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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.

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2 Answers 2

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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 $X$.)

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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 $0\le k<n$, then $U_k\cap U_n=\varnothing$, and conclude that the sets $U_n$ are pairwise disjoint.

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  • $\begingroup$ 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. $\endgroup$ Commented Sep 6, 2020 at 2:32
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    $\begingroup$ @CarlosJiménez: Yep, you’ve got it. $\endgroup$ Commented Sep 6, 2020 at 2:36

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