Understanding a proof of Axiom of Archimedes

Question

So I came across a proof of Axiom of Archimedes in my real analysis book. The proof is written as follows:

Axiom of archimedes: Given any real number $x$, there is an integer $n$ such that $x<n$.

Proof: If $x<0$, take $n=0$. Otherwise the set $S$ of integers $k$ such that $k \le x$ is nonempty. Since $S$ has the upper bound $x$, it has a least upper bound $y$. (This has been proved in preceding sections of the book) Since $y$ is the least upper bound for $S$, $y-\frac{1}{2}$ cannot be an upper bound for $S$, and so there is a $k \in S$ such that $k>y-\frac{1}{2}$. But $k+1>y+\frac{1}{2}>y$, and so $(k+1)\notin S$. Since $k+1$ is an integer not in $S$, we mucst have $k+1$ greater than $x$ by the definition of $S$.

Now, I understand this proof as a whole. The thing I don't understand is, why can't we just take $n=k+1$ after we have defined the set $S$? Why is it necessary to do this whole proof? I know this sounds stupid, but I'm not entirely sure what makes us not able to draw that conclusion. What way should I think in proving this theorem?

Any help is appreciated.

Answer 1

In the definition of $S$, $k$ is a dummy (or bound) variable; it is not a particular object. For instance, we can define $S$ by $$ S = \{ k \in \mathbb{Z} \mid k \leq x \}, $$ or we can define $S$ by $$ S = \{ m \in \mathbb{Z} \mid m \leq x \}. $$ (There's nothing special about the letter $k$ here.)

Now, it isn't until the fourth line of the proof above that $k$ denotes a particular object/number, for it's at that point that we read "there is a $k \in S$ such that".

In short, you wouldn't bring up $k+1$ right after defining $S$ because $k$ is not a particular object under consideration (at that point in the proof).

With all this in mind, it would be good to carefully go back through the proof and make note of when we're choosing/selecting particular objects/numbers.

Answer 2

The point here is that the proof only seem to be able to rely on the fact that an upper bound set of natural numbers has a least upper bound. In fact one could prove that such a set actually has a maximal element which would allow something similar to your proposal.

The difference is that the least upper bound ($\sup$) is not required to be in the set, while the maximal element ($\max$) is.

So with only the knowledge that there's a $\sup$ we can only proceed from there, but as the proof shows we can thereby find a $k$ large enough so that $n = k+1>x$.

If we had known the stronger statement that every upper bound non-empty set of integer has a maximal element OTOH we would let $k=\max S$ and conclude that $n=k+1\notin S$ and therefore $n=k+1 > x$

The proof of that every upper bound non-empty set $S\subset\mathbb Z$ has a maximal element is quite straight forward and similar. We basically need to prove that $\sup S\in S$. We now know that there is a $k\in S$ such that $k>\sup S-1/2$ and then show that if an integer $l>k$ then $l\notin S$. This way we have that $k\in S$ and if $j\in S$ we must have $j\le k$.

Answer 3

What you have is a non empty subset $S$ of $\mathbb{N}$ which is bounded above by some real number $x$. The problem would have been considerably simple if it had been bounded above by a natural number because then it would not require the use of Supremum principle.

Suppose we knew that it was bounded above by some rational $r = a/b$ with $a, b \in\mathbb{N}$. Then it is obvious that if $q$ is the quotient when $a$ is divided by $b$ then $q \leq r < q + 1$ and hence $S$ is bounded above by natural number $q + 1$. The desired number $n$ is thus $n = q + 1$.

The issue occurs only when the number $x$ is irrational. In that case we have either of the two approaches as follows:

• Treat irrational numbers (and in fact the whole of real numbers) as axiomatic objects satisfying certain axioms (including Supremum principle).
• Define them in relation to the already existing rational numbers. This helps us to find a rational number $r$ greater than $x$ and then a natural number $n > r$ does our job.

Most of the modern analysis textbooks prefer the first approach because it is easy and all the deep theorems of elementary analysis / calculus are obtained free of cost by the use of Supremum principle or some other equivalent principle.

The second approach is not so damn easy but it respects the principle that there is no free beer/lunch. More importantly one enjoys the deep theorems of analysis / calculus more when one has paid the due cost by taking this approach.

BTW your intuition is right. The set $S$ has a maximal element say $M$ and $n = M + 1$ is what we seek to find. But proving that $S$ has a maximum element is not possible unless we use one of the two approaches available.