# Definition of completely regular space.

A regular space is said to be completely regular space if for a closed set $A$ there exists a continuous function$$f: X \to [0,1]$$ such that$$f(A^c)=0, \quad f(A)=1.$$

Is this definition valid for completely regular space? If it is not correct then kindly state the correct one.

• – Angina Seng Apr 20 '18 at 4:43

A quantor over points not in $$A$$ is missing, and the space is not necessarily assumed to be regular:

A space $$X$$ is called completely regular iff

for all closed subsets $$A \subseteq X$$ and all $$x \notin A$$, there exists a continuous function $$f: X \to [0,1]$$ such that $$f(x) = 1$$ and $$f[A] \subseteq \{0\}$$.

We cannot ask for $$f[X\setminus A] = \{0\}$$! This would fail for almost all spaces, as it would imply that every closed $$A$$ is both closed and open.

A completely regular space is automatically regular (we can separate points and closed sets). We often assume $$X$$ is $$T_1$$ (points are closed) as well, and then the property can be denoted $$T_{3\frac12}$$. So if regular in your book implies points are closed, then demand that too for completely regular.

No. The definition is that for every closed set $A$ and point $x\notin A$, there is a continuous $f$ from $A$ to $[0,1]$ with $f(x)=0$ and $f(A)=1$.

What you give is the definition of a space in which every closed set is also an open set....i.e., a space in which each connected component is open and inherits the trivial (indiscrete) topology.