Aleph numbers, Beth numbers and the continuum hypothesis

Question

We have Aleph numbers which are the cardinals. I know $\aleph_0$ is the cardinality of the smallest infinite sets (countable sets). $\aleph_1$ is the lowest cardinality of sets having a cardinality bigger than $\aleph_0$ and so on.
On the other hand, we have Beth numbers. $\beth_0$ is equal to $\aleph_0$ and $\forall n\in\mathbb Z_{+},\beth_n=2^{\beth_{n-1}}$

If the continuum hypothesis (generalized) is proven, Aleph numbers and Beth numbers are the same and well-ordered obviously. Otherwise, some of Aleph numbers are not included in Beth numbers. Beth numbers remain well-ordered, but my question is how Aleph numbers are well-ordered without knowing the truth or falsehood of the continuum hypothesis (generalized). Or maybe why are Aleph numbers indexed with non-negative integers and not with non-negative real numbers?

Could someone explain it without AC or ZF or theorems like that? I am in middle school and I want to understand it without any ambiguity.

Answer 1

Define a sequence by transfinite recursion for all ordinals $\lambda$

1. $\omega_0 = \Bbb{N}$
2. $\omega_{\lambda+1} = \omega_\lambda^+$ for any ordinal $\lambda$, where $\omega_\lambda^+$ is the least ordinal bigger than $\omega_\lambda$ that has cardinality bigger than $\omega_\lambda$. Such an ordinal always exists by Hartogs' theorem. (It is not trivial - look it up if you like.)
3. $\omega_\lambda = \bigcup_{\gamma<\lambda} \omega_\gamma$ when $\lambda$ is a non-zero limit ordinal. (This is always an ordinal because the union of a set of ordinals is an ordinal.)

This definition requires other axioms of ZF, including most importantly Replacement, but it doesn't require AC, CH or GCH. The sequence $\omega_\lambda$ is strictly increasing. This means if $\beta<\gamma$ (as ordinals) then $\omega_\beta < \omega_\gamma$ (as ordinals.) $\aleph_\lambda$ is defined to be the cardinality class represented by $\omega_\lambda$ (practically $\omega_\lambda = \aleph_\lambda$). If $\beta < \gamma$ (as ordinals) then $\aleph_\beta < \aleph_\gamma$ as cardinals. The sequence of $\aleph$-s is well ordered because the ordinals are. If $\{\aleph_\beta: \beta \in I\}$ is a non-empty collection of $\aleph$-s indexed by a set of ordinals $I$ then take $\beta_0$ to the smallest ordinal in $I$ (the ordinals are well-ordered) and then $\aleph_{\beta_0}$ is the smallest cardinal in that collection of $\aleph$-s.

Now for $\beth$-s. Again, using transfinite recursion, define for all ordinals $\lambda$

1. $S_0=\Bbb{N}$
2. $S_{\lambda+1} = S_\lambda \cup P(S_\lambda)$ where $P$ denotes power set.
3. $S_\lambda=\bigcup_{\gamma<\lambda} S_\gamma$ when $\lambda$ is a non-zero limit ordinal.

Now define $\beth_\lambda$ to be the cardinality of $S_\lambda$. Notice that $S_\lambda\subset S_{\lambda+1}$ and by Cantor's theorem $\beth_\lambda < \beth_{\lambda+1}$ (as cardinalities.) From this it can be shown that for $\beta < \gamma$ (as ordinals), $\beth_\beta < \beth_\gamma$ as cardinals. Moreover, the $\beth$-s are well-ordered amongst themselves (where the order is cardinality) for the same reason that the $\aleph$-s are - which is that they're an increasing sequence (of cardinals) indexed by ordinals - and ordinals are well-ordered.

Notice however that the $\beth$-s being well-ordered amongst themselves - should not mislead you into believing they're comparable in any way to the $\aleph$-s. Without AC, it is possible that some, even all $\beth$-s (other than $\beth_0$) represent cardinalities of sets that are not well-orderable. If a $\beth$ is not well-orderable, then it may possibly be bigger than some $\aleph$-s (in cardinality) because it has well-orderable subsets - but it's not equal or smaller than any $\aleph$.