# Proving an equivalent definition for the supremum of a set

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.

• 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 ... Apr 16, 2020 at 18:54
• ... 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 $M\ne\sup S$.) Apr 16, 2020 at 18:57
• Great! Thanks for your response Brian. The second part is correct right? Please, do write your answer below so I can award you the points for your help! thanks! Apr 16, 2020 at 20:11
• Done! Yes, the second part if just fine. Apr 16, 2020 at 20:17

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.
As Luca says in his answer, the problem is with the last two sentences of the first paragraph. When you write ‘Therefore, $$M-\epsilon’, it looks like you’re talking about a specific $$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. 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 $$M−\epsilon.
I would actually have proved the contrapositive here: assume that for some positive $$\epsilon$$ there is no $$x\in S$$ such that $$M−\epsilon and observe that $$M−\epsilon$$ is then an upper bound for $$S$$, so $$M\ne\sup S$$.