Let $X$ an infinite $T_1$ space, then exist some subspace homeomorphic to $(\Bbb N,\tau)$ where $\tau$ is discrete or cofinite

Question

My attempt to prove the statement of the title: if $X$ is infinite and $T_1$ with topology $\tau_1$ then any basis for $X$ is infinite too.

If $X$ is $T_1$ then for any $x, y\in X$ with $x\ne y$ exists some open sets $U$ and $V$ such that $x\in U \land y\notin U$ and $x\notin V \land y\in V$.

$\color{red}{(1)}$ If you extend this analysis for a finite subset $F\subset X$ you can create a collection $\mathcal{U}$ of open sets where $\forall x_i,x_j\in F,\ i\ne j,\ \exists U_i\in\mathcal{U}:\ (x_i\in U_i)\land (x_j\notin U_i)$. Then $F\bigcap(\bigcap_{i=0}^{n} U_i)=\varnothing$ and $\mathcal {U}=\{U_i\}$ but $\bigcap_{i=0}^{n} U_i\in\tau_1$ due the definition of topology (and remember $F$ is finite).

Then:

1. If exists some countable infinite collection of disjoint open sets for the basis of $X$ then if we take some point belonging to any of these disjoint basic open sets then the resulting subspace is obviously homeomorphic to $\Bbb N$ with the discrete topology.

2. If doesnt exist an infinite collection of disjoint basic open sets we have that exist an infinite collection of non-disjoint open sets. If we take infinite countable points each one belonging to a different but non-disjoint open set and cause the space is $T_1$ we have that for every open set at most a finite number of points ($F$ in the expression $\color{red}{(1)}$) not belong to them cause the statement on $\color{red}{(1)}$, so the subspace is homeomorphic to $\Bbb N$-cofinite.

Question:

1. I feel my proof correct but not enough clear, maybe you can point how to clear it or if it lacks something?

2. The expression on $\color{red}{(1)}$ is enough understandable?

Thank you in advance.

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EDITION: the expression $\color{red}{(1)}$ is very hard to write it correctly to me by now, so Im going to describe in words: you can extend the definition of $T_1$-space not only to some $x,y\in X$ if not to $x_1,x_2,x_3,...,x_n\in X$. If you make all considerations then exist some $U_i$ for every $x_i$ where $x_i\in U_i$ but the others points does not belong to $U_i$. These collection of $U_i$ I call $\mathcal{U}$.

The intersection of every $U_i$ belongs to $\tau_1$ because is a finite intersection of open sets, and no one $x_i$ belongs to this intersection, obviously, and these $x_i$ are finite. This is what I wanted express on $\color{red}{(1)}$, sorry for the inconvenience :S

Answer 1

Personally, I don't understand the $2.$ in your approach. The expression $\color{red}{(1)}$ is understandable, but the second part of your proof I do not understand.

An attempt, which is quite possibly wrong, so please take it with a grain of salt:

1st case - there exists a countable infinite subset $A$, such that it has cofinite topology, in that case we are done.

2nd case - such a subset does not exist, therefore in each infinite countable subset, there exists an element and open neightborhood of that element, such that it is not cofinite.

When we now take any countable infinite subset $A$, we can pick an element that has a non-cofinite open neightborhood, and we can call these $a_1$ and $U_1$. Consider the set $A \setminus U_1$ - it is still an infinite, countable subset, so we can again pick an element $a_2$ and an open neightborhood $U_2$, such that $(A \setminus U_1) \setminus U_2$ is infinite. By taking $U_2' = U_2 \cap (X \setminus \{a_1\} )$ ($(X \setminus \{a_1\} )$ is an open set in $T_1$) and $U_1' = U_1$ we have open neightborhoods of $a_1$ and $a_2$ that do not contain the other respective element.

We repeat this process, and define in this way a sequence $\{a_i: i \in \mathbb{N} \}$ of distinct points in $X$. For any element $a_i$ in this sequence, we have an open neightborhood $U_i'$, such that $U_i'$ contains no prior elements in the sequence (because there's finitely many of them, we can get rid from $U_i$ by using $T_1$ property of $X$), and all the further elements are defined to be from $A\setminus U_i$, therefore it doesn't contain any of them either. So the sequence has discrete topology with respect to $X$ - therefore, we have found an infinite countable subset with discrete topology, and so we are done.

Answer 2

To address the questions you asked (sorry to be so late with this, but I've only had very limited time the last couple of days):

First, your expression $\color{red}{(1)}$ can be stated as:

Given a finite set $F\subseteq X$, for each $x \in F$, there is an open neighborhood $U_x$ of $x$ such that for all $y \in F, y \ne x \implies y \notin U_x$.

You can define $\mathcal U_F = \{U_x\ |\ x \in F\}$. I put the $F$ subscript on to remind you that $\mathcal U$ is defined from knowing $F$ first.

In point 2, you seem to be saying that in the case where $X$ does not contain an infinite collection of pairwise-disjoint empty sets, we can choose a countably infinite collection of open sets $\mathcal A$ and for each $A \in \mathcal A$, a point $x_A \in A$, such that if $A \ne B$, then $x_A \ne x_B$. However, like Jake1234, I don't see how $\color{red}{(1)}$ shows that $\{x_A\ |\ A \in \mathcal A\}$ has a cofinite subspace topology.