TITLE: How Can You Prove the Completeness of $\mathbb{N}$?
QUESTION [5 upvotes]: I am trying to self-study real analysis and I am finding it difficult to prove some statements even when I have intuition for it. One exercise in my book asks:

Prove that $\mathbb{N}$ is complete.

So I have intuition for the proof. My thinking is that if I have nonepty $A \subseteq \mathbb{N}$, where $A$ has an upper bound, then it should also have a maximum which is its supremum. That would imply that $\sup(A) \in A$, which would then also imply that $\sup(A) \in \mathbb{N}$ since $A \subseteq \mathbb{N}$. Then if my reasoning is correct, this would then show that $\mathbb{N}$ is complete. However I cannot figure out how to prove that a if $A \subseteq \mathbb{N}$ has an upper bound, it has a maximal element. How can one prove this statement? Also, are there other ways to show that $\mathbb{N}$ is complete?

REPLY [2 votes]: There are several different notions of “completeness”. From the context of your tags, it seems you mean Dedekind completeness (i.e. every upper-bounded set has a least upper bound). If you mean metric completeness, Chris Custer’s answer has you covered.
You could prove $\mathbb{N}$ is (Dedekind) complete from the well-ordering of $\mathbb{N}$. Let $$S = \{n \in \mathbb{N} : (\forall a \in A)(a \leq n)\} ,$$
i.e. $S$ is the set of all upper bounds on $A$. By the well-ordering of $\mathbb{N}$, the set has a minimum, which would be a least upper bound.
Alternatively, you could do a proof by induction. Assume for contradiction that $A$ has no least upper bound. Let $U$ be some upper bound. Define a sequence $f(n)$ by setting $f(1) = U$, and recursively choosing $f(n + 1)$ to be some upper bound that’s less than $f(n)$, which exists because no least upper bound exists. Then $f(n) \leq U - (n - 2)$, but this means that $f(U + 3) \leq -1$, which is impossible. Thus a contradiction.