

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer


share|cite|improve this question
up vote 8 down vote accepted



The proof can be found in



share|cite|improve this answer


share|cite|improve this answer

Suppose that $(X,\mathcal{T})$ is a topological space. Then define a relation $\prec$ on $\mathcal{T}$ by letting $U\prec V$ if and only if $\overline{U}\subseteq V$. Define a relation $\ll$ on $\mathcal{T}$ by letting $U\ll V$ if and only if there is some $(U_{r})_{r\in[0,1]\cap\mathbb{Q}}$ such that $U_{0}=U,U_{1}=V$ and $\overline{U_{r}}\subseteq U_{s}$ whenever $r<s$. Then

  1. the space $(X,\mathcal{T})$ is regular if and only if for each $U\in\mathcal{T}$, we have $U=\bigcup\{V\in\mathcal{T}|V\prec U\}$, and

  2. the space $(X,\mathcal{T})$ is completely regular if and only if for each $U\in\mathcal{T}$, we have $U=\bigcup\{V\in\mathcal{T}|V\ll U\}$.

This characterization of regular and completely regular spaces is the way that one typically defines regularity and complete regularity in point-free topology. In fact, in point-free topology it is much more convenient to use this characterization of complete regularity than it is to use the characterization of complete regularity in terms of frame homomorphisms from the real numbers to your frame.

Perhaps the most illuminating characterizations of complete regularity are the following characterizations.

Let $X$ be a topological space. Then the following are equivalent:

  1. $X$ is completely regular.

  2. $X$ has a compatible proximity.

  3. $X$ has a compatible uniformity.

Of course, proximities and uniformities impose additional external structure on topological spaces.

share|cite|improve this answer

Your Answer


