T1 Space is T0 Space
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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