“Every infinite set contains a countably infinite subset.”

In ZFC (ZF + the axiom of choice), there are many equivalent definitions of infinite sets... A set is Dedekind-infinite if and only if it has a countably infinite subset. However, without the axiom of choice, there may exist infinite sets that are not Dedekind-infinite; such sets need not contain a countably infinite subset.

Discussing constructive set theories and the axiom of choice, the entry notes that many familiar results about infinite sets use some form of choice: "Certain classical theorems, such as that every infinite set has a countably infinite subset, are classically equivalent to weak forms of the axiom of choice." It continues that in weaker systems without such choice principles, "it can fail that every infinite set is Dedekind-infinite or contains a countable subset."

Theorem. Every infinite set has a countably infinite subset. ... This theorem depends on the Axiom of Countable Choice. Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms. As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.

Definition 9.23. We say a set S is countably infinite when S has the same cardinality as N. ... Theorem 9.27. The set S of subsets of the set N of natural numbers is not countably infinite.

Theorem 1 If X is infinite there is an injective map from N to X. ... Theorem 3 An infinite subset of a countable set is countable. ... Theorem 4 Let S(X) be the set of all subsets of a set X. Then there is an injective map from X to S(X) but there is no surjective map from X to S(X). In particular there are infinite sets that are not countable. Proof. The map x ↦ {x} is an injective map from X to S(X).

In models of ZF where the axiom of choice fails, one can find infinite sets that are Dedekind-finite. These are sets which are infinite in the sense of having more than n elements for every natural number n, but which have no countably infinite subset and are not in bijection with any of their proper subsets.

The axiom of choice becomes important when one needs to prove the existence of a set with an arbitrarily chosen element from an infinite collection of other sets. Indeed, on the basis of the properties of set theory we have discussed up to now, it is not possible to prove this theorem. ... Therefore, one assumes that the axiom (or its weaker form) holds, and takes the existence of the set needed for the proof as granted.

Without the Axiom of Choice it is possible that there are infinite sets which are not Dedekind-infinite. Such Dedekind-finite infinite sets, if they exist, have no countably infinite subset. Various constructions due to Tarski and others show that the existence of Dedekind-finite infinite sets is consistent with ZF.

In the accepted answer, set theorist Andreas Blass writes: "It is consistent with ZF that there exists an infinite set with no countably infinite subset. In fact, the statement 'every infinite set has a countably infinite subset' is equivalent, over ZF, to the axiom of countable choice." He points to standard independence results from ZF showing that without some choice, there can be infinite sets all of whose countable subsets are finite (hence no countably infinite subset).

We say a set is “Dedekind-finite” if there is no bijection between it and a proper subset. A set is “Dedekind-infinite” if it is not Dedekind-finite. If we assume the Axiom of Choice, being Dedekind-finite is equivalent to being in 1–1 correspondence with a natural number. However, without the Axiom of Choice, there may exist infinite Dedekind-finite sets; Dedekind-finite sets that contain more than n elements for all n ∈ ℕ. 1. There is no 1–1 correspondence between X and a proper subset of X. 2. X has no countably infinite subset.

Definition. A set is A Dedekind infinite if A is equinumerous with a proper subset of itself, and Dedekind finite otherwise. Theorem. A set is Dedekind infinite if and only it is infinite. (This theorem is proved in the setting of ZFC, that is, ZF with the axiom of choice, where every infinite set has a countably infinite subset.)

A set is said to be countably infinite if it is infinite and countable, i.e., if its elements can be put into one-to-one correspondence with the natural numbers. ... Every infinite set contains a countably infinite subset; that is, there exists an injection from N into any infinite set.

A set is said to be Dedekind infinite iff it is equinumerous with a proper subset of itself. (Contrast this with the definition that a set is infinite if it is not equinumerous with any natural number.) In ZFC these two notions coincide, since every infinite set has a countably infinite subset. But in ZF without Choice there can be infinite sets with no countably infinite subset, i.e. infinite Dedekind-finite sets.

In ZF, ‘every infinite set is Dedekind-infinite’ is strictly weaker than the axiom of choice and is not provable. There are models of ZF with infinite Dedekind-finite sets; these are infinite sets which do not contain a countably infinite subset.

The question: "Does every infinite set have an infinite denumerable subset?" ... One answer: "Yes, but in ZF (without the axiom of choice) this is not provable in general. It is equivalent to the axiom that every infinite set is Dedekind-infinite, or to the axiom of countable choice. There are models of ZF in which there exists an infinite set with no countably infinite subset."

Question: "Is it consistent with ZF that there exists an infinite set with no countably infinite subset (i.e., no Dedekind-infinite subset)?" Accepted answer: "Yes. It is consistent with ZF that there exists an infinite set which is not Dedekind-infinite. In such a model, there is an infinite set all of whose subsets are finite, so in particular it has no countably infinite subset. This shows that 'every infinite set has a countably infinite subset' is not provable in ZF and is in fact equivalent to a weak form of choice."

In Theorem 6, Suber writes: "No infinite set has a smaller cardinality than a denumerable set (or, none is smaller than the set of natural numbers)." The proof: "Let S be a set of any infinite cardinality. Clearly we may take away one of its members, call it S1, without emptying S. We may take another, S2, and another, S3, and so on, each time without emptying S. In this way we may carve out a denumerable proper subset from S, namely, {S1, S2, S3...}. But S had any infinite cardinality. Hence all infinite sets have at least one denumerable proper subset."

It is consistent with ZF that there are infinite Dedekind-finite sets, i.e. infinite sets that do not contain any countably infinite subset. Such sets show that ‘every infinite set has a countably infinite subset’ cannot be proved in ZF alone.

In ZFC, it is a standard result in basic set theory that an infinite set is Dedekind-infinite if and only if it contains a countably infinite subset. Under the Axiom of Choice (or the weaker axiom that every infinite set is Dedekind-infinite), every infinite set has a countably infinite subset. However, the existence of models of ZF with infinite Dedekind-finite sets shows that the statement ‘every infinite set contains a countably infinite subset’ is not provable in ZF alone.

One answer (assuming some choice) gives the standard argument: "Let S be an infinite set. Then we can choose a0 in S, a1 in S \ {a0}, a2 in S \ {a0,a1}, and so on. Since S is infinite, this process never terminates, and {a0, a1, a2, ...} is a countably infinite subset of S." Another user comments: "Note that this proof uses a choice function; over ZF, the statement 'every infinite set has a countably infinite subset' is equivalent to the axiom of countable choice and thus not provable without some form of choice."

We defined that a set is Dedekind finite if it is not equinumerous with any proper subset, and otherwise it is Dedekind infinite. This classificatory scheme coincides with the usual notion of finite and infinite when one assumes the axiom of choice, since every infinite set then has a countably infinite subset. But in some models of ZF there are infinite Dedekind-finite sets, which are infinite yet have no countably infinite subset.

Question: "Every infinite set has a (a) countable subset (b) uncountable subset (c) both (d) none." Solution: "Quick reasoning: Every infinite set must have at least a countable subset, because you can always pick an infinite sequence of distinct elements from it, which can be listed like the natural numbers. So option (a) is correct."

In this lecture on elementary set theory, the instructor states (translated from Hindi commentary): "If we have an uncountable set, we can take it itself as a countably infinite subset? No, but we can always select elements one by one to form a sequence; every infinite set has at least one countably infinite subset." The video presents the standard step-by-step construction assuming we can repeatedly choose new elements from the infinite set, without discussing the role of the axiom of choice.