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Let be $m,n\in\mathbb{N}$ and $n\neq m$. Show that there exists no bijection between $[m]:=\{1,\dots,m\}$ and $[n]:=\{1,\dots, n\}$.

My proof

WLOG we assume $m<n$ and that there exists a bijection $\varphi:[m]\to[n]$. We know that each $i\in[m]$ has exactly one image namely $\varphi(i)\in[n]$. We eliminate $\varphi(1)$, then we eliminate $\varphi(2)$ and so on... This process stop after $m$ times. As $m<n$ there will be elements left in $[n]$ which have no preimage. This is a contradiction, so $\varphi$ can't be bijective.

This "proof" sounds too easy. Did I miss something or is it flawed?

Asaf Karagila
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Philipp
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    Related – Anne Bauval Oct 21 '23 at 22:15
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    What counts for a rigorous proof here depends heavily on what precise definitions you are using. – spaceisdarkgreen Oct 21 '23 at 22:16
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    Otherwise known as the pigeon hole principle.. – Porky Oct 21 '23 at 22:29
  • @spaceisdarkgreen why do I need precise definitions? I thought it is clear what I am doing, there is no ambiguity – Philipp Oct 21 '23 at 22:36
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    You need to use precise definitions because you're seeking a proof. And your argument has lots of ambiguity. What do you mean "eliminate"? What is "and so on..."? Why are there elements left at the end? None of this is sufficiently clear. – blargoner Oct 21 '23 at 22:48
  • @blargoner, "eliminate" means considering $[n]$ without $\varphi(1)$. The phrase "and so on" means that I eliminate all finite many $\varphi(i)$ with $1\leq i\leq m$. There are elements left because $m<n$. $[n]$ contains $n$-many elements and each of the $m$-many repetitions of the "process" eliminates exactly one element. I still don't understand what is unclear and why it needs more definitions? – Philipp Oct 21 '23 at 23:43
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    @Philipp My questions were rhetorical, not in search of answers. But if you think your argument is clear and complete as is, and anyone who would point out a flaw is wrong, why bother asking about it on this site? (Another rhetorical one.) – blargoner Oct 22 '23 at 00:08
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    If this problem is asked in the context of naïve set theory, your proof is certainly pointing to the right direction. (You would still want to formalize your recursive algorithm into the form of mathematical induction.) If it is asked in the context of a formal set theory such as ZF, you have to flesh out your proof because suddenly almost every part in your proof becomes nontrivial, including 'what is a finite set?', 'can we actually choose an element from set?', 'what it means by eliminating $\varphi(1)$' and so on. Their precise definition very much depend on the context and author. – Sangchul Lee Oct 22 '23 at 03:11

1 Answers1

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I remember hearing not long ago a claim (I'm paraphrasing) that if you took a class of beginning grad students and asked them to prove that a compact subset of a Hausdorff space is closed, you'd get mostly correct responses whereas if you asked them to prove the pigeonhole principle, you'd get a riot.

The difference is that for matters like the former question, we're well-trained to use the 'rigorous' part of our brains and clearly link up definitions with an argument to form a proof. Whereas for simple matters of finite combinatorics, we have always just used the 'pre-rigorous' channels of intuition, so we're in unfamiliar territory when asked to prove something of this sort.

Thus, what usually comes out is at best some hand-wavy explanation of why the pigeonhole principle is 'obviously true' that might make certain aspects precise, but is ultimately vague or circular. This is how I would categorize your attempt. You are essentially just explaining the intuition, drawing your mental picture with words and trying to make that come off as a proof. The circularity here is you're appealing to the idea that $[n]$ has a 'larger number' of elements than $[m],$ but usually we rigorously define 'larger number of elements' in terms of certain injections/surjections existing, so this is unsatisfying.

One related issue is that when we try to make these things precise, we have pay some attention to foundations... what exactly are we 'allowed' to assume about the natural numbers? (Apparently not the pigeonhole principle, though it's basic enough that I could imagine it being considered as an axiom.)

The customary foundational assumption useful in this kind of context is that the natural numbers obey the principle of induction, or the well-ordering principle. If you're trying to prove something 'rigorously' about the natural numbers and you aren't using one of these principles, you're probably doing it "wrong". So, with that in mind:

Let $n$ be the least natural number such that there is an $m<n$ with a surjection $[m]\to [n].$ Let $f$ be such a surjection.

Consider the bijection $[n]\to [n]$ that switches $f(m)$ and $n$ and compose with $f$ to get a surjection $f':[m]\to [n]$ such that $f'(m) = n.$ Then, observe that $f'\upharpoonright [m-1]$ must be a surjection onto $[n-1],$ which contradicts the minimality of $n$.

Aside from the innocuous fact that this proof is sloppy about the 'corner cases', we could probably still peck it to death with complaints about implicit assumptions. If we wanted to be more careful we could try to formalize more explicitly in set theory (taking the natural numbers to be the ordinals that don't exceed a limit ordinal) or use a proof assistant.

spaceisdarkgreen
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    This is why I taught my students this year that finite sets are those whose power set is well founded. Surprisingly enough, all the basic proofs go super easy, and since the definition is weird you can't rely on the pre rigorous part. – Asaf Karagila Oct 22 '23 at 07:18
  • I am not familiar with your notation $f'\upharpoonright [m-1]$. What exactly does it mean? Do you simply change $f':[m]\to[n]$ to $f':[m-1]\to [n-1]$? If so, why is the new $f'$ a surjection? – Philipp Oct 22 '23 at 11:17
  • @Philipp It's the restriction of the function. – spaceisdarkgreen Oct 22 '23 at 15:37