A Topology containing every infinite subset of a an infinite set is discrete.

Question

Is the following proof correct?

Proposition. Let $X$ be an infinite set and $\tau$ be a topology on $X$. If every infinite subset of $X$ is in $\tau$, prove that $\tau$ is a discrete topology.

Proof. Assume $x\in X$, since $X$ is infinite, surely there must exist an $a_1\in X$ such that $a_1\not\in\{x\}$, and by the same reasoning a $b_1\in X$ such that $b_1\not\in\{x,a_1\}$, continuing in this manner, we affirm that there exist $a_1,b_1,a_2,b_2,\dots,a_k,b_k,\dots$ in $X$ such that $$ \begin{cases} a_1\not\in\{x\}\\ b_1\not\in\{x\}\cup\{a_1\}\\ a_2\not\in\{x\}\cup\{a_1\}\cup\{b_1\}\\ b_2\not\in\{x\}\cup\{a_1,a_2\}\cup\{b_1\}\\ \vdots \\ a_k\not\in\{x\}\cup\{a_1,a_2\dots,a_{k-1}\}\cup\{b_1,b_2,\dots,b_{k-1}\}\\ b_k\not\in\{x\}\cup\{a_1,a_2\dots,a_{k-1},a_{k}\}\cup\{b_1,b_2,\dots,b_{k-1}\} \end{cases} $$ Now define $A = \bigcup_{r=1}^{\infty}\{a_r\}\cup\{x\}$ and $B = \bigcup_{r=1}^{\infty}\{b_r\}\cup\{x\}$, from hypothesis $A,B\in\tau$ and since $\tau$ is a topology on $X$, we have $A\cap B\in\tau$ as well, but from the above construction, it is evident that $\bigcup_{r=1}^{\infty}\{a_r\}\cap \bigcup_{r=1}^{\infty}\{b_r\} = \varnothing$, consequently $A\cap B = \{x\}$. In summary every $\{x\}\in\tau,\forall x\in X$, appealing to proposition $1.1.9$

$\blacksquare$

Note: $1.1.9$ is the result that any topology $\tau$ on a set $X$ containing $\{x\}\in\tau,\forall x\in X$, is a discrete topology.

Answer 1

It is fine. I would do it as follows:

Take $x\in X$. I want to prove that $\{x\}\in\tau$. To do so, I will consider two cases:

• $X$ is countable: Then $X\setminus\{x\}=\{a_1,a_2,a_3,\ldots\}$ for some bijective sequence $(a_n)_{n\in\mathbb N}$. Now, take$$A=\{x\}\cup\{a_{2n}\,|\,n\in\mathbb{N}\}\text{ and }B=\{x\}\cup\{a_{2n-1}\,|\,n\in\mathbb{N}\}.$$
• $X$ is uncountable: Then let $N$ be a countable subset of $X\setminus\{x\}$. Now, take$$A=\{x\}\cup N\text{ and }B=\{x\}\cup N^\complement.$$

In both cases, $A,B\in\tau$ (since both sets are infinite) and therefore $\{x\}=A\cap B\in\tau$.