# 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?

• (You may want to specify "countably infinite", since sometimes "countable" means "finite or bijectable with $\mathbb{N}$"). The statement that every infinite set contains an infinite countable set is equivalent to the Axiom of Countable Choice, so there is no way in which you can give an "explicit procedure" without invoking at least some version of Choice. – Arturo Magidin Aug 28 '11 at 4:34
• I don't know if there is a "simpler" way; I think yours is fine (unless you want to explicitly avoid all kinds of Choice). – Arturo Magidin Aug 28 '11 at 4:35
• (cont) There other axioms that are "weaker" than the Axiom of Choice, in the sense that if you assume the Axiom of Choice, then these statements are theorems; but if you do not assume the Axiom of Choice, and you do assume these theorems, then the Axiom of Choice is not a consequence. One of them is the Axiom of Countable Choice, which says that given a countable family of nonempty sets there is a way to choose one element from each set in the family. The Axiom of Countable Choice is equivalent, in ZF theory, to "every infinite set has an infinite countable subset." – Arturo Magidin Aug 28 '11 at 4:58
• (cont) Yes: it has been proven that the Axiom of Choice (and that the Axiom of Countable Choice) is independent of the other axioms; meaning it can neither be proven nor disproven. No, we don't simply take anything we cannot prove as an axiom. You may want to ask those questions separately (rather than in comments), so some of our resident logicians can point you to good introductions/sources. – Arturo Magidin Aug 28 '11 at 5:08
• Choosing a countably infinite set from another set does require at least the Axiom of Countable Choice. Most mathematicians don't care about using the Axiom of Choice (meaning, they use it whenever it is convenient), and most work in a set theory that does include the Axiom of Choice, so that you can assume that you can pick such a subset. But, because such a "choice" does in fact invoke the Axiom of Choice, this means you cannot give any "explicit way" of choosing the subset that will work for any set (to answer your misgivings about not giving such a procedure). – Arturo Magidin Aug 28 '11 at 5:16

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.

## protected by Zev ChonolesAug 22 '16 at 7:47

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