# A regular space has an infinite family composed by disjoint open sets.

As the title says, I need to prove thath if $$X$$ is an infinite $$T_3$$ (here $$T_3$$ is $$T_1$$ + regularity) topological space then there exist $$\mathcal{F}=\{U_n\mid n\in\mathbb{N} \}$$ such that for all $$n\in\mathbb{N}$$, the set $$U_n$$ is open and if $$n\neq m$$ then $$U_n\cap U_n=\emptyset$$.

My attempt:

First, take two different points $$x_1,x_2\in X$$. By Hausdorfness of $$X$$, there exist $$V_1, V_2$$ a disjoint open sets such that $$x_1\in V_1$$ and $$x_2\in V_2$$. Take $$x_3\in X\setminus\{x_1,x_2 \}$$. Then, by the regularity of $$X$$, there exist $$V_3$$ and $$V_4$$ a disjoint open sets such that $$x_3\in V_3$$ and $$\{x_1,x_2 \}\subseteq V_4$$. Then take $$U_1=V_4\cap V_1$$, $$U_2=V_4\cap V_2$$ and $$U_3=V_3$$. Therefore $$x_1\in U_1$$, $$x_2\in U_2$$ and $$x_3\in U_3$$ and moreover, $$U_1\cap U_2=\emptyset$$, $$U_1\cap U_3=\emptyset$$ and $$U_2\cap U_3=\emptyset$$ and all of them are open sets. This step is like the basis of the induction.

Now, suppose that we have constructed $$U_1,U_2,\dots,U_n$$ a family of mutually disjoint non-empty open sets. Following the later construction, we can take $$x_i\in U_i$$ for $$i\in\{1,\dots,n \}$$. For $$x_{n+1}\in X\setminus\{x_1,\dots,x_n \}$$, by regularity, there exist $$W_1$$ and $$W_2$$ disjoint open sets such that $$x_{n+1}\in W_1$$ and $$\{x_1,\dots,x_n \}\subseteq W_2$$. But from here I'm stuck. What can I do? Any suggestion? Thanks.

By a classic theorem due to Ginsberg and Sand (which is not very well-known), but was proved here, if $$X$$ is any infinite topological space then $$X$$ contains a subspace homeomorphic to one of the following five spaces:

1. $$\Bbb N$$ in the trivial (indiscrete) topology $$\tau_i:=\{\emptyset, \Bbb N\}$$.
2. $$\Bbb N$$ in the lower topology, $$\tau_l:=\{\emptyset, \{n\mid n \le m\}, m \in \Bbb N, \Bbb N\}$$.
3. $$\Bbb N$$ in the upper topology, $$\tau_u:=\{\emptyset, \{n\mid n \ge m\}, m \in \Bbb N, \Bbb N\}$$.
4. $$\Bbb N$$ in the cofinite topology, $$\tau_c:=\{\emptyset, \{\Bbb N\setminus F\mid F \subseteq \Bbb N \text{ finite }\}\}$$.
5. $$\Bbb N$$ in the discrete topology, $$\tau_d:= \mathscr{P}(\Bbb N)$$.

If $$X$$ is Hausdorff (or "better"), it cannot contain spaces 1-4, as these are not Hausdorff, so it has a countable discrete subspace, which implies the existence of the required $$U_n$$ to show every $$\{n\}$$ is open in the subspace. (with some minor modifications, we can also ensure that the $$U_n$$ are also disjoint on $$X$$.)

You actually need only that $$X$$ be Hausdorff. If $$X$$ has infinitely many isolated points, we’re done, so we may as well assume that $$X$$ has only finitely many isolated points. And given that, we may as well assume that $$X$$ has no isolated points. (Why?) Now let $$x_0$$ and $$x_1$$ be distinct points of $$X$$; there are disjoint open sets $$U_0$$ and $$V_0$$ such that $$x_0\in U_0$$ and $$x_1\in V_0$$. Choose $$x_2\in V_0\setminus\{x_1\}$$; there are disjoint open sets $$U_1$$ and $$V_1$$ such that $$x_1\in U_1\subseteq V_0$$ and $$x_2\in V_0$$. In general, given $$x_n\in U_n\subseteq V_n$$ and $$x_{n+1}\in V_n$$, choose $$x_{n+2}\in V_n\setminus\{x_{n+1}\}$$; there are disjoint open sets $$U_{n+1}$$ and $$V_{n+1}$$ such that $$x_{n+1}\in U_{n+1}\subseteq V_n$$ and $$x_{n+2}\in V_{n+2}$$. Clearly the recursive construction goes through to yield points $$x_n$$ and open sets $$U_n$$ for $$n\in\Bbb N$$ such that $$x_n\in U_n$$ for $$n\in\Bbb N$$. To finish the argument, show by induction on $$n$$ that if $$0\le k, then $$U_k\cap U_n=\varnothing$$, and conclude that the sets $$U_n$$ are pairwise disjoint.

• About your "why?" in the part where you assume that $X$ has no isolated points, to complete the proof, if $X$ has a finitely many isolated points then $A=\{ x\mid x \ \text{is an isolated point of$X$} \}$ is a closed set in $X$ by Hausdorfness. But $X\setminus A$ is an infinite open subspace of $X$. We can do the construction of the infinite family of mutually disjoint open sets over $X\setminus A$ and those open sets will be open too in $X$. And the proof is complete. Commented Sep 6, 2020 at 2:32
• @CarlosJiménez: Yep, you’ve got it. Commented Sep 6, 2020 at 2:36