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There are a lot of posts concerning how big infinite is, but I wonder how small infinite is.

One can clearly see (ignoring a few things) that$$\frac{\infty}2=\infty$$Which means that no matter how many times I try to shrink infinite, it won't give in to my methods.

Which means its really big?

But can infinite be small?

Sure, there are smaller infinities than others, but there must be a sort of smallest infinite.

One could argue that omega is the smallest infinite, but what happens when I try to make omega smaller?

If that doesn't make much sense, think about this way. Can't I make some infinite infinitely smaller while still having it infinitely large? And then can' I make that infinitely smaller?!

Perhaps, it might help to note that this must be larger than any real number in order to still be infinitely large.

From this point of view, I can't understand how there can even be a limit on how small infinite is!

Simply Beautiful Art
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    What do you mean by "try to make $\omega$ smaller"? – Cameron Buie Dec 22 '15 at 22:15
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    Your question has no sense. In cardinal theory, you have a notion of "small" and "big" infinites, but the symbol $\infty$, at least as you are using it, is not a number. – Ángel Valencia Dec 22 '15 at 22:15
  • @Sky Yeah. That's where the (ignoring a few things) came in. – Simply Beautiful Art Dec 22 '15 at 22:15
  • @CameronBuie I guess you just make it smaller, right? – Simply Beautiful Art Dec 22 '15 at 22:16
  • @ÁngelValencia Maybe it doesn't make sense, but I think its still a valid question. – Simply Beautiful Art Dec 22 '15 at 22:17
  • http://math.stackexchange.com/questions/10085/why-is-omega-the-smallest-infty – Asaf Karagila Dec 22 '15 at 22:19
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    I don't think this question deserves so many downvotes... It's a genuine question posed by someone trying to learn. – Eff Dec 22 '15 at 22:20
  • @Eff Thank you very much for supporting me. It is an interesting thought, isn't it? – Simply Beautiful Art Dec 22 '15 at 22:22
  • @AsafKaragila I've already tried to read that. However, it does not concern how small infinity can be. – Simply Beautiful Art Dec 22 '15 at 22:22
  • @SimpleArt, as I told you, it doesn't make sense. If a such "smaller" infinite "number" $\infty_0$ exists, then you have that $\infty_0/r$ must be a finite, real number for all real numbers $r\not=0$, or you would have a "smaller" infinite than $\infty_0$. But real numbers are closed under multiplication: $(\infty_0/r)\cdotp r$ must be a real number; thus it is nonsense. – Ángel Valencia Dec 22 '15 at 22:28
  • If we considering cardinal theory, you define the (cardinal) number aleph nought, $\aleph_0$, as the cardinality of the set $\mathbb{N}$ of natural numbers. It is considered the smallest infinite number, but it hasn't been satisfactorily stated because of the limitations of mathematical logic. – Ángel Valencia Dec 22 '15 at 22:31
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    @ÁngelValencia What do you mean "it hasn't been satisfactorily stated because of the limitations of mathematical logic"? – Noah Schweber Dec 22 '15 at 22:32
  • @SimpleArt As stated on the page Asaf linked, "Yes, there is a proof that $\omega$ is the smallest infinite cardinality." That is, you cannot get smaller and still be infinite. It seems to me that is very relevant to "how small infinity can be." – David K Dec 22 '15 at 22:34
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    @NoahSchweber Cantor-Bernstein theorem can be used to stated that $\aleph_0$ is the smallest infinite cardinal number, but it depends of the Axiom of Choice. If we omit it, we might have other consequences. I think it is an explanation of what I said below. – Ángel Valencia Dec 22 '15 at 22:35
  • @SimpleArt: How would you propose to do that? Typically, $\omega$ is a very specific set. How would one make the set smaller without making it a different set? – Cameron Buie Dec 22 '15 at 22:36
  • @CameronBuie Well, what should I know? And this different set, is it still infinite? Don't we need to set conditions or something in order to determine if it becomes finite or infinite? – Simply Beautiful Art Dec 22 '15 at 22:40
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    @Ángel: Actually the Cantor-Bernstein theorem does not depend on the axiom of choice. – Asaf Karagila Dec 22 '15 at 22:59
  • @AsafKaragila I didn't know it, thank you. – Ángel Valencia Dec 22 '15 at 23:07
  • Yes, we do. I imagine that your idea is to consider a subsey of $\omega.$ The interesting thingnis that all subsets of $\omega$ are either finite or are the same size as $\omega$. In fact, most subsets of $\omega$ are as big as $\omega,$ itself! Hence, there is no infinite set that is smaller than $\omega.$ As Asaf's answer points out, though, this doesn't mean that $\omega$ need be the smallest infinite set, just that it is minimal. – Cameron Buie Dec 22 '15 at 23:10
  • @CameronBuie Interesting. But how does that define the size of $\omega$? It makes it both smaller and larger than itself. and the same size? – Simply Beautiful Art Dec 22 '15 at 23:14
  • Two sets $X$ and $Y$ are the same size as $\omega$ in the sense of cardinality if and only if there is a function $f:X\to Y$ that is one-to-one and onto. It is a peculiar quirk of infinite sets that they may have proper subsets of the same size. – Cameron Buie Dec 22 '15 at 23:52
  • @CameronBuie That does make sense actually. Heh, makes me think of the infinite hotel paradox where we can make infinite larger by mapping it into a larger version of itself. – Simply Beautiful Art Dec 23 '15 at 00:27

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Actually, in cardinal arithmetic without the axiom of choice, it is not the case that there must be a "smallest" infinity!

It is consistent with ZF that there is an infinite Dedekind-finite set - that is, a set which is infinite, but which is not in bijection with any of its proper subsets. In particular, if $X$ is such a set, then if $Y\subseteq X$ is infinite, then any infinite proper subset of $Y$ is not in bijection with $Y$, that is, is of strictly smaller cardinality. So in this sense, there are infinite sets which can be "made smaller" without end.

$\aleph_0$ (the cardinality of the set of natural numbers), however, can't - any infinite subset of $\mathbb{N}$ has the same cardinality as the whole. So $\aleph_0$ is a minimal infinite cardinal. Basically, without choice, we don't even need the infinite cardinals to be comparable in size, and all sorts of weird behavior is possible.

EDIT: As to the edited question, you should read the question and answers that Asaf Karagila has linked to (see below). Basically, the proof that $\aleph_0$ is a smallest possible infinite cardinal, and (assuming choice) is the smallest infinite cardinal, is quite accessible. The proof that without choice we can have "shrinkable infinities" is unfortunately very complicated, but I'll link to a good explanation if I can find one.

EDIT EDIT: You might try https://mathoverflow.net/questions/200003/independence-of-the-countable-axiom-of-choice. It's not exactly the same statement, but the proof structure is the same.

Community
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Noah Schweber
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Aleph numbers would be examples dealing with cardinalities that some may interpret as a way to view infinity, e.g. the cardinality of the Natural numbers, the Real numbers, etc. $\aleph_0$ would be the smallest infinite cardinal number in this case according to Wikipedia.

Point at infinity would be an example where there aren't sizes but the concept of infinity is used to create various spaces that may be interesting to consider here.

JB King
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