# A completely regular space is a Hausdorff space

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?

## 1 Answer

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 $T_2$.

• how does that imply Hausdroff? we need to show that two distinct points have disjoint open neighbourhoods right? – Abishanka Saha Feb 16 '14 at 10:54
• 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 ${1}$. – Thomas Feb 16 '14 at 10:58
• Nice one, that explains it. – Abishanka Saha Feb 16 '14 at 11:00