T4 Property Preserved in Closed Subspace

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Theorem

Let $T = \displaystyle \left({S, \tau}\right)$ be a topological space.

Let $T_K$ be a subspace of $T$ such that $K$ is closed in $T$.


If $T$ is a $T_4$ space then $T_K$ is also a $T_4$ space.


That is, the property of being a $T_4$ space is weakly hereditary.


Corollary

If $T$ is a normal space then $T_K$ is also a normal space.

That is, the property of being a normal space is weakly hereditary.


Proof

Let $T = \left({S, \tau}\right)$ be a $T_4$ space.

Then:

$\forall A, B \in \complement \left({\tau}\right), A \cap B = \varnothing: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

That is, for any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.


We have that the set $\tau_K$ is defined as:

$\tau_K := \left\{{U \cap K: U \in \tau}\right\}$

where $K$ is closed in $T$.


Let $A, B \subseteq K$ be closed in $K$ such that $A \cap B = \varnothing$.

From Intersection with Subset is Subset‎ we have that $A \subseteq K \iff A \cap K = A$.

As $K$ is itself closed in $T$, it follows that so is $A \cap K = A$ from Topology Defined by Closed Sets.

Similarly, so is $B \cap K = B$.


Because $T$ is a $T_4$ space, we have that:

$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

As $A, B \subseteq K$ such that we have that:

$A \subseteq U \cap K, B \subseteq V \cap K: \left({U \cap K}\right) \cap \left({V \cap K}\right) = \varnothing$

From the definition of topological subspace, both $U \cap K$ and $V \cap K$ are open in $K$.

Thus the $T_4$ axiom is satisfied in $K$.

$\blacksquare$


Also see

The above proof hinges on the fact that as $K$ is closed in $T$, then $A \cap K = A$ and $B \cap K = B$ are also closed in $T$.

If $K$ is not closed in $T$, there is no such guarantee of closedness of $A$ and $B$.


Sources