# Regular open sets and semi-regularization.

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)

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:

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

and

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

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

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

Here monotonicity of $$\varrho_T$$ and $$\varrho_T(\operatorname{cl}_T(M)) = \varrho_(M)$$ as well as $$U \subset \varrho_T(U)$$ for $$U\in T$$ (where $$T$$ is an arbitrary topology) 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. $$\mathcal{R}_\tau \subset \mathcal{R}_{\tau'}$$.

• 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 $\varrho_\tau(U)$? – Nino 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! – Nino Jul 7 '13 at 16:39
• @MadHatter Currently I don't see why that should be obvious. Either I saw something then that I don't see now, or I jumped ahead then. The inclusion $\varrho_{\tau}(S) \subset \varrho_{\tau'}(\varrho_{\tau}(S))$ is all we need at that place, and that follows almost immediately from the $\tau'$-openness of $\varrho_{\tau}(S)$ (since $M \subset \overline{M}$, openness of $M$ gives $M \subset \varrho(M)$ in every topology). The reverse inclusion, i.e. $\varrho_{\tau'}(\varrho_{\tau}(S)) \subset \varrho_{\tau}(S)$ for $\tau'$-regular open $S$, is then an immediate consequence of the whole… – Daniel Fischer Oct 6 at 20:23
• …modified line (since the outer terms are equal, all ${\subset}$s are actually ${=}$s). Alternatively, it follows from what I prove afterwards (independently), that the $\tau$-closure and the $\tau'$-closure coincide for $\tau$-open sets [here $\varrho_{\tau}(S)$ is even $\tau'$-open]. But I think replacing the ${=}$ in question with ${\subset}$ is the clearest thing. Thanks. – Daniel Fischer Oct 6 at 20:24