Short version: Does the Loewner order have the least-upper-bound property?
Long version:
Suppose I have a finite set of finite-dimensional positive semidefinite matrices $\mathcal{M}$. That is, we have $M\ge 0$ for all $M\in\mathcal{M}$. Here "$\ge$" is the Loewner order, which is a partial order on matrices where we say $A \ge B$ when $v^*(A-B)v \ge 0$ for all vectors $v$ (and where "${}^*$" is the adjoint).
An "upper bound" on the set $\mathcal{M}$ is a matrix $A$ satisfying $A \ge M$ for all $M\in\mathcal{M}$.
For any set $\mathcal{M}$, is there necessarily a matrix $\bar{M}$ that provides a least upper bound in the sense that $\bar{M} \le A$ for all upper bounds $A$?
It seems relevant that the space of upper bounds is convex, i.e., if $A$ and $B$ are upper bounds then so is $C_\lambda = \lambda A + (1-\lambda)B$ for any $\lambda\in [0,1]$. However, it's easy to check that for any two upper bounds $A$ and $B$ there need not be a choice $\lambda$ such that $C_\lambda \le A$ and $C_\lambda \le B$.
No. A complete explanation can be found here for which I quote a relevant counterexample: Let $M_1$ and $M_2$ be $2\times 2$ and project onto the respective basis elements: $$M_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \qquad M_2=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$ Then it's clear that the identity $I$ is an upper bound ($I \ge M_1, I\ge M_2$). Furthermore, you can't choose a matrix strictly smaller than $I$, i.e., no upper bound $A$ on the set $\mathcal{M}=\{M_1,M_2\}$ could satisfy $A\le I$ (except $A=I$).
But one can also check that the matrix $$B=\begin{pmatrix}1+1/\sqrt{2}&1\\1&1+1/\sqrt{2}\end{pmatrix}$$ satisfies $B\ge M_1$ and $B\ge M_2$, yet $B$ and $I$ are incomparable. Therefore, no least upper bound exists.