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Suppose the sets $\{1,...,n\}$ and $\{1,...,p\}$ are in bijection. Why is it true: $n=p$? Intuitively clear, but can't write a proof

NOTE please don't use cardinalities or anything advanced because this is very basic

thank you

  • Let p < n and suppose f: maps {1,2... n} to {123.....p} bijectively. Consider $f^{-1} $ of all {1,2... p}. These are p items and form a subset of {1,2... n}. Remove these from {1,2,3... n}. You will hap p-n items. These were not mapped to {1,2,3... p} because you removed them. So f didnvt map them all. A contradiction. You can do this directly if you want. – fleablood Aug 07 '16 at 05:56
  • See also http://math.stackexchange.com/questions/239566/subset-of-a-finite-set-is-finite/239567#239567 – Asaf Karagila Aug 07 '16 at 06:19
  • I think the hardest part is keeping track of what are axioms and what are propsitions. One of the answers below use cardinality of of image sets which is the most direct way. But I don't know that is an axiom or an equivalent proposition. I used induction but I wasn't sure if we could assume induction principal. No matter how you look at it there is a direct correlation between natural numbers and set ordinal it and I'm not sure that this isn't simply a restatement of basic natural number definitions. (Whatever axioms you particularly use.) – fleablood Aug 07 '16 at 06:36
  • How does one not use cardinality here? – StubbornAtom Aug 07 '16 at 14:46
  • @StubbornAtom: Instead one unpacks the definitions of cardinality and the basic theorems about cardinals (and specifically finite cardinals) into the proof. – Asaf Karagila Aug 07 '16 at 15:38

2 Answers2

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The standard proof goes something like this.

Let's assume that the two sets are $\mathbb{N}_n$ and $\mathbb{N}_p$, as abbreviations.

Suppose there is a bijection $f: \mathbb{N}_n \to \mathbb{N}_p$.

Then $f$ is a surjection. Let $|\cdot|$ denote cardinality of a set. Since $f$ is a surjection, it follows that $|\mathbb{N}_p| = p\leq|\mathbb{N}_n| = n$, i.e., $n \geq p$.

Since $f$ is a bijection, it has an inverse which is also bijective, say $f^{-1}: \mathbb{N}_p \to \mathbb{N}_n$. Now apply the same theorem as above, and you're finished.

Clarinetist
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Suppose you have a bijective map $f $ from {1,2,3... p} to {1,2,3.... n}. Remove f (p) from {1,2,3....n} and remove p from {1,2,3... p}. You now have a set { 1,2,3... ,p-1} with p-1 elements and another subset of {1,2,3,.... n} with n-1 element. $f $ restricted to this to sets is a bijection.

Repeat $\min (p,n)$ times. You will now have the empty set and a set with |p-n| elements. $f $ will be a bijective map between them. If the set with |p-n| elements has any elements there will be corresponding elements in the empty set. There are not so the set with |p-n| elements has no elements.

So |p-n|=0. So p=n.

fleablood
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  • Or, instead of hiding the induction in the "repeat $\min(p,n)$ times", just properly formulate the proof using an induction hypothesis on $n$ (this requires a slight reordering of the bijection $f$, namely that you can assume that $f(p)=n$). – Asaf Karagila Aug 07 '16 at 06:16