1 answer
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| whybecause | (no comment) | Dec 30, 2021 at 22:47 |
By Cantor's diagonal argument, there can be no bijection between the power set of any set and the set itself. This is true for finite and infinite sets. So taking the power set ($\mathcal P$) of a set of some infinite cardinality always gives you a set with a strictly larger infinite cardinality.
The cardinality of $\mathbb R$ is the same as the cardinality of $\mathcal{P}(\mathbb{N})$, but $\mathcal{P}(\mathbb{R})$, $\mathcal{P}(\mathcal{P}(\mathbb{R}))$, $\mathcal{P}(\mathcal{P}(\mathcal{P}(\mathbb{R})))$, etc., are an infinite sequence of sets each of which has an infinite cardinality larger than all of the previous. (These cardinalities are the first ‘several’ beth numbers.)
If you assume the generalized continuum hypothesis, which is not a conjecture waiting to be proven or disproven, but rather an axiom that can be included or excluded from the particular set theory you choose to use, then the above construction basically describes all the kinds of infinities there can be, with sufficiently clever formalization of concepts like ‘applying $\mathcal P$ an infinite number of times’. (In other words, according to the GCH, every infinite cardinality is a beth number.) If you don't assume the GCH, then the situation is complicated; there are other cardinalities that must be described by different means, and their precise nature depends on the details of the other assumptions you do or don't make with your set theory. For example, the set of all countable ordinal numbers is defined to have cardinality $\aleph_1$; this set and this cardinality exist independent of the GCH and even independent of the axiom of choice, another axiom you can take or leave. However, $\aleph_1$ might be equal to the cardinality of $\mathbb R$, or it might be smaller, or even incomparable in set theories without the axiom of choice.
0 comment threads
0 comment threads