A characterization of suprema

Question

The following fact seems to show up all over real analysis and measure theory:

Proposition. Consider an above-bounded non-empty subset $X$ of the real line and an element $x \in \mathbb{R}$. Then the following are equivalent:

• $x$ is a least upper bound of $X$
• $x$ is an upper bound of $X$, and $\forall(\varepsilon > 0)\exists(y \in X)(|y-x|<\varepsilon)$

Question. Is there an easy way to see that these are equivalent?

Preferably the proof should work for $\mathbb{Q}$ as well, since this seems to hold there. Perhaps it is true for all ordered fields (I don't know).

Answer 1

This is true in any totally ordered abelian group. The proof is very simple. Suppose $x$ is an upper bound of $X$. If there exists $\epsilon>0$ such that there is no $y\in X$ with $|y-x|<\epsilon$, then $x-\epsilon$ is also an upper bound for $X$, since there is no $y\in X$ in the interval $(x-\epsilon, x]$ (and there is also no element of $X$ which is greater than $x$). So $x$ is not the least upper bound.

Conversely, if $x$ is not the least upper bound, there is some smaller upper bound $x'<x$. Letting $\epsilon=x-x'$, we see that there is no $y\in X$ such that $|y-x|<\epsilon$, since such a $y$ would be greater than $x'$.

Answer 2

Proof:

"$\Rightarrow$"

By contraposition! $$\lnot(\forall\epsilon>0\exists y\in X:|y-x|<\epsilon) \Leftrightarrow \exists \epsilon >0 \forall y \in X:|y-x|\ge \epsilon$$ This is equal to saying that: $$y \le x - \epsilon \tag 1$$ which means that $x$ can't be the least upper bound since we found one even lower!

"$\Leftarrow$"

By contraposition! We know that, since know $x$ is not the least upper bound, that for any other upper bound $x'$ of $X$ we can write: $$x>x' \Leftrightarrow x-x'>0$$ so we can define $$\epsilon :=x-x' > 0$$ Then we use the definition of an upper bound: $$ \forall y \in X: y \le x \Leftrightarrow x-y\ge0$$ Now we can write $$\exists \epsilon > 0\forall y \in X: |x-y|\ge \epsilon \Leftrightarrow x-y \ge x-x' \Leftrightarrow y \le x'$$ which is true, since $x'$ is an upper bound of X! $\square$

Now I want to adress some further issues. The existence of a least upper bound is justified by $X$ being a subset of $\mathbb{R}$ which is a complete set with a total order. Note that not every set has this property, e.g. $\mathbb{Q}$

$(1)$ surely right for $x\ge 0$, for $x<0$ we know that $y<0$, so we still arrive at this inequality!