Why do we need a "lengthy proof" to show $\mathbb{N}$ is not bounded above in $\mathbb{R}$? I have been reviewing my intro to analysis course and stumbled on a rather easy proof. The proof is standard, suppose $\mathbb{N}$ is bounded above and reach a contradiction using the Completeness Axiom. My question is why do we even need to do that? Consider the following proof:
To show $\mathbb{N}$ is not bounded above is the same as showing that given  $n_1\in\mathbb{N}$, there exits $n_2$ s.t. $n_2>n_1$ and $n_2\in\mathbb{N}$. Letting $n_2=n_1+1$ completes the proof. Why do we even need to use the completeness axiom if we already know $\mathbb{N}\subset \mathbb{R}$. 
 A: Maybe a picture will help. $\;\!\;\!$

So the purpose of the "lengthy proof" is to show that complete ordered fields are never too lengthy. Now that's irony!
A: Saying that $\Bbb N$ is not bounded above in $\Bbb R$ is not the same as saying that for all $x\in \Bbb N$ there exists $y\in \Bbb N$ such that $y>x$. In fact, there are several subsets of $\Bbb R$ which satisfy this property and are bounded above, namely the interval $[0,1)$, which has $1$ as an upper bound. What you must prove is that for all $x\in\Bbb R$ there exists $y\in \Bbb N$ such that $y>x$, which is different.
A: Your proof shows that there is no biggest element of $\Bbb{N}$. However, just because there is no biggest element of $\Bbb{N}$ does not mean there is no upper bound.
Take $(0, 1)$, for example. If you have some biggest element $m \in (0, 1)$, then we know $m < 1$, so $1-m > 0$ and thus $m+\frac{1-m}{2}=\frac{1+m}{2} > m$. Clearly, since $m < 1$, $1+m < 2$, so $\frac{1+m}{2} < 1$, meaing $\frac{1+m}{2} \in (0, 1)$. Thus, we have found a bigger element than the biggest element of $(0, 1)$, so there is no biggest element of $(0, 1)$.
However, we can not deduce from the fact that $(0, 1)$ has no biggest element that it is not bounded above. Clearly, $1 \in \Bbb{R}$ is an upper bound of $(0, 1)$ even though $(0, 1)$ has no biggest element.
Thus, even though it seems obvious that $\Bbb{N}$ has no upper bound, we need to use the Completeness Axiom in order to prove it because that tells us about the real numbers themselves. In fact, there are some models of the real numbers like the hyperreals where $\Bbb{N}$ is actually bounded above because the Completeness Axiom does not hold. However, your proof that $\Bbb{N}$ has no biggest element would still hold in the hyperreals, but that doesn't mean it has no upper bound.
