T1 Space is T0 Space

Theorem

Let $\struct {S, \tau}$ be a Fréchet ($T_1$) space.


Then $\struct {S, \tau}$ is also a Kolmogorov ($T_0$) space.


Proof

Let $\struct {S, \tau}$ be a $T_1$ space.

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

From the definition of $T_1$ space:

Both
$\exists U \in \tau: x \in U, y \notin U$
and:
$\exists V \in \tau: y \in V, x \notin V$


From the Rule of Simplification:

$\exists U \in \tau: x \in U, y \notin U$


From the Rule of Addition:

Either
$\exists U \in \tau: x \in U, y \notin U$
or:
$\exists V \in \tau: y \in V, x \notin V$

which is precisely the definition of a Kolmogorov ($T_0$) 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
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