# If each subset of $L$ with an upper bound has a least upper bound, then each subset with a lower bound a a greatest lower bound

Proposition: Let $L$ be a lattice in which every subset with an upper bound has a least upper bound. Then every subset with a lower bound has a greatest lower bound.

Attempt

Definition: A lattice, $L$, is a partially ordered set where given any two elements of $L$, $a$ and $b$, the set $\{ a,b \}$ has a least upper bound and a greatest lower bound.

Denote a subset of $L$ by $S$. If $S$ has an upper bound, then $S$ has a least upper bound, or more precisely, given $u \in L$, $\forall s \in S$, $s \le u$ and given any upper bound, $v$, of $S$, $u \le v$.

Suppose $S$ has a lower bound, $w$. Then $\forall s \in S$, $w \le s$.

My question

From here, I think I need to rely on the definition of a lattice. However, the definition only applies to a two element subset, where the proposition provides any subset.

Two candidates I thought that may lead to progression is the transitivity axiom for a partially ordered set, or a previous proposition I provided that any chain is a lattice. Presently, I am unsure on how to figure these in.

I would appreciate some assistance on what I need to do in order to complete this proof.

Source: Kaplansky, I. (1972). Set Theory and Metric Spaces.

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I think you want to specify non-empty sets.... Anyway, this just follows from the definitions. Note if $A$ is non-empty and bounded below, then the set of lower bounds of $A$ is bounded above (by any element of $A$). Consider the least upper bound of this set. –  David Mitra Jun 7 '13 at 22:14

Suppose that $S$ is has a lower bound, and denote the set of lower bounds by $B$. Since $B$ consists values that for each $b \in B$, $b \le s$, $\forall s \in S$. Thus $S$ must consist of upper bounds of $B$, and thus $B$ is bounded above. Therefore by the proposition $B$ has a least upper bound which corresponds the the greatest lower bound of $S$.
Strictly speaking, you should check that the lub of $B$ is (not just "corresponds to" but "is") the glb of $S$. It would also be good to make sure you understand where you've tacitly used an assumption of nonemptiness; see the comments of David Mitra and user14111 on the question. –  Andreas Blass Jun 8 '13 at 15:55
@tmpys The way to deal with this is exactly what I wrote in my previous comment --- one must check that the lub of $B$ is the glb of $S$. That implies, in particular, that the lub of $B$ is in $B$, so that the possibility that you were worried about doesn't actually arise. –  Andreas Blass Oct 15 '14 at 13:41