T5 Space iff Every Subspace is T4

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Theorem

Let $T = \struct {S, \tau}$ 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 $\struct {S, \tau}$ is $T_5$.

Take an arbitrary subspace $\struct {Y, \tau_Y}$ of $S$.

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

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

From Closure of Subset in Subspace we have that:

$\paren {A \cap Y}^- = A^- \cap Y$

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 $A \cap B = \O$, we obtain:

$A^- \cap B = \O$

and similarly:

$A \cap B^- = \O$

Thus we now have:

$A, B \subset S$

and:

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

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:

$V_0 \cap V_1 = \O$

and trivially:

$A \subset V_0$

and:

$B \subset V_1$

Thus $Y$ is $T_4$.

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

$\Box$


Sufficient Condition

Suppose every subspace of $\struct {S, \tau}$ is $T_4$.

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

Then because every subspace of $\struct {S, \tau}$ is $T_4$, the subspace:

$\struct {S \setminus \paren {A^- \cap B^-}, \tau_{S \setminus \paren {A^- \cap B^-} } }$

is $T_4$.


Because we have $A^- \cap B = B^- \cap A = \O$, from Set is Subset of its Topological Closure we see that:

$A \cap B = \O$

Also:

$A \subset S \setminus \paren {A^- \cap B^-}$

and:

$B \subset S \setminus \paren {A^- \cap B^-}$

Then, because:

$\struct {S \setminus \paren {A^- \cap B^-}, \tau_{S \setminus \paren {A^- \cap B^-} } }$

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


Because:

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

and similarly:

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

we find that:

$A \subset V_0$

and:

$B \subset V_1$

and:

$V_0 \cap V_1 = \O$

Thus $\struct {S, \tau}$ is a $T_5$-space.

$\blacksquare$


Sources

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