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From what (little) I understand of this Wikipedia article, where $\omega$ denotes the ordinal "identified with" $\mathbb{N}$, and $\aleph_0$ is the cardinality of $\mathbb{N}$, and $\mathfrak{c}$ is the cardinality of $\mathbb{R}$:

$$\lim_{n\to\infty}n=\lim_{n\to\infty}an+b,\,\forall a,b\in\mathbb{N}$$ $$\aleph_0=a\aleph_0+b,\,\forall a,b\in\mathbb{N}$$ $$\mathfrak{c}=2^{\aleph_0}\gg\aleph_0$$ $$\omega=1+\omega\lt\omega+1\lt\omega+2\lt\dots$$

These are all infinite, or "transfinite" quantities. From algebra and calculus, I've learned to respect infinity as not a real number, and to treat it only in the sense of limits, where infinity $+,-,\times,\div$,[insert essentially any operation here] with anything else is still infinity, as otherwise paradoxes arise. I understand the bijection argument for why the cardinality of the reals is much greater than the cardinality of the naturals, and I understand the origin of the exponentiation base two, but that seems incongruous with the identities that show $\aleph_0$ remains $\aleph_0$ when operated on: e.g. $\aleph_0^2=\aleph_0\neq2^{\aleph_0}$, somehow (???) - how can we be certain that the power set has a different cardinality? - and I wonder if this is justified by the hyperreal or the surreal number system: is the $\omega$ ordinal the same $\omega$ seen in the hyperreals?

I feel like there is a formal underpinning here that I'm completely missing, that rigorously and logically treats these infinities, or these "trans"finities in different and sensible ways.

Additionally, in this article, $\omega+1>\omega=1+\omega$ is justified with by taking the union of two sets, say $S,T$, and defining this (seemingly arbitrarily?) as equivalent to finding the union of $\{0\}\times S$ and $\{1\}\times T$. I assume this is the Cartesian product, so assuming $S,T$ are copies of the naturals, T denoted by a dash, this would be the union of $\{\{0,0\},\{0,1\},\{0,2\}\dots\}$ and $\{\{1,0^{\prime}\},\{1,1^{\prime}\},\{1,2^{\prime}\}\dots\}$ which is apparently well-ordered as $\{\{0,0\},\{0,1\},\dots\{1,0^{\prime}\},\{1,1^{\prime}\}\dots\}$ and each subsequent element is strictly greater than the one before it, and so this has ordinal $\omega+\omega=\omega\cdot2>\omega$, which is apparently justified as $\{1,0^{\prime}\}$ has "no direct predecessor", implying the ordinals of the rest of $\mathbb{N}^{\prime}$ begin at $\omega$. What if we had defined $T$ as also being $\{0\}\times T$? How come the set product is valid, necessary and meaningful/non-arbitrary?

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    $\begingroup$ You seem to be mixing three different notions of infinity (and the arithmetic operations on them): Ordinal numbers, cardinal numbers, and limit-directions in real analysis. Avoid mixing these, or at least be aware that you're mixing three different things and there's no reason to expect facts or intuitions about one to apply to another. $\endgroup$ Jun 29, 2021 at 12:14
  • $\begingroup$ @AndreasBlass I am aware and I intentionally mix them to show the personal paradox for me: I would like to know how these different notions can all co-exist as valid infinities. I appreciate they are all different, with different roots and motivations, but that's basically my question - why does this all work, even given the fact that they are different notions? Do we simply have a different meaning/definition of $+\times\dots$ for the different notions? $\endgroup$
    – FShrike
    Jun 29, 2021 at 12:17
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    $\begingroup$ Yes, the definitions of the operations differ in the three cases. $\endgroup$ Jun 29, 2021 at 12:20

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As it stands, your question is a bit difficult to answer because it spans a lot of subjects, and there is no way to do them all justice in an answer box. Let me try to give you some useful pointers, and then do my best to answer your questions concretely but very concisely.

First, the main thrust of your question: yes, all the equations/inequalities that you've listed in your question are true and rigorous when interpreted in the right context. The apparent inconsistencies stem from the fact that the same notation ($+, \times$, exponentiation) is used for different operations. There are various definitions for these operations that turn out to be the same on natural numbers (or: finite sets), but different on infinite sets. For example, we could define $n^m$ by saying $n^0 = 1$ and $n^{m+1} = n \times n^{m}$, or we could define $n^m$ by saying that it is the number of functions from $\{0,\ldots, m-1\}$ to $\{0, \ldots, n-1\}$. This gives the same result, but they do not generalize in the same way.

There are three "main" ways of talking about infinity:

  • The $\infty$ from analysis/calculus, like $\lim_{n \to \infty} \frac{1}{n} = 0$. I find it most useful to either: think about this $\infty$ as a fiction for talking about what happens at very large numbers; or think about this geometrically, where the $\infty$ is a point "atop" the real numbers, like $1$ is a point atop the open interval $(0, 1)$. This one doesn't have much to do with the other two.
  • Ordinals. This is the one you're talking about when you see a lot of $\omega$s, which is the ordinal corresponding to the natural numbers, and the smallest ordinal. Informally, the ordinals come from two operations: the successor ('taking the next one'), and the limit ('taking everything we have so far'). So ordinals start at 0, and we take successors: $0, 1, 2, 3, \ldots$. Then we add them all up into $\omega$, and then we add successors again, giving $$ 0, 1, 2, 3, \ldots, \omega, \omega + 1, \omega + 2, \ldots, $$ and then we add a new number, which we end up calling $\omega \cdot 2$ (because once we have multiplication, it turns out to be equal to $\omega$ multiplied by $2$, and so on. Ordinals on Wikipedia.
  • Cardinals. These are the sizes of sets, and you know you're talking about cardinals when you see $\aleph$s. In fact (assuming ZFC), every cardinal is as a set is also an ordinal, and you can "number" all the cardinals using ordinals. This makes them closely related. However, the kind of questions you typically ask about them are very different. In particular, the operations are very different from the ordinal operations. Cardinal numbers on Wikipedia.

Apart from these two, you mention the hyperreal and surreal numbers: these are interesting objects, but also unrelated to the three main ways of talking about infinity that I described above. As far as I know (but an expert may disagree), these are a bit closer to mathematical curiosities than to standard ways of thinking about infinity.

e.g. $\aleph_0^2 = \aleph_0 < 2^{\aleph_0}$, somehow (???) - how can we be certain that the power set has a different cardinality?

Indeed. These are alephs, so we are talking about set cardinalities. $\aleph_0^2 = \aleph_0$ essentially says that the sets $\mathbb N \times \mathbb N$ and $\mathbb N$ have the same number of elements. The proof for this is pretty similar to the standard proof that $\mathbb Q$ is countable. The inequality says that there are more functions $\mathbb N \to \{0, 1\}$ than there are elements of $\mathbb N$: this is Cantor's theorem.

Additionally, in this article, $ω+1>ω=1+ω$ is justified with by taking the union of two sets, say $S,T$, and defining this (seemingly arbitrarily?) as equivalent to finding the union of $\{0\}×S$ and $\{1\}×T$.

This is not really arbitrary: when talking about ordinals, we thinking about orderings (specifically: well-orderings) of sets. It turns out to be a natural operation to "concatenate" two orders. (In fact, on the finite ordinals i.e. natural numbers, this is just addition.) The construction you mentioned just makes this formal.

What if we had defined $T$ as also being $\{0\}×T$?

The reason why we do $\{0\} \times$ and $\{1\} \times$ at all, is to avoid problems when $S, T$ have elements in common. If you already know that $S \cap T = \emptyset$, you might as well use $S \cup T$. (You would still give it the ordering where each element of $S$ comes before each element of $T$, though.)

How come the set product is valid, necessary and meaningful/non-arbitrary?

The only real way to learn why a particular definition is meaningful and non-arbitrary is to learn more about it, and to understand that these definitions allow us to concisely ask interesting questions.

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    $\begingroup$ "The only real way to learn why a particular definition is meaningful and non-arbitrary is to learn more about it, and to understand that these definitions allow us to concisely ask interesting questions." Yes, and sometimes even to answer interesting questions. $\endgroup$ Jun 29, 2021 at 13:31
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    $\begingroup$ @IdioticShrike The link between $2^{\aleph_0}>\aleph_0$ while $\aleph_0\times \aleph_0=\aleph_0$ is that we have a proof that a bijection between $\{0,1\}^{\Bbb N}$ and $\Bbb N$ cannot exist (Cantor's diagonal proof), and we have an explicit construction of a bijection $\Bbb N\times \Bbb N\to \Bbb N$. On cardinals $|A|\leq |B|$ is defined as the existence of an injection $A\to B$, thus apart from Cantor's diagonal proof and the construction, the rest is simply definitons. $\endgroup$
    – Vsotvep
    Jun 29, 2021 at 13:56
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    $\begingroup$ @IdioticShrike That any $(0,s)\in \{0\}\times S$ is smaller than any $(1,t)\in\{1\}\times T$ is how we define $<$ on $\{0\}\times S\cup \{1\}\times T$. That it is well-ordered is provable (that it's a total order is easy, that it is well-ordered follows from $S$ and $T$ being well-ordered). $\endgroup$
    – Vsotvep
    Jun 29, 2021 at 13:59
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    $\begingroup$ @IdioticShrike Ordinals could be defined as follows: $\varnothing$ is an ordinal, if $\alpha$ is an ordinal, then $\alpha\cup\{\alpha\}$ is an ordinal, if $A$ is a set of ordinals, then $\bigcup A$ is an ordinal. One can then show that $0=\varnothing$ is an ordinal, that $1=\{0\}$, $2=\{0,1\}$, ... $n=\{0,1,\dots,n-1\}$ are ordinals, and thus that the set $\bigcup \Bbb N = \omega = \{0,1,2,\dots\}$ is an ordinal. There are many ordinals without direct predecessor (e.g. $\omega+\omega$), thus this does not define $\omega$ uniquely. $\endgroup$
    – Vsotvep
    Jun 29, 2021 at 14:03
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    $\begingroup$ The meaning of $\{0,1\}^{\Bbb N}$ is the set of all functions from $\Bbb N$ to $\{0,1\}$. In cardinal arithmetic, $X^Y$ is defined as the cardinality of the set of functions from $B$ to $A$ where $X=|A|$ ($X$ and $A$ are bijective) and $Y=|B|$. Thus $2^{\aleph_0}$ is the cardinality of the set of functions from a set of cardinality $\aleph_0$, such as $\Bbb N$, to a set of cardinality $2$, such as $\{0,1\}$. You could prove that the $A$ and $B$ that you choose as representatives does not matter (basically because bijective functions concatenate to become another bijective function) $\endgroup$
    – Vsotvep
    Jun 29, 2021 at 14:06

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