Let $\alpha>0$ and $X$ be an $\alpha$-normal (meaning, for $x,y\in X$,
$0\leq x\leq y$ implies $\|x\|\leq\alpha\|y\|$) ordered Banach space
with closed generating cone $X\_{+}$. If $X$ is reflexive, then every
pair of elements from $X$ has a minimal upper bound, i.e., for $x,y\in X$,
there exists some $z\in X$ with $\{x,y\}\leq z$, such that $\{x,y\}\leq w\leq z$ implies
$w=z$. One can show this using Zorn's lemma and the fact that order
intervals are weakly compact (since they are convex, closed and norm
bounded, and the space is reflexive).



> 
> My question is whether there exists a non-reflexive space $X$ where
>  not every pair of elements has a minimal upper bound, or can the
>  reflexivity requirement be dropped?
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I couldn't see any way of attacking this without reflexivity, so I started thinking about counter examples. Of course, all Banach lattices are immediately disqualified from potential counterexamples. The best hope I had for an example was something like $X:=\mathbb{R}\oplus\_{1}C\_{0}(\mathbb{R})$
with the (ice-cream) cone $X\_{+}:=\{(t,f)\in X:t\geq\|f\|\_{\infty}\}$,
but every pair of elements turns out to have a minimal upper bound...



Does every $\alpha$-normal ordered Banach space have minimal upper bounds?
The answer is that the reflexivity assumption cannot be dropped.


The following simple example (due to Tony Wickstead) is a 1-normal non-reflexive space with closed and generating cone, where there exists a pair of elements without minimal upper bound:



> 
> Let $X$ be the space of all convergent sequences where $x\_n\to (x\_1+x\_2)/2$, ordered by the standard cone (pointwise). Then it is easily seen that the elements $0\in X$ and $(1,-1,0,0,0,\dots)\in X$ has no minimal upper bound.
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