Tychonoff Space is Urysohn Space

Theorem

Let $\struct {S, \tau}$ be a Tychonoff space.

Then $\struct {S, \tau}$ is also an Urysohn space.

Proof

Let $T = \struct {S, \tau}$ be a Tychonoff space.

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

From the definition of Tychonoff space:

$\struct {S, \tau}$ is a $T_{3 \frac 1 2}$ space

$\struct {S, \tau}$ 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 $\set x$ and $\set y$ are disjoint.

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

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

This is exactly the definition of an an Urysohn space.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties