# Is this proof of the Archimedean Property valid?

Archimedean property: The set of natural numbers $\mathbb{N}$ is not bounded above.

Proof :

Suppose $\mathbb{N}$ is bounded above. Then, by the supremum property, there exits a lowest upper bound "$s$" for all $n \in\mathbb{N}$. Call "$k$" the biggest natural number, which is smaller than $s$. Then $k+1>s$ and since $k+1$ is a natural number, $s$ is not an upper bound of $\mathbb{N}$. Thus, we have reached a contradiction and can conclude that $\mathbb{N}$ is not bounded above.

• No, finding k requires that N is unbounded. – William Elliot Sep 17 '17 at 20:21
• @William Ellios, I don't see that. Can you go into a little more detail? – stollenm Sep 17 '17 at 20:26
• Having a supremum does not mean a maximum is ever achieved. Example $(0,1)$ has a supremum. But I can not then say "Call 'k' the biggest real number that is smaller than $\sup(0,1)$". There is no such number and there is no reason to think there should be one. – fleablood Sep 17 '17 at 20:46
• "Bounded above" in $\mathbb N$ or in $\mathbb Q$ or in $\mathbb R$ or...? – DanielWainfleet Sep 18 '17 at 5:21

For every real number $s>0$, the set $\{ n\in\mathbb N \mid n<s\}$ contains a largest element.

I don't know exactly which axiomatic facts abour $\mathbb R$ (and its relation to $\mathbb N$) you have available, so it might be possible for you to prove this without already having the Archimedean property. But it doesn't really sound likely.

(The proof suggested by William Elliot gets around this problem).

• Thank you @Henning Makholm. That's what I wanted to understand. – stollenm Sep 17 '17 at 20:43

Assume N subset R is bounded above.
So there is a least upper bound of N. Call it s.
If n in N, then n + 1 in N. Hence n + 1 <= s, n <= s - 1.
Thus s - 1 is a smaller upper bound of N, a contradiction.

• Cute way of putting it. I've never seen that before. – fleablood Sep 17 '17 at 20:59

"Call "k" the biggest natural number, which is smaller than s."

You have no reason to believe such a thing exists.

But you can prove it.

If $S \subset \mathbb N$ that is bounded above then $s \sup S$ exists.

Let $j = s-1$. Then $j$ is not a upper bound of $S$. So there exists a natural $k$ so that $j=s-1 < k \le s$.

Thus $s = j+1 < k+1 \le s+1$ and $k+1$ is not in $S$. Nor are any natural numbers that are greater than $k$. So $k$ is the greatest natural number in $S$.

So every upper bounded set of natural numbers will have a greatest element.

....

But this also proves that for any upper bounded set $S$ of natural numbers there exists natural numbers not in $S$ (namely all the natural numbers larger than $k$).

Thus we no longer have anything left to prove. (No upper bound set of natural numbers contain all natural numbers, so $\mathbb N$ itself which does contain them all, can not be bounded above.)

.....

So actually, "every upper bounded set of natural numbers has a maximum element" and "the set of natural numbers are not bounded above" are equivalent statements both with the same proof.
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