Tychonoff Space is Regular, T2 and T1

Theorem

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


Then $\struct {S, \tau}$ is also:

a regular space
a $T_2$ (Hausdorff) space
a $T_1$ (Fréchet) space.


Proof

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

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.

We have that a $T_{3 \frac 1 2}$ space is a $T_3$ space.

From the definition, a regular space is:

a $T_3$ space
a $T_0$ (Kolmogorov) space.

So a Tychonoff space is a regular space.

Then we have a regular space is a $T_2$ (Hausdorff) space.

Then we have a $T_2$ (Hausdorff) space is a $T_1$ (Fréchet) 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: Completely Regular Spaces
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