9 Theorem A set $X$ is order-complete relative to an ordering if and only if each non-void subset which has a lower-bound has an infimum.

Suppose that $X$ is order-complete and $A$ is non-void subset which has a lower bound. Let $B$ be the set of all lower bounds of $A\,.$ Then $B$ is non-void and surely every member of the non-void set $A$ is an upper bound for $B\,.$ Hence, $B$ has a least upper bound, say, $b\,.$ ...

This is excerpted from General Topology's Orderings by John Kelley; here he proves the theorem that a set is order-complete iff all its non-void subsets which have lower bounds have infimum.

I did stumble a little and couldn't comprehend the statement "Hence, $B$ has a least upper bound $b\,.$"

Is it a trivial conclusion from the fact that all elements included $A$ are upper bounds of $X\,?$

But how does it guarantee that $A$ indeed contains the lowest upper bound $b$ of $B\,?$

I couldn't get the author's use of the word "Hence"; I'm not seeing it obvious how $A,$ albeit containing elements acting as upper bounds to $B,$ contains the lowest upper bound $b$ too.

Could anyone shed light on why the author used "Hence" and thus meant that the fact that $A$ contains $b,$ the lowest upper bound of $B\,?$


In the paragraph leading up to that theorem, Kelley defines X to be order-complete if every non-empty subset of X with an upper bound has a least upper bound.

"Hence" is where that definition gets applied: B is non-empty (because we know A has a lower bound), has an upper bound (any member of A, which is non-empty, will do), and is a subset of X. So since X is order-complete, B has a least upper bound b.

You're 100% right that b might not be contained in A! But the rest of the proof doesn't rely on b being in A. It goes on to show that, because of how we defined B, b is a lower bound of A, and that it's in fact the greatest lower bound of A.

(One aspect of the terminology that you might be finding confusing - a set can "have" a least upper bound or greatest lower bound without the l.u.b. or g.l.b being in the set. An example is the interval (0,1): 0 is its greatest lower bound and 1 is its least upper bound, but neither are in the set.)


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