Take the 2-minute tour ×

Prove that if X is topological space, and S is a subset of X such that has no limit points ( in particular all it´s points are isolated) then there exist for any s$\in$ S a special neighborhood $V_s$ of s that not only is disjoint from S, also disjoint from $V_y$ where $y \in S$ and distinct of s. Is this result true for topological spaces? Or at least metric spaces, or in the worst of the cases for $R^n$.

I know that this result it´s true for R, but my proof uses the order so it´s no valid in the other cases :/ .

share|improve this question

3 Answers 3

up vote 1 down vote accepted

Such a set $S$ is a closed, discrete subset of $X$. A space $X$ with the property that whenever $S\subseteq X$ is closed and discrete, there is a family $\{V_s:s\in S\}$ of pairwise disjoint open sets such that $s\in V_s$ for each $s\in S$ is said to be collectionwise Hausdorff. This is a weakening of the property of collectionwise normality, which says that every discrete family of closed subsets (not just points) can be expanded to pairwise disjoint open sets. Every paracompact space is collectionwise normal and therefore collectionwise Hausdorff, and every metric space is paracompact, so every metric space is collectionwise Hausdorff. However, there are some quite nice spaces that are not collectionwise Hausdorff. One is Bing’s so-called Example G, which is normal but not collectionwise Hausdorff.

A simpler example that isn’t quite so nice can easily be obtained from the Sorgenfrey line, whose underlying set is $\Bbb{R}$ and whose topology has as a base the set of left-closed, right-open intervals $[a,b)$ such that $a,b\in\Bbb{R}$ with $a<b$. Call this space $S$. $S$ is a GO-space, or generalized ordered space, a property equivalent to (and sometimes defined as) being a subspace of a LOTS (linearly ordered topological space). Every GO-space is hereditarily collectionwise normal, and $S$ is in addition first countable and Lindelöf, so it’s a pretty nice space. Now let $X=S\times S$, sometimes called the Sorgenfrey plane. $X$ is still pretty nice: it’s first countable and completely regular, for instance. However, $X$ is neither normal nor collectionwise Hausdorff: the antidiagonal, $\Delta=\{\langle x,-x\rangle:x\in S\}$, is a closed discrete set that can’t be expanded to a collection of pairwise disjoint open sets, and $Q=\{\langle x,-x\rangle\in\Delta:x\in\Bbb{Q}\}$ and $\Delta\setminus Q$ are disjoint closed sets that can’t be separated by disjoint open sets.

share|improve this answer

I think you want to say $V_s$ contains no point of $S$ other than $s$.

If I understand you correctly, you're asking if you can choose the sets $V_s$ so that $V_s\cap V_r=\emptyset$ for $r\ne s$?

This would not be true in general topological spaces. Let $X$ have the topology with open sets $\emptyset$ and any set containing a fixed $x$ in $X$. Then the set $A=X\setminus\{x\}$ has no limit points (if $y\in X$ consider the open set $\{y,x\}$ ). But any two nonempty open sets in $X$ have the point $x$ in common.

share|improve this answer
Thanks! For metric spaces? –  Evolution Mar 6 '12 at 1:19

The result is false in general, for example the Moore plane is completely regular space for which the property fails. On the other hand, The result is true for metric spaces.

share|improve this answer

Your Answer


