# How to view beth number as ordinal?

The following is the snippet of the paragraph given in my lecture note:

(Note: $\beth$ is the beth-number.)

(When I introduced $\beth_\alpha$ I told you that it was to be a particular size of infinity; specifically, the cardinality of $S_\alpha$. Now I am suggesting that you can think of $\beth_\alpha$ as an ordinal (and, in particular, an initial ordinal). Isn’t there something weird going on here? In fact, there’s nothing to worry about. It’s just that initial ordinals can be used to do double-duty: although they are ordinals, they can also be used to represent the cardinalities of sets (and, in particular, the cardinalities of ordinals). So, for instance, $\omega$ is an ordinal, but it can also be used to represent a particular cardinality, and when it’s playing that role it is often called $\beth_{0_o}$, rather than $\omega$.)

If we think of the beths as ordinals, we can start characterizing some seriously big infinities. $\beth_{\beth_\omega}$, for example, is much bigger than anything we had talked about so far. And $\beth_{\beth_{\beth_\omega}}$ is bigger still.

My question is what does $\beth_{\beth_\omega}$ and $\beth_{\beth_{\beth_\omega}}$ correspond to? I can't really understand how to view beth as ordinals.

You can view $\beth_0$ as the ordinal $\omega$, but it is more difficult to understand what other beth numbers look like. (One could argue that seing what $\omega$ looks like isn't that easy either, but looking like is not a matematical notion after all.)

For instance, $\beth_1 = 2^{\aleph_0}$ is hard to describe. We know that $\omega_1 \leq \beth_1$ so $\beth_1$ contains every countable ordinal, but it is hard to say more because $\omega_1 = \beth_1$ is undecidable in ZFC (it is the continuum hypothesis). In fact, it is consistent with ZFC that all aleph numbers up to (and excluding) $\aleph_{2^{\omega}}$ are in $\beth_1$.

The situation with $\beth_{\beth_{\omega}}$ and $\beth_{\beth_{\beth_{\omega}}}$ should be even more complicated.

So the answer in my opinion is we don't know what beth-numbers other than $\beth_0$ look like as ordinals; we can only get some insight if we study a specific model of ZFC where CH is settled in some way.

In any case, in mathematics, it isn't always possible to visualize properties, and one has to rely on the language instead.

• if $\beth_1$ contained all $\aleph_\alpha$ for $\alpha < 2^{\omega}$ then it would contain $\aleph_{2^\omega}$ as well. It is consistent with ZFC that $\beth_1$ contains all aleph numbers up to and including $\aleph_{2^\omega}$, but we can't exclude the limit. – Deedlit Aug 16 '15 at 22:48
• @Deedlit No, that's incorrect - it's very easy to show that $\aleph_{2^\omega}\ge\beth_1$, so in fact $\beth_1$ can never contain $\aleph_{2^\omega}$. However, it is consistent that $\beth_1=\aleph_{2^\omega}$, in which case $\beth_1$ contains all the $\aleph$s up to $\aleph_{2^\omega}$, but not $\aleph_{2^\omega}$ itself. – Noah Schweber May 6 '16 at 0:33
• Oh right - I must have confused "contains" with "is greater than or equal to". – Deedlit May 6 '16 at 6:32