Tychonoff Space is Regular, T2 and T1
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Theorem
Let $\struct {S, \tau}$ be a Tychonoff space.
Then $\struct {S, \tau}$ is also:
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:
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