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So, I have a (maybe trivial but I do not know it) interesting question. Since it deals with set theory, I assume natural numbers are God given and Axiom of Choice, together with its equivalent forms, as trichotomy of ordinals, Zorn's lemma and so on, if needed. I am familiar with any set is bijective to an ordinal.

A set is finite if it is bijective to {1,2,...,n} for some natural number $n \in \mathbb{N}$.

A set $X$ is infinite if you have a proper subset $A \subsetneq X$ bijective to $X$.

These are my definitions of finite and infinite sets. Now, a question is:

Theorem (maybe this doesn't hold) If a set is not finite, then it is infinite.

So, if there is no bijection between a given set $X$ and $\{1,2,...,n\}$ for any $n$, then I can find a proper subset $A \subsetneq X$ of $X$ that is bijective to it.

Is this known? How is this proved? I am very curious. Sowwy if trivial or non-interesting and thankieees!

Dibidus
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  • Kronecker also claimed that the natural numbers are "god-given" , but we cannot check whether this is valid. A non-empty set $S$ is finite iff there is a natural number $n$ such that there is a bijection between $S$ and {$1,2,3,\cdots ,n$} That means if there is no such $n$ (that is the set does not have $n$ elements for some natural number $n$) , then the set must be infinite. This is well-known. – Peter May 31 '24 at 11:38
  • https://en.wikipedia.org/wiki/Dedekind-infinite_set – GhostAmarth May 31 '24 at 11:42
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    This is known and indeed "Dedekind-infiniteness" is what you want to search. I think it's easiest to first establish that "$X$ is not finite" implies "$X$ has a subset that bijects with $\omega$". You can prove this easily using trichotomy of cardinals (which I think you meant), but also by directly using Choice to inductively construct an infinite sequence of distinct elements of $X$. – Izaak van Dongen May 31 '24 at 11:55
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    well you are accepting of the axiom of choice you say. So consider for for every $n\in \mathbb N$ we select an $x_n\in A$, $A$ not finite where $x_n\ne x_{j:j< n}$. (If this were not possible then there'd have to be an $m \in \mathbb N$ where $A={x_k|k\le m}$ and thus $A$ is finite). Now let $B = A\setminus {x_0}$. Let $f:A\to B$ via $f(x_k)=x_{k+1}$ for all $x_k\in {x_n}$ and $f(x)=x$ for all $x \not \in {x_n}$. That's a bijection.... we can probably do that without the AoC .... – fleablood May 31 '24 at 11:57
  • @fleablood you can write that as an official answer, you got a like! – Dibidus May 31 '24 at 12:40
  • Thanks but I think GhostAmarth's and Izaak van Dongen's comments about Dedekind-infinite and the link to wikipedia are more thorough and authoratative. This isn't an area I know all that much about and the idea that without AoC these definitions need not be equivalent (apparently with AoC there can exist and infinite set with no countably infinite subsets) is more relevant and has more depth. – fleablood May 31 '24 at 14:59

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So the concept of finite/infinite can be approached in multiple ways:

  1. A set is called finite if it is equinumerous with a finite ordinal. Otherwise it is called infinite.
  2. A set $X$ is called Dedekind-infinite if there exists a proper subset of $X$ equinumerous with $X$. Otherwise it is called Dedekind-finite.

And then there are tens of different variants of those definitions (see the wiki article).

In ZF (without choice) one can prove that every finite set is Dedekind-finite. This is quite simple application of induction.

But without the Axiom of Choice the statement "Dedekind-finite implies finite", which is equivalent to "infinite implies Dedekind-infinite" cannot be proved. And in fact there are models of ZF where it is false.

With the Axiom of Choice it goes like this: we take an infinite set $X$, and we well-order it. Let $\theta\in X$ denote the smallest element in $X$. Next we define inductively: $\theta_0=\theta$ and $\theta_{n+1}$ to be the successor of $\theta_n$. This is a well defined sequence, because otherwise $\theta_m$ would be the greatest element in $X$ for some $m$. And then $X$ would have to be a finite ordinal through the $(\theta_n)$ sequence itself.

Then we define

$$f:X\to X$$ $$f(x)=\begin{cases} \theta_{n+1} &\text{if }x=\theta_n\text{ for some }n \\ x &\text{otherwise} \end{cases}$$

This is an injective function, but not surjective ($\theta_0$ does not belong to the image), and therefore $f(X)$ is a proper subset of $X$ equinumerous with $X$ through $f$ itself. Which completes the proof.

freakish
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  • I'd like to see (if possible) some models where it is false. It's not immediately obvious to me how or even if we can demonstrate an infinite set with no proper subsets bijective to it. (Or is it just a a matter of saying it theoretically no impossible but no example can be shown?) – fleablood May 31 '24 at 15:03
  • @fleablood you may want to read this: https://math.stackexchange.com/questions/3063329/example-of-a-set-of-real-numbers-that-is-dedekind-finite-but-not-finite without AoC the answer is never easy unfortunately. – freakish May 31 '24 at 16:32
  • Thank you. That was about what I expected. – fleablood May 31 '24 at 16:59