1

Please check if my proof is fine or it contains any error!

Theorem: Let $I_m=\{i\in\mathbb N\mid i\leq m\},I_n=\{i\in\mathbb N\mid i\leq n\},\text{ and } f:I_m\to I_n$ be injective. Then $m\leq n$.

Let $T=\{m\in\mathbb N\mid \forall n\in\Bbb N(f:I_m\to I_n\text{ be injective }\implies m\leq n)\}$. Since $0\leq n$ for all $n\in\mathbb N,0\in T$. Assume $k\in T$, then $\forall n\in\Bbb N(f:I_k\to I_n\text{ is injective }\implies k\leq n)$. I will prove $k+1\in T$ as follows.

Let $g:I_{k+1}\to I_n$ be injective. $I_{k+1}$ contains at least two points and $g$ is injective, so does $I_n$, or equivalently $n=t+1$. Let $\tau:I_n\to I_n$ be a mapping that transposes $g(k+1)$ and $n$, and leaves the other elements of $I_n$ fixed. Then $\tau\circ g:I_{k+1}\to I_n$ is injective, and $\tau\circ g(k+1)=n$. As a result, ${\tau\circ g}_{|I_k}I_k\to I_t$ is injective $\implies k\leq t$ [Since the theorem is true for $m=k$] $\implies k+1\leq t+1\implies k+1\leq n$. Thus $k+1\in T$.

By principle of induction, $T=\mathbb N$, and consequently the theorem is proved. $\blacksquare$

Akira
  • 17,367
  • 2
    Your proof is fine. – Christian Blatter May 18 '18 at 08:30
  • Thank you so much @ChristianBlatter! I've recently posted a proof of Inclusion–exclusion principle at https://math.stackexchange.com/questions/2783980/inclusion-exclusion-principle-proof-verification, but receive no comment and answer. It will be great if you can have a check on my proof. – Akira May 18 '18 at 09:14

1 Answers1

1

A Proof by contradiction:

If possible suppose $m > n$, then $I_m$ must have at least two points which have the same image. (Piegon Hole principle)

And thus, $f: I_m \to I_n$ is not injective, which is a contradiction.

User
  • 2,938
  • In my textbook, the author uses this theorem to prove Piegonhole principle, so it's wrong to use Piegonhole principle to prove this theorem. By the way, have you seen any error in my proof? – Akira May 18 '18 at 06:40
  • @CrazyGuy, Yes In the very first step, you are claiming $0 \in T$ but $T \subset \mathbb N$ – User May 18 '18 at 07:12
  • I'm unable to understand what's wrong with "$0 \in T$ but $T \subset \mathbb N$". Could you please be more specific? – Akira Sep 03 '18 at 01:36
  • @LeAnhDung, $0$ is not a natural number. – User Sep 03 '18 at 09:33
  • 1
    From https://math.stackexchange.com/questions/283/is-0-a-natural-number, is it true that $0$ is not a natural number is opinion-based? – Akira Sep 03 '18 at 09:37