Tao's Analysis I 4th ed has the following exercise 3.6.3:
Let $n$ be a natural number, and let $f:\{i \in \mathbb{N}:i \leq i \leq n\} \to \mathbb{N}$ be a function. Show that there exists a natural number $M$ such that $f(i) \leq M$ for all $1 \leq i \leq n$. (Hint: induct on $n$. You may also want to peel at Lemma 5.1.14.). Thus finite subsets of the natural numbers are bounded. Use this to give an alternate proof of Theorem 3.6.12 that does not use Lemma 3.6.9.
Showing that a finite subset of $\mathbb{N}$ is bounded can be done by induction, and isn't too difficult (unless I am in error).
The final part of the exercise asks us to prove Theorem 3.6.12 which is:
The set of natural numbers $\mathbb{N}$ is infinite.
Question: I can't see how to start with a statement that a finite subset of $\mathbb{N}$ is bounded, and conclude the infinitude of the natural numbers $\mathbb{N}$. I'd welcome some guidance.
Discussion
The best candidate for an argument I can find is the contrapositive:
If a set is not bounded, then it is infinite (in cardinality).
Then I can show that for any number $n \in \mathbb{N}$ I can find a larger one using the Peano axiom about each natural having a successor. Therefore, this set $\mathbb{N}$ is not bounded. And therefore, using the above contrapositive statement, it must have infinite cardinality.
Is this correct? It feels wrong.
Tao's book defines a finite set as one with $n \in \mathbb{N}$ cardinality, and infinite otherwise. He defines a set having cardinality $n$ if there is a bijection with the set $\{1, \ldots, n\}$. We don't use this in the above argument.
Best Answer
If $\mathbb{N}$ was finite, then it would be bounded by the first part of the exercise (it would be a finite set containing the first $M$ natural numbers; you can choose f to be the inclusion map). In particular the maximum $M$ of this set would exist (you can always find the maximum in finite sets), and it would definitely be a natural number. As you said, though $M+1$ is also a natural number which is larger than the maximum, a contradiction. What you claimed looks correct to me. Anyway, I think you made a typo at the beginning of your question: $f:\{1 \leq i \leq n\} \to \mathbb{N}$.