I know the statement above is true because of the Archimedian Theory. Would the following proof make sense to prove it?
This is a proof by contradiction.
If the set of natural numbers does have an upper bound, then it has a least upper bound. Let $x = \sup \mathbb{N}$, supposing that each does exist as a finite real number. Then there is not a natural number $n > x$. Therefore $n \leq x$. Then, $n \leq x – 1$ cannot be true for all natural numbers.
There is some natural number $m$ such that $m > x – 1$. Since $m$ is a natural number, $m + 1$ must also exist, so $m + 1 > x$. But this cannot be so since we defined $x$ as the largest number.
Therefore, by contradiction, the above statement stands true.
Best Answer
You need to use some specific properties of the real numbers. Otherwise, your proof might work for non-standard number systems which are not Archimedean.
If you construct your reals as Dedekind cuts of rationals then you only need the Archimedean property of the rationals: just take any rational in the set of upper bounds. If you use Cauchy's construction, pick a point beyond which all elements are at distance at most one (say), and again use the Archimedean property of the rationals.
It remains to prove the Archimedean property of the rationals. Assuming the construction as equals class of fractions $p/q$ with $p$ positive, we know that $p/q \leq p$.