I have two questions, actually. The first is as the title says: how do we know there exists an infinite cardinal such that there exists no other cardinals between it and ℵ0? (We would have to assume or derive the existence of such an object before we label it something like ℵ1.)

My second question is, can we say for certain if there’s any limit to the number of cardinals existing between ℵ0 and continuum (i.e. 2ℵ0)? I mean, how could we know that there’s not an infinite number of cardinals between the two – perhaps even more than ℵ0?

**Answer**

To prove the existence of ℵ1 we use the concept of Hartogs number. The question asks, really, why are there uncountable ordinals, since ℵ1 is by definition the least ordinal which is not countable.

Take a set of cardinality ℵ0, say ω. Now consider all the orders on ω which are well-orders, and consider the order isomorphism as an equivalence relation. The collection of all equivalence classes is a set.

Fact:If (A,R) is a well-ordered set, then there exists a unique ordinal α such that (A,R)≅(α,∈).

Map every equivalence class to the unique ordinal which is order isomorphic to the members of the class. We now have a set and all its members are ordinals which correspond to possible well-ordering of ω.

Fact:The union of a set of ordinals is an ordinal, it is in fact the supremum of the elements in the union.

Let α be the union of the set defined above. We have that α is an ordinal, and that every ordinal below α is a possible well-ordering of ω (and therefore countable).

Suppose α was countable too, then α+1 was also countable (since α+1=α∪{α}), and therefore a possible well ordering of ω. This would contradict the above fact that α is greater or equal to all the ordinals which correspond to well-orderings of ω, since α<α+1.

This means that α is uncountable, and that it is the *first* uncountable ordinal, since if β<α then β can be injected into ω, and so it is countable. Therefore we have that α=ω1=ℵ1.

Note that the above does not require the axiom of choice and holds in ZF. The collection of all well-orders is a set by power set and replacement, so is the set of equivalence classes, from this we have that the collection of ordinals defined is also a set (replacement again), and finally α exists by the axiom of union. There was also no use of the axiom of choice because the only choice we had to do was of "a unique ordinal" which is a definable map (we can say when two orders are isomorphic, and when a set is an ordinal - without the axiom of choice).

With the axiom of choice this can be even easier:

From the axiom of choice we know that the continuum is bijectible with some ordinal. Let this order type be α, now since the ordinals are well-ordered there exists some β≤α which is the *least ordinal* which cannot be injected into ω (that is there is no function whose domain is β, its range is ω and this function is injective).

From here the same argument as before, since γ<β implies γ is countable, β is the first uncountable ordinal, that is ω1.

As to why there is no cardinals strictly between ℵ0 and ℵ1 (and between any two consecutive ℵ-numbers) also stems from this definition.

- ℵ0=|ω|, the cardinality of the natural numbers,
- ℵα+1=|ωα+1|, the cardinality of the least ordinal number which cannot bijected with ωα,
- ℵβ=⋃α<βℵα, at limit points just take the supremum.

This is a function from the ordinals to the cardinals, and this function is strictly increasing and continuous. Its result is well-ordered, i.e. linearly ordered, and every subset has a minimal element.

This implies that ℵ1 is the first ℵ cardinal above ℵ0, i.e. there are no others between them.

Without the axiom of choice, however, there are cardinals which are not ℵ-numbers, and it is consistent with ZF that 2ℵ0 is not an ℵ number at all, and yet there are not cardinals strictly between ℵ0 and 2ℵ0 - that is ℵ0 has *two* distinct immediate successor cardinals.

For the second question, there is no actual limit. Within the confines of a specific model, the continuum is a constant, however using forcing we can blow up the continuum to be as big as we want.

This is the work of Paul Cohen. He showed that you can add ω2 many subsets of ω (that is ℵ2≤2ℵ0), and the proof is very simple to generalize to any higher cardinal.

In fact Easton's theorem shows that if F is a function defined on regular cardinals, which has a *very* limited set of constraints, then there is a forcing extension where F(κ)=2κ, so we do not only violate CH but we violate GCH (2ℵα=ℵα+1) in a very acute manner.

**Attribution***Source : Link , Question Author : anon , Answer Author : Asaf Karagila*