# Does every $\alpha$-normal ordered Banach space have minimal upper bounds?

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?

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...

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.