# T5 Space iff Every Subspace is T4

## Theorem

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

Then $T$ is a $T_5$ space if and only if every subspace of $T$ is a $T_4$ space.

## Proof

### Necessary Condition

Suppose $\left({S, \tau}\right)$ is $T_5$.

Take an arbitrary subspace $\left({Y, \tau_Y}\right)$ of $S$.

Let $A, B \subset Y$ be closed sets with $A \cap B = \varnothing$.

Let $R^-$ denote the closure of $R \subset Y$.

From Closure in Subspace we have that:

- $\left({A \cap Y}\right)^- = A^- \cap Y$

But as $A \subseteq Y$ we have from Intersection with Subset is Subset that $A \cap Y = A$, and so:

- $A^- = A^- \cap Y$

Since $A, B \subset S$, we look at $A^- \cap B$.

Since $A, B \subset Y$, we have:

- $A^- \cap B = A^- \cap B \cap Y \subset A^- \cap Y = A^-$.

But from Closed Set Equals its Closure we have $A^- = A$, since $A$ is closed in $Y$.

Thus we obtain:

- $A^- \cap B \subset A$

and similarly:

- $A \cap B^- \subset B$

Since $A \cap B = \varnothing$, we obtain:

- $A^- \cap B = \varnothing$

and similarly:

- $A \cap B^- = \varnothing$

Thus we now have $A, B \subset S$ and $A^- \cap B = B^- \cap A = \varnothing$.

Thus, since $S$ is $T_5$, there exist $U_0, U_1$ disjoint open sets in $S$, such that $A \subset U_0$ and $B \subset U_1$.

Then $V_i = U_i \cap Y$ is an open set in $Y$ for $i = 0, 1$.

Also $V_0 \cap V_1 = \varnothing$, and obviously $A \subset V_0$ and $B \subset V_1$.

Thus $Y$ is $T_4$.

Since $Y \subset S$ was arbitrary, we have found that every subspace of $\left({S, \tau}\right)$ is $T_4$.

$\Box$

### Sufficient Condition

Suppose every subspace of $\left({S, \tau}\right)$ is $T_4$.

Let $A, B \subset S$ with $A^- \cap B = B^- \cap A = \varnothing$.

Then since every subspace of $\left({S, \tau}\right)$ is $T_4$, the subspace:

- $\left({S \setminus \left({A^- \cap B^-}\right), \tau_{S \setminus \left({A^- \cap B^-}\right)}}\right)$

is $T_4$.

Since we have $A^- \cap B = B^- \cap A = \varnothing$, from Set is Subset of its Topological Closure we see that $A \cap B = \varnothing$.

Also:

- $A \subset S \setminus \left({A^- \cap B^-}\right)$

and:

- $B \subset S \setminus \left({A^- \cap B^-}\right)$

Then, since

- $\left({S \setminus \left({A^- \cap B^-}\right), \tau_{S \setminus \left({A^- \cap B^-}\right)}}\right)$

is $T_4$, there exist $U_0, U_1 \subset S \setminus \left({A^- \cap B^-}\right)$ disjoint open sets in $S \setminus \left({A^- \cap B^-}\right)$ such that $A \subset U_0$ and $B \subset U_1$.

Since:

- $U_0 = V_0 \cap \left({S \setminus \left({A^- \cap B^-}\right)}\right)$ for some open set $V_0 \subset S$ in $S$

and similarly:

- $U_1 = V_1 \cap \left({S \setminus \left({A^- \cap B^-}\right)}\right)$ for some open set $V_1 \subset S$ in $S$

we find that $A \subset V_0$ and $B \subset V_1$ and $V_0 \cap V_1 = \varnothing$.

Thus $\left({S, \tau}\right)$ is a $T_5$-space.

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 2$: Functions, Products, and Subspaces