Let $n$ be a natural number, and let $f : ${$i ∈ N :1≤ i ≤ n$}→$N$ be a function. Show that there exists a natural number $M$ such that $f(i) ≤ M$ for all $1 ≤ i ≤ n$.Thus finite subsets of the natural numbers are bounded. (Hint induct on $n$.)
By the hint lets induct on $n$.
If $n=0$ then $f: \emptyset \to N$. Suppose there exist some $i$ for which $f(i) > M$. If so then it will mean that there $\exists i \in \emptyset$ which is not true. Therefore, the proposition holds for $n=0$
Suppose it is true for $f : ${$i ∈ N :1≤ i ≤ n$}→$N$; $\exists M \in N$ such that $f(i) ≤ M$ for all $1 ≤ i ≤ n$. Show that for $f : ${$i ∈ N :1≤ i ≤ n+1$}→$N$; $\exists M \in N$ such that $f(i) ≤ M$ for all $1 ≤ i ≤ n+1$. We know that $f(n+1) = k \in N$. Either $k \le M$ in which case $f(n+1) \le M$ or $k > M$ in which case set $M = k$ and we would have $f(i) ≤ M$ for all $1 ≤ i ≤ n+1$.
A good proof! I just have some pointers regarding notation.
In your inductive hypothesis (second paragraph), rather say "Suppose it is true for some $n \in N$." You can go on to clarify what this means (that there exists some $M \in N$...).
Be careful in the the $n+1$ case, you don't want to confuse the bound $M$ from the previous case with the bound for $n+1.$ Since the bound $M$ is defined by the $n$ case, you don't want to say $M = k,$ rather introduce a new variable for the bound for the case $n+1.$ Also, it is not necessary to introduce the variable $k,$ you can use $f(n+1)$ as is.
Just to check, are you taking the natural numbers to include 0? Since this will influence your base case.