@PhilX, I resurrected this months-old thread because you have a lot of similar questions about transfinite cardinals in this and other threads and I wanted to unpack some of these issues if you're still interested.
The question of whether every infinite set is cardinally equivalent to an Aleph is actually the Axiom of Choice (AC), not the Continuum Hypothesis (CH).
First, two quibbles.
(Quibble 1) "The null set" is sometimes used as another name for the empty set. And in measure theory, a null set is any set of measure zero. I believe by "the null set" you intend to mean Aleph_0. Your usage is misleading. In no way is Aleph_0 "the" null set nor is it "a" null set.
(Quibble 2) It's also not true that all the Alephs are indexed by natural numbers. In fact they're indexed by ordinals. So after Aleph_0, Aleph_1, Aleph_2, ... we have Aleph_w, where w (lower-case omega in the literature) is the ordinal corresponding to the cardinal Aleph_0. The Alephs keep going forever, and the entire collecton of Alephs is a proper class.
Back to your question.
Every Aleph is an ordinal, by definition. An Aleph is the least ordinal that's cardinally equivalent to a given set. In other words if we have a set X, then if X is cardinally equivalent to any ordinal at all, there must be some least ordinal with that property (since any nonempty set of ordinals is well-ordered). We name that least ordinal the cardinal of that set. Since the ordinals are well-ordered, so are the cardinals: Aleph_0, Aleph_1, etc.
There is no cardinal strictly between two consecutive Alephs, by the very definition of the Alephs.
Repeating this point, which is the crux of the matter: Every cardinal is an ordinal, and the ordinals are well-ordered. So the cardinals are well-ordered. There's a first one, a second one, and so forth. Nothing in between.
If you accept AC, then any set whatsoever may be well-ordered. Therefore every infinite set is cardinally equivalent to some Aleph.
If you don't accept AC, then there is some set that can't be well-ordered, hence it is not cardinally equivalent to any ordinal, hence it's not an Aleph.
In other words if you deny AC, then there is some infinite set that can't possibly have any cardinality assigned to it at all! It's not less than some cardinal, it's not greater than some cardinal. It's just out there on the side somewhere with no sensible way of assigning it a size. People are always complaining that AC leads to counterintuitive results (Banach-Tarski, the Hat paradox, etc); but so does the negation of AC!
That's the short answer. To provide a longer explanation I'd need to write up a tutorial on ordinals, cardinals, and Alephs, I'd better ask first if this is still of interest to you or if you've sorted it all out since you wrote your post a few months ago.
Here are some Wiki links of interest. I'd suggest reading them in this order.
https://en.wikipedia.org/wiki/Well-order
https://en.wikipedia.org/wiki/Ordinal_number
https://en.wikipedia.org/wiki/Aleph_number
https://en.wikipedia.org/wiki/Cardinali ... _continuum
I also wanted to mention that you are misunderstanding CH. CH says that between Aleph_0 and 2^Aleph_0 there are no other Alephs. In other words CH says that 2^Aleph_0 = Aleph_1.
It's easy to show that the cardinality of the real numbers is 2^Aleph_0. [Ask me for the proof if you're curious.] But the question is, which Aleph is this? It could be Aleph_1 or it could be Aleph_47 or for all we know it could be Aleph_(Aleph_47).
Note that your belief (expressed in some of your other posts) that the reals have cardinality Aleph_1 is in fact a restatement of CH. It's an open question, if one is a Platonist. It's a meaningless question, if one is a formalist. Paul Cohen, who proved that AC and CH are independent of ZF, was on record as thinking that the cardinality of the reals must be far larger than Aleph_1, because taking power sets is such a powerful operation.
https://en.wikipedia.org/wiki/Paul_Cohe ... ematician)