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I am asked to show if, $X$ be an ordered set with g.l.b property. Let $E\subseteq X$ be non empty and bdd. above. Prove that $\sup E$ exists in $X$.

I tried like this:

since $X$ has g.l.b property, so $\inf X$ exists , call it $l$, now since $E$ is bounded so $|x|\le R\forall x\in E$ for some $R$ . Suppose $\sup E$ does not exists in $X$...what will be the contradiction? am I going in right track? Thanks.

Myshkin
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    Does the g.l.b. property mean that every non-empty subset bounded from below has the greatest lower bound? – Adayah Aug 10 '17 at 17:07
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    Hint: Let $B \subset X$ be the set ${b | \forall a \in A, b \geq a}$. Can you pick up from there? – platty Aug 10 '17 at 17:08
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    @platty You probably wish $B$ to be non-empty, so the inequality should be $b \geqslant a$. – Adayah Aug 10 '17 at 17:10
  • each nonempty subset of S which is bounded from below has a greatest lower bound – Myshkin Aug 10 '17 at 17:12
  • No, I can not, could you please elaborate a bit – Myshkin Aug 10 '17 at 17:13
  • Sketch: show that $\inf B$ exists and is an upper bound of $A$. Then show that no smaller upper bound exists; by definition, this makes $\inf B = \sup A$. – platty Aug 10 '17 at 17:15

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At first, why did you suppose that $\inf X$ does exist? Think of $\mathbb{R}$, which is unbounded below.

Now, nack to our proof. Let $E$ be a bounded from above, non-empty subset of $X$. Since we only know that $X$ has the g.l.b. property, we need to construct a set that is bounded from below and, of course is non-empty. So, since $E$ is bounded from above, there exists an $a\in X$ such that: $$x\leq a\ \forall x\in E$$ Now, consider $$F:=\{x\in X:x\geq y,\ \forall y\in E\}$$ So, it is clear that $a\in F$, so $\varnothing\neq F\subseteq X$. Moreover, since $E$ is non-empty, there exists a $b\in E$, so, it is clear by $F$'s definition, that: $$b\leq x,\ \forall x\in F$$ so $F$ is bounded from below. Due to our hypothesis, there exists an $s\in X$ such that: $$s=\inf F$$ Now, we claim that $s\in X$ is an upper bound of $E$.

Let us suppose that there exists a $x\in E$ such that $$x>s$$ Then, by $F$'s definition, there should exist no $y\in F$ such that $$s\leq y\leq x$$ since for every $y\in F$ we have that $y>x$ So, $x$ is a greater than $s$ lesser bound for $F$, which is a contradiction. So $$x\leq s, \forall x\in E$$

Now, if $s\in E$, then it is clear that $s=\sup E$. If $s\not\in E$, let $t$ be another upper bound of $E$ with $$t<s$$ Then, since $t$ is an upper bound it is clear that $t\in F$, which is a contradiction, since, then: $$s=\inf F\leq t<s$$ So, $$s=\sup E$$

Vassilis Markos
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