Sequence of Implications of Separation Axioms
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Theorem
Let $P_1$ and $P_2$ be separation axioms and let:
- $P_1 \implies P_2$
mean:
- If a topological space $T$ satsifies separation axiom $P_1$, then $T$ also satisfies separation axiom $P_2$.
Then the following sequence of separation axioms holds:
Perfectly Normal | $\implies$ | Perfectly $T_4$ | $\implies$ | $T_4$ | |||||||||
$\Big\Downarrow$ | |||||||||||||
Completely Normal | $\implies$ | $T_5$ | |||||||||||
$\Big\Downarrow$ | $\Big\Downarrow$ | ||||||||||||
Normal | $\implies$ | $T_4$ | |||||||||||
$\Big\Downarrow$ | |||||||||||||
$T_{3 \frac 1 2}$ | $\impliedby$ | Completely Regular (Tychonoff) | $\implies$ | Urysohn | |||||||||
$\Big\Downarrow$ | $\Big\Downarrow$ | $\Big\Downarrow$ | |||||||||||
$T_3$ | $\impliedby$ | Regular | $\implies$ | $T_{2 \frac 1 2}$ (Completely Hausdorff) | |||||||||
$\Big\Downarrow$ | $\Big\Downarrow$ | ||||||||||||
Semiregular | $\implies$ | $T_2$ (Hausdorff) | |||||||||||
$\Big\Downarrow$ | |||||||||||||
$T_1$ (Fréchet) | |||||||||||||
$\Big\Downarrow$ | |||||||||||||
$T_0$ (Kolmogorov) |
Proof
The relevant justifications are listed as follows:
- Perfectly Normal implies Perfectly $T_4$ implies $T_4$ by definition.
- Completely Normal implies $T_5$ by definition.
- Completely Regular (Tychonoff) implies $T_{3 \frac 1 2}$ by definition.
- Semiregular implies $T_2$ (Hausdorff) by definition.
$\blacksquare$
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