Proving an equivalent definition for the supremum of a set

Question

Here $S$ is a subset of an ordered field $F$.

Attempt:

First, put $M = \sup S$. Let $\epsilon > 0$ be given. We must find and $x \in S$ such that $M - \epsilon < x$. Suppose this is false, that is, suppose we can find that $M - \epsilon \geq x$ for all $x \in S$. This would imply that $M- \epsilon$ is an upper bound. But then $M$ cannot be the least!! this is a contradiction. Therefore, $M - \epsilon < x$ and $x \leq M$ is true by hypothesis. Therefore, $\boxed{ M- \epsilon < x \leq M }$

Conversely, suppose for all $\epsilon > 0$, one can find $x \in S$ so that $M - \epsilon < x \leq M$. Let $u$ be an arbitrary upper bound for $S$. Thus $x \leq u$. If $M \leq u$ then we are done. If on the contrary $M > u$, then $M-u > 0$ and if we put $\epsilon = M - u$, we can furnish and $x_0 \in S$ so that

$$ M - (M- u) = u < x_0 \leq M$$

A contradiction and we thus obtain that $M = \sup S$

Question:

I have been told that $ \implies $ part of my proof is flawed, but I cant seem to see where my mistake is. Is my proof rigorous enough? I would love some criticism.

Answer 1

You are right when you say that if $M-\varepsilon > x$ for all $x \in M$ imply that $M$ would not be the least upper bound. After it is not so correct: you must say that there exists (at least) on $x$ such that $M-\varepsilon \leq x \leq M$. From your sentence its seems that every $x$ must have this properties. In few words you have to use a quantifier.

Answer 2

As Luca says in his answer, the problem is with the last two sentences of the first paragraph. When you write ‘Therefore, $M-\epsilon<x$’, it looks like you’re talking about a specific $x$, but in fact at this point $x$ has not been assigned a specific value: you’ve simply shown that it cannot be the case that all members of $S$ are less than or equal to $M−\epsilon$. You need to specify an $x$ about which to make these statements. After you get your contradiction you should say something like this: Therefore there is an $x\in S$ such that $M−\epsilon<x$. Now you’re talking about a particular object $x$, and you can go on to assert that it is less than or equal to $M$ by hypothesis and therefore satisfies the inequality $M−\epsilon<x\le M$.

I would actually have proved the contrapositive here: assume that for some positive $\epsilon$ there is no $x\in S$ such that $M−\epsilon<x\le M$ and observe that $M−\epsilon$ is then an upper bound for $S$, so $M\ne\sup S$.

The second part of your proof is fine.