Problem on a set which is a subset of a finite set

[issacnewton]

Homework Statement

Prove that if $n \in \mathbb{N}$ and $A \subseteq I_n$, then $A$ is finite and $|A|\leq n$. Furthermore, if $A \ne I_n$, then $|A| < n$. We are given that
$$ I_n = \big \{ i \in \mathbb{Z}^+ | i \leq n \big \} $$

Relevant Equations

Definition of a finite set and definition of bijections

Here is my attempt. Since we have to prove that $A$ is finite, we need to prove that there exists some $m \in \mathbb{N}$, such that there is a bijection from $A$ to $I_m$. And hence we have $A \thicksim I_m$. Now, since there are $n$ elements in $I_n$, number of elements in $A$ are less than or equal to $n$. So, we have $|A|\leq n$. Let, $|A| = m$. Now, we can construct a function from $A$ to $I_n$, such that $m$ elements in $A$ are paired with first $m$ members in $I_n$. And, we see that the first $m$ members in $I_n$ is the set $I_m$. So, the function would be one to one and onto and hence we have $A \thicksim I_m$. Which proves that $A$ is a finite set. Furthermore, if $A \ne I_n$, then $A$ is a strict subset of $I_n$ and $A$ would have less number of elements than $I_n$. So, we have $|A| < n$. Is the proof valid ?

Thanks

 

[WWGD]

I don't know if you're allowed to use this but I think inclusion/exclusion would do it since it applies to finite sets.

 

[issacnewton]

WWGD, thanks, but is the proof presented here valid ?

 

[WWGD]

This is a tricky issue because it depends on all the definitions, results and theorems available to you. Still, from what I have read, it seems you're assuming what you want to prove: ( bold is mine).

"Here is my attempt. Since we have to prove that A is finite, we need to prove that there exists some m∈N, such that there is a bijection from A to Im. And hence we have A∼Im. Now, since there are n elements in In, number of elements in A are less than or equal to n"