> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. $X$ is **completely regular/Tychonoff/$T_{3\frac{1}{2}}$** if $X$ is $T_1$, and for any $A$ [[Closed Set|closed]] and $x \not\in A$, there exists a [[Continuity|continuous]] function $f \in C(X, [0, 1])$ such that $f(x) = 1$ and $f(A) = \bracs{0}$. > [!theorem] > > Let $X$ be a completely regular space, then $X$ is also [[Regular Space|regular]]. > > *Proof*. Let $A$ be closed, $x \not\in A$, and $f \in C(X, [0, 1])$ such that $f(x) = 1$ and $f(A) = \bracs{0}$. Then $f^{-1}((0.5, \infty))$ and $f^{-1}((-\infty, 0.5))$ are disjoint open sets that separate $x$ from $A$.