Tychonoff Space is Urysohn Space

Theorem
Let $\left({S, \tau}\right)$ be a Tychonoff space.

Then $\left({S, \tau}\right)$ is also an Urysohn space.

Proof
Let $T = \left({S, \tau}\right)$ be a Tychonoff space.

Let $x, y \in S: x \ne y$.

From the definition of Tychonoff space:


 * $\left({S, \tau}\right)$ is a $T_{3 \frac 1 2}$ space
 * $\left({S, \tau}\right)$ is a $T_0$ (Kolmogorov) space.

From Tychonoff Space is Regular, $T_2$ and $T_1$ we use the fact that $T$ is a $T_1$ (Fréchet) space.

Clearly $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are disjoint.

From the definition of $T_1$ (Fréchet) space, we have that $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are closed.

So by definition of a $T_{3 \frac 1 2}$ space, there exists an Urysohn function for $\left\{{x}\right\}$ and $\left\{{y}\right\}$.

This is exactly the definition of an an Urysohn space.