Countable subspace of uncountable set is indiscrete or $T_0$

Question

I'm solving the problems from "Topology Without Tears"

Let $(X,\mathcal{T})$ be a topological space, where $X$ is an infinite set. Prove that $(X,\mathcal{T})$ has a subspace homeomorphic to $(\mathbb{N},\mathcal{T}_1)$, where either $\mathcal{T}_1$ is the indiscrete topology or $(\mathbb{N},\mathcal{T}_1)$ is a $T_0$-space.

My half-proof:

$X$ can be countable or uncountable. Number of open sets in $\mathcal{T}$ can be countable or uncountable.

There are 4 cases:

a) $X$ - infinite, countable. $\mathcal{T}$ - finite

b) $X$ - infinite, countable. $\mathcal{T}$ - infinite, countable

b2) $X$ - infinite, countable. $\mathcal{T}$ - infinite, uncountable

c) $X$ - infinite, uncountable. $\mathcal{T}$ - countable

d) $X$ - infinite, uncountable. $\mathcal{T}$ - uncountable

Cases:

c) Let's consider the case when $X$ - uncountable and $\mathcal{T}$ - countable.

1. We can build a partition $\mathcal{P}$ (not necessarily open sets) of $X$ such that for any $p \in \mathcal{P}$ and for any open set $s \in \mathcal{T}$, the intersection $p \cap s = p$ or $\emptyset$. Also number of elements of $\mathcal{P}$ is less than or equal to $\mathcal{T}$, hence $\mathcal{P}$ is countable.

2. Since $X$ is uncountable (by definition) and $\mathcal{P}$ has countable number of elements, then one element $p \in \mathcal{P}$ is an uncountable subset of $X$.

3. Subset $p$ is uncountable and for any open set $s \in \mathcal{T}$, $p \cap s = p$ or $\emptyset$. Then there is an infinite countable subset $n \subset p$ and subspace $(n, \mathcal{T}_n)$ with topology induced by $\mathcal{T}$ which is homeomorphic to $(\mathbb{N},\mathcal{T}_1)$. And $\mathcal{T}_n$ is indiscrete.

a) $X$ - infinite, countable. $\mathcal{T}$ - finite.

We can build a indiscrete subspace using algorithm similar to case c). If $X$ is infinite and $\mathcal{T}$ is finite. Then at least one open set is infinite and we can find a indiscrete subspace.

b2) X - infinite, countable.  - infinite, uncountable

My intuition tells me that $X$ has an infinite countable subset with discrete topology => we can find $T_0$ subspace.

I don't know how to prove b) and d). If $X$ is a $T_0$ space than it's easy to prove. The counter example is a product of R*R with usual and indiscrete topology (which is not $T_0$) - in this case if we take one point from every indiscrete R then we will get $T_0$ subspace.

Is my overall reasoning correct? May be there is an easy proof?

Answer 1

For $x\in X$ let $\mathcal N(x)=\{U\in\mathcal T:x\in U\},$ i.e., $\mathcal N(x)$ is the set of all open neighborhoods of $x.$

Call two points $x,y\in X$ equivalent if $\mathcal N(x)=\mathcal N(y).$ This is clearly an equivalence relation. We consider three cases.

Case 1. Some equivalence class $A$ is infinite.

The induced topology on $A$ is indiscrete. Since $A$ is infinite, it has a subset $B$ which is countably infinite. The subspace $B$ is homeomorphic to $\mathbb N$ with the indiscrete topology.

Case 2. There are infinitely many equivalence classes.

We get an infinite $\text T_0$-subspace $A$ of $X$ by choosing one point from each equivalence class. A countably infinite subspace of $A$ will be homeomorphic to $\mathbb N$ with a topology making it a $\text T_0$-space.

Case 3. Each equivalence class is finite, and the number of equivalence classes is finite.

Then $X$ is finite, contradicting our assumption that $X$ is infinite.

Answer 2

Notation: $[S]^2$ is the set of $2$-member subsets of $S,$ for any set $S.$

The closure bar will denote closure in $X.$

There is an elementary theorem in infinitary combinatorics which set-theorists write as $$\omega\to (\omega)^2_2$$ which means that for any infinite $X$ and any function $f:[X]^2\to \{0,1\}$ there is a countably infinite $Y\subset X$ such that $$\{f(u):u\in [Y]^2\}=\{0\} \lor \{f(u):u\in [Y]^2\}=\{1\}.$$

For $u=\{p,q\}\in [X]^2$ let $f(u)=0$ if $(p\in \overline {\{q\}} \land q\in \overline {\{p\}}).$ Otherwise let $f(u)=1. $

Let $Y$ be a countably infinite subset of $X$ such that $\{f(u):u\in [Y]^2\}=\{j\},$ where $j=0$ or $j=1.$

If $j=0$ then $p\in \overline {\{q\}}$ for all $p,q\in Y$ so $Y$ is an indiscrete subspace of $X.$

If $j=1$ then for every $\{p,q\}\in [Y]^2$ we have $(\;p\not \in \overline {\{q\}} \lor q\not \in \overline {\{p\}}\;),$ so $Y$ is a $T_0$ subspace of $X$.

Answer 3

I'm going to attempt a less elegant proof, so please excuse me. Since I'm also working through Topology Without Tears, I'm going to evade using equivalence classes, as that hadn't been brought up in the text thus far.

Let ($X$,$\tau$) be a topological space, where $X$ is infinite. We are required to prove that ($X$,$\tau$) has a subspace homeomorphic to ($\mathbb N$,$\tau$$_1$), where $\tau$$_1$ is either indiscrete or $T$$_0$.

Case 1:

There exists finite pairs $x$,$y$ $\in$ $X$, such that $x$ $\in$ $U$ and $y$ $\notin$ $U$, for some open subset $U$. This topology is not $T$$_0$. But since the space is infinite we can find an induced topology on a countable infinite subspace $S$ $\cap$ $\tau$ $=$ $S$, which is indiscrete.

Case 2:

There exists infinite pairs $x$,$y$ $\in$ $X$, such that $x$ $\in$ $U$ and $y$ $\notin$ $U$, for some open subset $U$. This topology is $T$$_0$. We can find an induced topology on a countably infinite subspace that is is also $T$$_0$. This subspace is homeomorphic to $\mathbb N$.

Thus we have shown $\tau$$_1$ is either indiscrete or $T$$_0$, as required.