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In a Hausdorff space $(X,\tau)$, we can generate a coarser topology, say $\tau'$, by taking its base to be the family of regular open sets in $(X,\tau)$. (Semi-regularization of $(X,\tau)$)

Given that it's already proven, how to we proceed to prove that the "regular open sets in $(X,\tau)$ are same as the regular open sets in $(X,\tau')$"?

NOTE: A set $S$ is called 'regular open' if $S = \mathrm{Int}(\overline{S})$, where $\overline{S}$ denotes the closure of the set.
(Question 1.7.8 - General Topology by Engelking)

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up vote 2 down vote accepted

The key is that $\varrho \colon \mathfrak{P}(X) \to \mathfrak{P}(X);\; \varrho(M) = \overset{\circ}{\overline{M}}$ is idempotent (in every topology).

To see that:

$$\varrho(M) \in \tau \Rightarrow \varrho(M) \text{ is an open subset of } \overline{\varrho(M)} \Rightarrow \varrho(M) \subset \varrho(\varrho(M))$$


$$\varrho(M) \subset \overline{M} \Rightarrow \overline{\varrho(M)} \subset \overline{M} \Rightarrow \varrho(\varrho(M)) \subset \varrho(M).$$

Now, let us index the regularisation operation $\varrho$ by the topology with respect to which it is done.

Let $\mathcal{R}_T = \{M \subset X \colon \varrho_T(M) = M\}$. So we want to show $\mathcal{R}_\tau = \mathcal{R}_{\tau'}$.

Let first $S \in \mathcal{R}_{\tau'}$. Since $S$ is $\tau'$-open, and $\tau' \subset \tau$, it is also $\tau$-open. Hence

$$S \subset \varrho_\tau(S) = \varrho_{\tau'}(\varrho_\tau(S)) \subset \varrho_{\tau'}(\overline{S}) \subset \varrho_{\tau'}(\operatorname{cl}_{\tau'}(S)) = \varrho_{\tau'}(S) = S.$$

Here monotnoictiy of $\varrho$ and $\varrho(\overline{M}) = \varrho(M)$ have been used, these properties are obvious or easily verifiable. Thus we have shown $\mathcal{R}_{\tau'} \subset \mathcal{R}_\tau$.

Now let $S \in \mathcal{R}_\tau$. By definition of $\tau'$, that means $S$ is $\tau'$-open, hence $S \subset \varrho_{\tau'}(S)$. For the reverse inclusion, we first show that $\operatorname{cl}_{\tau'}(S) = \operatorname{cl}_\tau(S)$ (for $\tau$-open $S$, hence in particular for $S \in \mathcal{R}_\tau$, but not in general, of course!). Since $\tau' \subset \tau$, the $\supset$ inclusion is clear.

Now let $x \notin \operatorname{cl}_\tau(S)$. By definition, that means there is a $U \in \mathcal{V}_x$ such that $S \cap U = \varnothing$. $S$ is open, hence also $S\cap \overline{U} = \varnothing$, and, since $\varrho_\tau(U) \subset \overline{U}$, a fortiori $S \cap \varrho_\tau(U) = \varnothing$. But $\varrho_\tau(U)$ is $\tau'$-open, hence $x \notin \operatorname{cl}_{\tau'}(S)$, so $\complement \operatorname{cl}_\tau(S) \subset \complement \operatorname{cl}_{\tau'}(S)$, and therefore $\operatorname{cl}_{\tau'}(S) \subset \operatorname{cl}_{\tau}(S)$.

And then we have

$$S \subset \varrho_{\tau'}(S) = \operatorname{int}_{\tau'}(\operatorname{cl}_{\tau'}(S)) = \operatorname{int}_{\tau'}(\operatorname{cl}_{\tau}(S)) \subset \varrho_\tau(S) = S$$

for $S \in \mathcal{R}_\tau$, hence $S \in \mathcal{R}_{\tau'}$, i.e. $\mathcal{R}_\tau \subset \mathcal{R}_{\tau'}$.

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Thanks! I get most of it but, I don't understand the part where you proved that, $\operatorname{cl}_{\tau'}(S) \subset \operatorname{cl}_{\tau}(S)$. How is it possible that $S \cap U = \varnothing$ implies $S \cap \varrho_\tau(U) = \varnothing$? Is not $U$ a subset of $\varrho_\tau(U)$? – Lubashan Lakshitha Jul 7 '13 at 16:24
It works for open $S$ (with respect to $\tau$). Sure, $U \subset \varrho_\tau(U)$, but, since $S$ is open, $S\cap U = \varnothing \Rightarrow S\cap \overline{U} = \varnothing$. And from that you get $S\cap \varrho_\tau(U) = \varnothing$ because $\varrho_\tau(U) \subset \overline{U}$. – Daniel Fischer Jul 7 '13 at 16:28
Got it. Thanks a million. And the comment is much appreciated. Thank you, again! – Lubashan Lakshitha Jul 7 '13 at 16:39

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