_\epsilon$ with $|x_1 - x_2| < \frac \epsilon 4$, then $U_{x_1} \cap U_{x_2}\ne \emptyset$.

To answer your addional question in the comment below: Let $X$ be regular and $A =\{a_n \mid n \in \mathbb N\}$ closed, discrete and countable. For each $n \in \mathbb N$ there are by regularity and closedness of all subsets of $A$ in $X$ disjoint open sets $U_n \ni x_n$, $V_n \supseteq A \setminus \{x_n\}$. Now let $W_n = U_n \cap \bigcap_{i < n} V_i$, then $W_n$ is an open set containing $x_n$ (as finite intersection of such). Now let $n < m$, then $W_n \subseteq U_n$, $W_m \subseteq V_n$, so $W_n \cap W_m = \emptyset$. Hence $X$ is (don't know if this is a common term, but IMHO it describes the property) countable collectionwise Hausdorff.

  • $\begingroup$ I see; to prove it by the baire property of the $R$. However, if for countable closed discrete subspace, it is true? $\endgroup$
    – Paul
    Jul 4, 2012 at 12:59
  • $\begingroup$ Hm ... have to think about it ... $\endgroup$
    – martini
    Jul 4, 2012 at 13:10
  • $\begingroup$ @John: Added sth. above, does this work? $\endgroup$
    – martini
    Jul 4, 2012 at 14:59
  • $\begingroup$ @martini: For the added bit you can also construct the $W_n$ recursively. Choose a nbhd $W_0$ of $a_0$ whose closure is disjoint from $\{a_n:n>0\}$. Given $W_k$ for $k<n$, choose a nbhd $W_n$ of $a_n$ whose closure is disjoint from $\bigcup_{k<n}\operatorname{cl}W_k\cup\{a_k:k>n\}$. $\endgroup$ Jul 4, 2012 at 17:43
  • $\begingroup$ @martini, it really help me:) $\endgroup$
    – Paul
    Jul 5, 2012 at 0:31

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