A completely regular space is a $T_1$ space $X$ with the property that if $x\in X$ and $F$ is any closed subspace of $X$ which does not contain $x$ then there exists a function $f\in\mathcal{C}(X,\mathbb{R})$, such that $f(x)=0$ and $f(F)=1$. (Here $\mathcal{C}(X,\mathbb{R})$ is the class of all bounded continuous real functions on $X$).

Though it is intuitively clear, how does this imply that every completely regular space is a Hausdorff space?


In a $T_1$ space sets consisting of single points are closed sets. Just use such a set for your $\cal C$.

Edit, to reply to a comment: Now choose a continuous function $f$ like the one which is guaranteed by the regularity assumption. Then look at the sets $f^{-1}((3/4, 5/4))$ and $f^{-1}(-1/4, 1/4)$, these are disjoint open neighbourhoods of the points under consideration. Hence $X$ is $T_2$.

  • $\begingroup$ how does that imply Hausdroff? we need to show that two distinct points have disjoint open neighbourhoods right? $\endgroup$ – Abishanka Saha Feb 16 '14 at 10:54
  • 1
    $\begingroup$ Choose a continous function $f$ which separates the points (which exists by the regularity assumption) and consider $f^{-1}$ applied to disjoint open neighbourhoods of ${0}$ and ${1}$. $\endgroup$ – Thomas Feb 16 '14 at 10:58
  • $\begingroup$ Nice one, that explains it. $\endgroup$ – Abishanka Saha Feb 16 '14 at 11:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.