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.

  • $\begingroup$ By definition $\aleph_0$, $\aleph _1$, etc. represent the smallest infinite cardinal number, the second smallest infinite cardinal number, etc. This means they have an order. Whether or not the continuum hypothesis is true is irrelevant to this, because we are defining the aleph numbers to be ordered (by construction) $\endgroup$
    – FD_bfa
    Commented Jun 28, 2023 at 11:27
  • $\begingroup$ The part that requires the continuum hypothesis (CH) is establishing a correspondence between the aleph and beth numbers. For example, knowing that $\aleph _1 = 2^{\aleph _0}$ is not guaranteed without assuming CH $\endgroup$
    – FD_bfa
    Commented Jun 28, 2023 at 11:30
  • $\begingroup$ The reason we need CH in order to establish this link is that $2^{\aleph _0}$ is defined to be the cardinality of the power set of another set with cardinality $\aleph _0$ and so it is not straightforward to show which cardinal number this corresponds. We are thinking about how big this set really is which cannot be established without this additional assumption $\endgroup$
    – FD_bfa
    Commented Jun 28, 2023 at 11:34
  • 3
    $\begingroup$ "If the continuum hypothesis is proven, Aleph numbers and Beth numbers are the same" Actually you need the generalized continuum hypothesis for this; it's consistent that $\aleph_1=\beth_1$ but $\aleph_2<\beth_2$, for example. $\endgroup$ Commented Jun 28, 2023 at 14:26
  • $\begingroup$ The fact that the aleph numbers are well-ordered is basic; it has to be proved, but it has nothing to do with the continuum hypothesis etc. The fact that the beth numbers are well-ordered follows from the fact that the aleph numbers are well-ordered, because the beth numbers are in order-preserving correspondence with the aleph nbumbers. (Also, assuming the axiom of choice, they are a subclass of the aleph numbers.) $\endgroup$
    – user14111
    Commented Jun 29, 2023 at 6:46

1 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 $\aleph$.


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