How small is infinite? [closed]

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!

closed as unclear what you're asking by Travis, Chris Godsil, Harish Chandra Rajpoot, colormegone, LeucippusDec 23 '15 at 2:40

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• What do you mean by "try to make $\omega$ smaller"? – Cameron Buie Dec 22 '15 at 22:15
• 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
• 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
• @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
• @Ángel: Actually the Cantor-Bernstein theorem does not depend on the axiom of choice. – Asaf Karagila Dec 22 '15 at 22:59

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.

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.

• Yes, so there are no infinities smaller than that? – Simply Beautiful Art Dec 22 '15 at 22:17