What is the link between the " intuitive" idea of infinity and the technical definition in set theory?

Question

The fact that a set S has a proper subset T such that there is a bijection from T to S is amazing in the same way as the fact that a set is infinite.

I know that, actually, the first property is the same as the second.

However, I do not really see the link between the intuitive idea of an infinite set and the technical definition.

What are the motivations that led mathematicians to adopt such a technical definition.

How were they led from the intuitive idea of an " endless" or " inexhaustible" set ( this may express the intuitive idea) to the official definition using " proper subset" and " bijection".

Answer 1

First, I want to repeat that any set having the bijection-with-a-proper-subset-of-itself property is indeed pretty unintuitive. Unlike, I would say, the notion of an infinite set. So, the link between the two is indeed not immediately clear ... first one would like to see how any set could possibly have that weird property.

But, this is what Cantor showed is possible. And, once it was shown to be possible, we can try and make the link to infinite sets. And it is here that you can get a bit more intuition.

In particular, I would say that it is immediately intuitively clear that any finite set does not have the property of having a bijection between itself and a proper subset. Hence, any set with the latter property has to be infinite. So, I would say that direction should be pretty intuitive.

However, why would any infinite set have this bijection-with-proper-subset property? That is still less intuitive ... but I would say the intuition here is that since the set is infinite, you can put one element aside and the resulting set remains infinite ... and hence one ought to be able to create a bijection between any infinite set and a proper subset with one element removed (along the lines of Hilbert's Hotel).

As such, we have the intuition behind the equivalence. And, with that, something we can use as a nice technical definition. Indeed, I think it is important to realize that it is not as if we started out as thinking of infinite sets as exactly those sets with this weird property: we had the concept of an infinite set far before we were thinking about bijections. However, now that we have an equivalence with a nice and hard technical property, we can use that as a 'hard' technical definition.

Answer 2

The definition you cite is due to Dedekind, and it can be shown to be equivalent to the following:

$A$ is Dedekind-infinite if and only if there is an injective function from $\Bbb N$ to $A$.

If you agree that:

1. $\Bbb N$ is inexhaustible,
2. If $A$ is inexhaustible and $f\colon A\to B$ is injective, then $B$ is also inexhaustible.

Then every Dedekind-infinite set is inexhaustible.

Let me point out assuming the axiom of choice, this is indeed a complete characterization of infinite sets. But without choice we define finite as equipotent with a bounded set of natural numbers, and infinite simply means "not finite", and it is consistent that there are infinite sets which are not Dedekind-infinite.

Nevertheless, the exhaustion argument is really about finite sets here. Namely, if $X$ is infinite, then whenever $n\in\Bbb N$ and $x_0,\dots,x_{n-1}\in X$, there is some $x\in X$ such that $x\neq x_i$ for all $i<n$.

This means that what you're really asking is why does Dedekind-infinite implies infinite. But this is trivial from the above equivalence.

Answer 3

A couple of historical remarks, and a bit more about the difference between "Dedekind finite" and "finite".

Dedekind proposed the definition in his 1888 essay, "Was sind und was sollen die Zahlen?" ("What are numbers and what should they be?", or less literally, "The Nature and Meaning of Numbers"). He just plops the definition down on the page without motivation, but in a footnote he says:

... I submitted the definition of the infinite which forms the core of my whole investigation in September, 1882, to G. Cantor and several years earlier to Schwarz and Weber. All other attempts that have come to my knowledge to distinguish the infinite from the finite seem to me to have met with so little success that I think I may be permitted to forego any criticism of them.

Cantor's most mature exposition of set theory appeared in the memoir "Beiträge zur Begründung der tranfiniten Mengenlehre" ("Contributions to the founding of the theory of tranfinite numbers", 1895-1897). He defines "infinite" as "not finite":

Aggregates with finite cardinal numbers are called "finite aggregates", all others we will call "transfinite agggregates"...

and he defines finite basically as what you get by adding in elements a finite number of times:

To a single thing $e_0$ ... corresponds the cardinal number 1... By addition of new elements we get the series of aggregates $$E_2=(E_1,e_2), E_3=(E_2,e_3),\ldots,$$ which give us successively, in unlimited sequence, the other so-called "finite cardinal numbers"...

In other words, he doesn't provide a formal definition of "infinite".

Modern axiomatic set theory (e.g., Zermelo-Fraenkel) offers a definition of finite that is probably closer to Cantor's intuitive conception. You can find the details in many places.

As Asaf Karagila remarked, without the axiom of choice (AC), Dedekind-infinite is not equivalent to this other definition. The author Lorenz Halbeisen (in Combinatorial Set Theory) uses the term "transfinite" to mean "Dedekind-infinite", and "infinite" for the other definition, a terminology I like. (Although I guess it's unhistorical.)

Without AC, transfinite is stronger than infinite. Briefly:

• $A$ is infinite iff $(\forall n\in\mathbb{N})$ there is an injection from $n$ into $A$;
• $A$ is transfinite iff there is an injection from $\mathbb{N}$ into $A$.

Or symbolically:

• $A$ is infinite iff $(\forall n\in\mathbb{N})|n|\leq|A|$;
• $A$ is transfinite iff $\aleph_0\leq|A|$.

Without going into detail, let me suggest why you need AC to show that infinite implies transfinite. We're trying to construct an injection from $\mathbb{N}$ into $A$. We're given the existence of injections from $n$ into $A$, for each $n\in\mathbb{N}$. Now, if we had a sequence of injections $f_n:n\rightarrow A$, with $f_{n+1}$ extending $f_n$ for each $n$, then we could "take the limit" and get an injection $F:\mathbb{N}\rightarrow A$. (Simply set $F(n)=f_{n+1}(n)$ for all $n$.) With a little cleverness, it's possible to obtain such a sequence of injections, if we're merely given a sequence of injections $g_n:n\rightarrow A$. (In other words, the "extension" requirement can be finessed.) And we know that for every $n$, there's a non-empty set of injections $\{g_n:n\rightarrow A\}$. The problem is to choose such a $g_n$ for each $n$. Without AC, this may not be possible.