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Cantor's Theorem shows that there are an infinite number of distinct infinite set cardinalities, as there is at least one infinite set, and it provides a method for producing a set with a larger cardinality from another set that works even for infinite sets.

The Continuum Hypothesis (which has been shown to be neither provable nor disprovable, but for the purposes of this question let us assume that it is true) claims that there are no sets with a cardinality between that of integers and the real numbers.

Furthermore there is proof that the cardinality of the integers is the smallest of the infinite cardinalities (Infinite sets with cardinality less than the natural numbers).

And the increment provided by Cantors Theorem (the powerset) happens to take the integers and create a set with the same cardinality as the reals.

Does all this imply an infinite sequence of discrete cardinalities exist? Has this already been studied? Does this give us the potential to have "infinity numbers" With infinity 1 denoting the cardinality of the integers and infinity 2 denoting the cardinality of its powerset, and so on rather than just countable and uncountable infinities?

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    $\begingroup$ Actually, the axiom of choice lets us know that we can always find the "next higher infinity," whether the continuum hypothesis is true or not. $\endgroup$ – Thomas Andrews Jun 29 '17 at 17:13
  • $\begingroup$ @Thomas, The continuum hypothesis was included to satisfy the discrete part of the question, not that there are infinitely many infinities. Cantor's theorem implies that as well. $\endgroup$ – DaggerOfMesogrecia Jun 29 '17 at 17:17
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    $\begingroup$ I think you've misread my comment. They are already discrete without the continuum hypothesis. You can always find the next highest infinity from any infinity. $\endgroup$ – Thomas Andrews Jun 29 '17 at 17:19
  • $\begingroup$ @Thomas, Ah I will have to read more into this. I don't intuitively see how the axiom of choice implies this. Thanks for the tip though. $\endgroup$ – DaggerOfMesogrecia Jun 29 '17 at 17:21
  • $\begingroup$ Basically, if you have an infinite set $X$, you well-order $P(X)$ (which you can do with the axiom of choice) and then take the smallest $Y\in P(X)$ in that ordering such that $U=\{Z\leq Y\}$ has cardinality greater than $X$. Then $U$ is the smallest cardinality greater than the cardinality of $X$. $\endgroup$ – Thomas Andrews Jun 29 '17 at 17:25
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There's quite a few questions here, so I'm going to take them one at a time.

Does all this imply an infinite sequence of discrete cardinalities exist?

Yes. This follows just from CT's. You know that $|\mathbb{Z}|$ is infinite, and that $|P(\mathbb{Z})| > |\mathbb{Z}|$. Applying CT again you get that $|P(P(\mathbb{Z}))| > |P(\mathbb{Z})| > |\mathbb{Z}|$, and so on ad infinitum.

Has this already been studied?

Yes, extensively. Study of these infinite numbers has lead us to some really powerful discoveries about very foundational questions in mathematics.

Does this give us the potential to have "infinity numbers" With infinity 1 denoting the cardinality of the integers and infinity 2 denoting the cardinality of its powerset, and so on rather than just countable and uncountable infinities?

This one is a little trickier. The short answer is yes. Normally, these "infinity numbers" are just called the cardinal numbers. If we assume the continuum hypothesis, that is, there is no set with cardinality greater than $\mathbb{Z}$ but less than $\mathbb{R}$, your formulation is exactly correct. We will call the first infinite number $\aleph_0$ (read as "aleph null" or "aleph 0" (aleph is the first letter of the Hebrew alphabet)). Then, again assuming the CH (assuming the CH is extremely important to this) we will define $\aleph_1$ as $|P(\mathbb{Z}|$, and so on. However, the fun doesn't just stop there! Consider, for example, the following set:

$$\bigcup_{i = 0}^\infty \aleph_i$$

What is cardinality of this set? It has cardinality greater than every $\aleph_i$ for $i \in \mathbb{Z}$, so it can't be on our list. So, we will define it as a new cardinal number, $\aleph_\omega$. We can continue patterns like this, building bigger and grander infinities, which eventually get so large that they start to hold consequences for logic and foundational mathematics.

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  • $\begingroup$ Thanks for your answer! One question. From my understanding |ℵ0 ⋃ ℵ1| would be equal to |ℵ1|. So wouldn't the expression from the last part of your answer just have a cardinality equal to ℵ∞? $\endgroup$ – DaggerOfMesogrecia Jun 29 '17 at 17:26
  • $\begingroup$ Yes, but $\aleph^\infty$ is bad notation. After we've constructed $\aleph_\omega$ like I have, we can then consider $|P(\aleph_\omega)| = \aleph_{\omega + 1}$. In a similar way, we can make $\aleph_{\omega + i}$ for all $i$. Then, we make a big union like before, and get $\aleph_{2\omega}$. $\aleph_{2\infty}$ would be confusing, as people would want to turn that just back into $\aleph_{\infty}$, but this number is bigger than $\aleph_{\omega}$. e: Another reason using $\omega$ is good is that it allows you to easily see that the ordinals and the cardinals stand in bijection* *sort of $\endgroup$ – Duncan Ramage Jun 29 '17 at 17:34
  • $\begingroup$ It seems to me that something like ℵℵ0 would be more intuitive than ℵω but I suppose in the end it's just notation. $\endgroup$ – DaggerOfMesogrecia Jun 29 '17 at 17:40
  • $\begingroup$ $\aleph_{\aleph_{0}}$ does see a fair amount of use too. It tends to be used in proofs and constructions where you recursively build cardinals from cardinals, but it's fine. I think it's unused generally because $\omega$ is one stroke to $\aleph_0$'s four strokes, but otherwise the notation is the same. $\endgroup$ – Duncan Ramage Jun 29 '17 at 17:46
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You seem to be talking about the beth numbers. Assuming the generalized continuum hypothesis, these are the same as the aleph numbers.

There is a proper class of all such numbers; e.g. in addition to the sequence $\beth_0, \beth_1, \beth_2, \ldots$ of cardinalities you get by iterating power sets, these numbers continue on to the even larger cardinal number

$$ \beth_\omega = \sup_{n \in \mathbb{N}} \beth_n $$

and then $\beth_{\omega + 1}$ is the cardinality of the power set of $\beth_\omega$, and so forth.

The indexes for these sequences is the collection of all ordinal numbers.


Incidentally, "uncountable" is an adjective that refers to all cardinalities greater than that of the natural numbers. So "countable" and "uncountable" exhaust all possible cardinalities, but there are many distinct uncountable cardinal numbers.

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  • $\begingroup$ Ah yes! I continued my research after asking my question and stumbled across the Aleph numbers which really showed me a lot about what I was looking for. But I hadn't found the Beth numbers yet, thanks! $\endgroup$ – DaggerOfMesogrecia Jun 29 '17 at 17:19
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See cardinal numbers, and probably also the ordinal numbers. This is a rich subject and your question is much too vague, but these are a good start.

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