User:Cjhanrahan

This is the sandbox for the user cjhanrahan.

Proof
Let $\tau$ be the topology on $T$.

Let $\mathbb K = \left\{{K \in \tau: K \subseteq H}\right\}$.

Then:

By the definition of closed set, $K$ is open in $T$ $T \setminus K$ is closed in $T$.

Also, from Set Complement inverts Subsets we have that $T \setminus K \supseteq T \setminus H$.

Now consider the set $\mathbb K'$ defined as:
 * $\mathbb K' := \left\{{K' \subseteq T: \left( T \setminus H \right) \subseteq K', K' \text { closed in } T}\right\}$.

From the above we see that $K \in \mathbb K \iff T \setminus K \in \mathbb K'$.

Thus:


 * $\displaystyle T \setminus H^\circ = \bigcap_{K' \mathop \in \mathbb K'} K' = \left( T \setminus H \right) ^-$

The other result can be demonstrated similarly.