If a topology contains all infinite subsets, then it is the discrete topology

I am solving question $8$ of exercises in section 1.1. of chapter 1 in "Topology without tears". The question reads as follows.

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 the discrete topology

My idea was to show that every singleton set is in the topology and thereby the topology is discrete.

Choose an infinite set $A$ from $X$ such that $A^c$ is also infinite. $A$ and $A^c$ belong to $\tau$ since they are infinite sets. Now consider any $x \in X$. Note that $A \cup \{x\}$ and $A^c \cup \{x\}$ belong to $\tau$ since they are infinite sets. Hence, their intersection which is nothing but $\{x\} \in \tau$ for every $x \in X$.

The problem I have is I don't know how to prove the first sentence in the previous paragraph. These are my line of thoughts to prove them.

• If $X$ is a countably infinite set, then I can list the elements are let the odd numbered elements fall into $A$. This guarantees $A$ and $A^c$ are both infinite.
• If $X$ is uncountable, then I can choose a countably infinite subset and call it $A$.

Are the above arguments rigorous? I am not especially happy with my second argument of choosing a countably infinite subset from an uncountable set since I do not give an explicit procedure of constructing the set $A$.

Also, is there any other simpler way of answering the original question?

Solution 1:

The argument is perfect, and Arturo already has explained you that if you use the (countable) axiom of choice you can always ensure that there is such a set $A$. Also, your reasoning for the existence of such a set $A$ is correct when formalized in ZFC, using the axiom of choice (but of course, in the second case, you must show using AC that every uncountable set has a countably infinite subset).

Let me note here that what you have done is the best possible, that is, there is no way to get rid of AC, and moreover that in order to prove it, you must have a subset A of X such that both $A$ and $A^c$ are infinite.

It's consistent with ZF that there exists an amorphous set X (that is, an infinite set X such that for every subset $A\subseteq X$ either $A$ or $X-A$ is finite). Then you can have a topology over X such that all infinite subsets of X are open, but the topology is not the discrete topology.

Let $\tau$ be the topology on X consisting exactly of the infinite subsets of X, toghether with the empty set.

Let's see that this is a topology. Indeed, $X$ and $\emptyset$ are in $\tau$ by definition (since X is infinite), and union of open sets is open, for the union of an infinite set with anything else is also infinite.

Suppose now that A and B are infinite subsets of X. Suppose $A \cap B$ is finite. Then $X-(A \cap B) = (X-A) \cup (X-B)$ must be infinite, since X is infinite. But since X is amorphous and $A$,$B$ are infinite, both $X-A$ and $X-B$ are finite, hence $(X-A)\cup(X-B)$ must also be finite, a contradiction. Thus if $A,B$ are infinite so is $A \cap B$ and this proves that intersection of two open sets is open.

Then, (X,$\tau$) is a topological space with all infinite subsets of X open, yet no finite subset of X open (except the empty set), and in particular not discrete.