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Prove the following implication by induction on $m$:

If there exists an injection $N_m \rightarrow N_n$, then $m \le n$.

$N_n$ and $N_m$ are sets with $n$ and $m$ elements, respectively.

I know how to do proofs by induction, I'm just struggling with the inductive step. So far I have the predicate P(n) = If there exists an injection $N_m \rightarrow N_n$, then $m \le n$.

I proved the base case "P(1) = If there exists an injection $N_1 \rightarrow N_n$, then $1 \le n$" sloppily by saying no element in $N_n$ could be assigned to more than one element of $N_1$ because there is only a single element in $N_1$. There must at least one element in $N_n$ in order for it to be a function, so $1 \le n$.

So I know the inductive hypothesis is P(k) = If there exists an injection $N_k \rightarrow N_n$, then $k \le n$. Now I need to prove P(k+1) = If there exists an injection $N_{k+1} \rightarrow N_n$, then $k+1 \le n$, but I'm clueless about how to do this step.

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I'm curious as to what you've tried. Perhaps start with providing definitions for what being an injection means, and what "base case" you can use. Also, please clarify whether you are struggling with the notion of an injective function, or how to proceed with a proof by induction on $m$, or both... –  amWhy Nov 4 '12 at 1:55
Sorry amWhy, I edited the post with what I've tried. –  user1356910 Nov 4 '12 at 2:10
Great! That helps us help you! Thanks for being responsive to suggestions! –  amWhy Nov 4 '12 at 2:18
Yes, you're correct. You need to establish that, assuming P(k) is true, then P(k+1) is true. Just a typo perhaps, but you need to show that if there exists an injection $N_{k+1} \to N_n$ then $k+1 \leq n$. (I say that because what you've got in your edit is $N_k + 1 \to N_n$. For subscripting more than one character, you need to include, all characters in { }, like this $N_{k+1} \to N_n$.) Note that you can use take as given that P(k) is true. You need only rely on the definition of an injection... –  amWhy Nov 4 '12 at 2:25
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2 Answers

For any $f: N_m \rightarrow N_n$, if $m > n$ then by the Pigeonhole Principle (at least) two elements of $N_m$ must be sent to the same element of $N_n$. Then $f$ is not injective. Now take the contrapositive.

Factoid: (which I think is not well-known?) the Pigeonhole Principle is very similar to the Principle of Mathematical Induction. See, for example, this article. However, criticisms can be found here as well as in The complexity of the Pigeonhole Principle by M. Ajtai, in which PHP is shown to be (in some sense) stronger than PMI.

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Well I certainly didn't know that fact. Thank you for sharing! –  EuYu Nov 4 '12 at 2:29
@EuYu: My sense is that very few are aware of this equivalence; so few that I should probably give that article (which I rooted out some time ago) a careful re-reading/scrutinizing... –  Benjamin Dickman Nov 4 '12 at 2:46
Sadly I can't access that article. :( –  EuYu Nov 4 '12 at 2:48
Indeed, I think my worries are founded. See, for example, mathforum.org/kb/message.jspa?messageID=125747 –  Benjamin Dickman Nov 5 '12 at 19:42
Hmm, very interesting. It seems that problem is deeper than it appears. –  EuYu Nov 5 '12 at 20:02
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The statement is true for $m=1$: If there is a map $f:\{1\}\to N_n$ then necessarily $n\geq1$.

Assume that the statement is true for $m\geq1$ and that we are given an injective map $f:\ [m+1]\to [n]$. We have to prove that $m+1\leq n$. We distinguish two cases:

(a) $\quad f(r)\ne n$ for all $r\in[m]$.

In this case restricting $f$ to $[m]$ provides an injective map $f':\ [m]\to[n-1]$. By the induction hypothesis it follows that $m\leq n-1$. Therefore we have $m+1\leq n$ in this case.

(b) $\quad f(r)=n$ for a unique $r\in[m]$.

In this case $f(m+1)<n$. Therefore the new function $$f'(k):=\cases{f(k) &$k\in[m], \ k\ne r$\cr f(m+1) &$(k=r)$\cr}$$ is an injective function $f':\ [m]\to[n-1]$. By the induction hypothesis it again follows that $m\leq n-1$, or $m+1\leq n$.

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