Urysohn's Lemma Converse
Lemma
Let $T = \struct {S, \tau}$ be a topological space.
Let there exist an Urysohn function for any two $A, B \subseteq S$ which are closed sets in $T$ such that $A \cap B = \O$.
Then $T = \struct {S, \tau}$ is a $T_4$ space.
Proof
Let $A$ and $B$ be arbitrary closed sets of $T$ and let $f$ be an Urysohn function for $A$ and $B$.
Let $C = \hointr 0 {\dfrac 1 4}$ and $D = \hointl {\dfrac 3 4} 1$.
Since:
- $C = \openint {-1} {\dfrac 1 4} \cap \closedint 0 1$
- $D = \openint {\dfrac 3 4} 2 \cap \closedint 0 1$
each is open in $\closedint 0 1$ under the subspace topology by definition.
Hence, $\map {f^{-1} } C$ and $\map {f^{-1} } D$ are open in $T$.
Furthermore, by definition of Urysohn function, $A \subset \map {f^{-1} } C$ and $B \subset \map {f^{-1} } D$.
Also, from Preimage of Intersection under Mapping:
- $C \cap D = \O \implies \map {f^{-1} } C \cap \map {f^{-1} } D = \O$
Therefore, $T$ is a $T_4$ space.
$\blacksquare$
Also see
Source of Name
This entry was named for Pavel Samuilovich Urysohn.
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