I'm trying to prove the following proposition:

Let $X$ be a $T_1$ space such that for any $U$,$V$ open subsets of $X$ it holds that $cl(U\cap V) = cl(U) \cap cl(V)$ Show that $X$ is $T_3$ if and only if $X$ has a basis consisting of sets that are open and closed.

I have these two results that I'm trying to apply:

A. Let $X$ be $T_1$, then the following are equivalent:

  1. $X$ is$T_3$

  2. For any $x \in X$ and any open $U \subset X$ such that $x \in X$, there exists and open set $V$ such that $x \in V \subset cl(V) \subset U$

  3. Every $x \in X$ has a neighborhood basis consisting of closed sets.

B. Let $X$ be $T_1$, then $X$ is $T_3$ if and only if for every closed $F \subset X$ there exist a family of open sets $\mathcal{B}$ such that for every $U\in \mathcal{B}, F\subset U$ and $F = \bigcap \{cl(U) |U\in \mathcal{B} \}$

It was easy to show that if such a basis exists then $X$ is $T_3$ from (A.3).

But I'm having a lot of trouble showing the other implication, especially, I don't even have a candidate for the basis. I tried "stiching together" all the neighborhood basis in (A.3) and I also tried stiching the $\mathcal{B}$ in B.

I also found this thread, which has a similar property for intersections, the difference being that I only have $cl(U\cap V) = cl(U) \cap cl(V)$ for open sets while the thread has it for any sets. If I could prove that $X$ has the discrete topology I would also have the result I'm after.

So if someone could hint be towards the basis I'm looking for (I can probably show that it satisfies the properties needed) or how to use the intersection property that would be great.

Thanks in advance.


Let $x\in X$ and $N$ a closed neighbourhood of $x$. We need to show that there is a clopen set $W$ with $x \in W \subset N$. Let $U = \operatorname{int} N$ and $V = X\setminus N$. Then $U$ and $V$ are disjoint open sets, hence $\overline{U} \cap \overline{V} = \varnothing.$ Clearly $x \in \overline{U}\subset N$, so if we can show that $\overline{U}$ is open, we are done. But if $y\in \overline{U}$, then $X\setminus \overline{V}$ is an open neighbourhood of $y$, and $X\setminus \overline{V} \subset X\setminus V = N$, so $y \in \operatorname{int} N = U$. Thus $\overline{U} \subset U$, and since the inclusion in the other direction always holds, we have $U = \overline{U}$ a clopen set.


X is extremally disconnected, ie for all open U, cl U is open.
Proof. If U open, then empty set
. . = cl(U cap X\cl U) = cl U cap cl X\cl U = cl U cap X(int cl U)
. . int cl U subset cl U subset int cl U.

Zero dimensional spaces (has base of clopen sets) are regular.
Proof is simple.

Extremally disconnected, regular spaces are zero dimensional.
A base of clopen sets for X is B = { cl U | U open }.
Proof. Assume x in open U.
. . Thus some open V with x in V, cl V subset U
. . and cl V is in B


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