A topological space $(X,\mathcal T)$ is said to be divisible iff for each neighborhood $U$ of the diagonal $\Delta=\{(x,x)\mid x\in X\}$ in $X\times X$, there is a symmetric neighborhood $V$ of the diagonal such that: $$V\circ V\subseteq U$$

Is every $T_4$ topological space divisible?

  • $\begingroup$ Where is the problem from? $\endgroup$ – Jonas Meyer Mar 29 '13 at 5:58
  • $\begingroup$ every hausdorff compact space is divisible. I think it may be true for every $T_4$ space. $\endgroup$ – user59671 Mar 29 '13 at 6:01
  • $\begingroup$ Neat. Do you know an example of a space that is not divisible? $\endgroup$ – Jonas Meyer Mar 29 '13 at 6:07
  • $\begingroup$ @Janas: Every $T_1$ topological space that is not $T_{3\frac{1}{2}}$ cannot be divisible. $\endgroup$ – user59671 Mar 29 '13 at 12:43
  • $\begingroup$ ({1,2},{$\emptyset$,{1},{1,2}}) is a $T_0$ divisible topological space which is not completely regular. But every $T_1$ divisible space is also $T_{3\frac{1}{2}}$ $\endgroup$ – user59671 Mar 29 '13 at 13:12

Unfortunately not, but (the way I know how) to demonstrate this is not entirely simple

Definition: A T$_1$-space $X$ is called collectionwise normal if for every discrete family $\{ F_i : i \in I \}$ of closed subsets of $X$ there is a pairwise disjoint family $\{ W_i : i \in I \}$ of open subsets of $X$ such that $F_i \subseteq W_i$ for all $i \in I$.

Theorem (H.J. Cohen, 1952): Every (completely regular) divisible space is collectionwise normal.

proof. Let $\{ F_i : i \in I \}$ be a discrete family of closed subsets of $X$. For each $i \in I$ let $U_i = X \setminus \bigcup_{j \neq i} F_j$, and let $U = \bigcup_{i \in I} ( U_i \times U_i )$. Clearly $U$ is an open neighbourhood of the diagonal $\triangle \subseteq X \times X$, and so by divisibility there is a symmetric neighbourhood $V$ of $\triangle$ such that $V \circ V \subseteq U$. For each $i \in I$ define $$W_i = \bigcup_{y \in F_i} \{ x \in X : \langle x , y \rangle \in V \}.$$ It can be shown that the family $\{ W_i : i \in I \}$ is as required. $\hspace{0.5cm}$$\Box$

(Perhaps for this reason divisible spaces are sometimes called strongly collectionwise normal.)

And there are normal spaces which are not collectionwise normal, for example Bing's $G$.

[Cohen's theorem is from his article Sur un problème de M. Dieudonné, C. R. Acad. Sci. Paris 234, (1952). 290–292.]

  • $\begingroup$ nice.‌ thanks‌‌‌‌. $\endgroup$ – user59671 Mar 29 '13 at 7:02

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