Discrete Space satisfies all Separation Properties
Jump to navigation
Jump to search
Theorem
Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ fulfils all separation axioms:
- $T$ is a $T_0$ (Kolmogorov) space
- $T$ is a $T_1$ (Fréchet) space
- $T$ is a $T_2$ (Hausdorff) space
- $T$ is a semiregular space
- $T$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space
- $T$ is a $T_3$ space
- $T$ is a regular space
- $T$ is an Urysohn space
- $T$ is a $T_{3 \frac 1 2}$ space
- $T$ is a Tychonoff (completely regular) space
- $T$ is a $T_4$ space
- $T$ is a normal space
- $T$ is a $T_5$ space
- $T$ is a completely normal space
- $T$ is a perfectly $T_4$ space
- $T$ is a perfectly normal space
Proof
We have that a Standard Discrete Metric induces Discrete Topology.
Then we use Metric Space fulfils all Separation Axioms.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $6$