The set of self-adjoint operators over a Hilbert space doesn't form a lattice How can someone prove, that over a Hilbert space $\mathcal{H}$ with $dim \mathcal{H} \geq 2$, the set of all self-adjoint operators doesn't form a lattice?
I found a result from Kadison, that if two self adjoint operators $A, B$ are not comparable, then they don't have a least upper bound. In 2 dimensions, I got two upper bounds of the (standard) projections $P_1, P_2$, which are not comparable. I think its not enough, but I can't go any further. 
Please help!
 A: Already s.harp has posted a good answer. However, the counterexample matrix $B$ in their post seems to appear out of nowhere like magic. For intuition about what is going on and how to construct examples like this (and indeed to understand the nature of all possible positive definite counterexamples), it is useful to consider the geometric underpinnings of the (Loewner) matrix partial order.
Specifically, there is a direct correspondence between Hermitian positive definite matrices and ellipsoids. Whether or not one h.p.d. matrix is greater than another is determined by whether one matrix's ellipsoid contains the other matrix's ellipsoid.
In detail, let $E_A$ and $E_B$ be the images of the unit ball under the action of Hermitian positive semidefinite matrices $A$ and $B$, respectively. that is,
\begin{align}
E_A :=& \{Ax : ||x|| \le 1\} \\
E_B :=& \{Bx : ||x|| \le 1\} \\
\end{align}
Geometrically, $E_A$ and $E_B$ are (hyper-)ellipsoids. 

Theorem.


*

*If $E_B \subset E_A$, then $B \le A$.

*If $E_A \subset E_B$, then $A \le B$.

*If neither $E_A$ nor $E_B$ completely contain the other, then $A$ and $B$ are not comparable.

Pictorially, the case where $A > B$ would look something like this:

The blue and red ellipsoids on the right represent the image of the unit ball in the domain (shown in black/gray on the left) under the mappings $A$, and $B$, respectively. The case where $A$ and $B$ are incomparable would look something like this:


Proof.
To prove this geometric interpretation, we simultaneously diagonalize the matrices $A$ and $B$ by solving the generalized eigenvalue problem. This will allow us to reduce the theorem to the case of diagonal matrices, for which the result is trivial. Specifically, let
$$A P = B P \Lambda.$$
be the solution to the generalized eigenvalue problem of $A$ and $B$, where $P$ is a (non-orthogonal) matrix of eigenvectors, and $\Lambda$ is the corresponding diagonal matrix of eigenvalues.
The eigenvector matrix $P$ then diagonalizes $A$ and $B$ simultaneously (see here for details):
$$\begin{cases}
P^T A P = \Lambda \\
P^T B P = I.
\end{cases}$$
Now, instead of comparing $A$ and $B$ to see if their difference is positive definite, we can instead compare $P^TAP$ and $P^TBP$. This transformation does not change whether one ellipsoid contains the other, since the same transformation is applied to both ellipsoids. Furthermore, the condition for a matrix $M$ being positive definite, 
$u^T M u > 0\quad \forall u$, is equivalent to the transformed version, $v^T P^T M P v > 0 \quad \forall v$, by the redefinition of variables $u:= Pv$.
But the transformed difference,
$$P^T A P - P^T B P = \Lambda - I,$$
is diagonal, and for diagonal matrices the theorem is trivial (The ellipsoid of a diagonal matrix contains that of another diagonal matrix iff all the diagonal entries of the first matrix are larger than all of the diagonal entries of the second matrix). $\Box$

Now with this machinery in place, we can return to the original question and easily find an answer. In order to show that self-adjoint matrices do not form a lattice, we need to find two self adjoint matrices that do not have a least upper bound. 
The idea of choosing projectors from the original question is a good one, let's go with that. Let $P_1$ be the projector onto the $y$-coordinate and $P_2$ be the projector onto the $x$-coordinate. The ellipsiods $E_{P_1}$ and $E_{P_2}$ associated with projectors $P_1$ and $P_2$ are vertical and horizontal line segments, respectively. They form a "cross" shape centered at the origin. 
The ellipsoid associated with the identity matrix is the unit disc. Since the unit disc contains both line segments of the cross, the identity matrix is an upper bound for both of the projectors. There cannot be another upper bound that is strictly less than the identity, since any ellipsoid strictly within the unit disc cannot contain the full cross. So, either the identity is the least upper bound, or there is no least upper bound.
But the identity cannot be the least upper bound, since we can find another upper bound, $B$, that is incomparable to the identity. All we need to do is find another ellipsoid, $E_B$, that contains the cross, but neither contains, nor is contained by, the unit disc. Of course this is easy to do - just squash the disc along some nonaxial direction, while simultaneously stretching it in the perpendicular direction. 
This is shown in the following image:

Hence, the there can be no least upper bound of $P_1$ and $P_2$, so the self adjoint matrices cannot form a lattice.
In general, this technique provides a complete geometric picture of how to construct incomparable upper bounds. You just take the ellipsoid associated with one upper bound, and simultaneously squash/stretch in some directions that are not principal axes of the ellipsoids associated with the matrices being bounded.
A: We'll do it in $M_{2\times2}$ by finding out that the identity is minimal in $\mathcal A:=\{A\mid A≥P_1, A≥P_2\}$, meaning that if $A≤\mathbb1$ and $A\in \mathcal A$ one has $A=\mathbb1$.
By finding an element $A$ of $\mathcal A$ that is incomparable with $\mathbb1$ one shows that there can exist no least upper bound of $(P_1,P_2)$, since $B≤\mathbb1$, $B\in\mathcal A$ imply $B=\mathbb1$, whereas a least upper bound must also satisfiy $B≤A$.
So let $A≥P_1,P_2$, $A≤\mathbb 1$. An element $B$ in $M_{2\times2}$ is positive iff $\mathrm{Tr}(B)≥0$ and $\mathrm{det}(B)≥0$.
So from $A≤\mathbb1$ one gets:
$$a_{11}+a_{22}≤2\qquad (1-a_{11})(1-a_{22})≥a_{12}a_{21}=|a_{12}|^2$$
Whereas $A≥P_1,P_2$ gives that $A$ must be positive (so both $a_{11},a_{22}≥0$) and
$$a_{11}+a_{22}≥1\qquad a_{11}(a_{22}-1)≥|a_{12}|^2\quad a_{22}(a_{11}-1)≥|a_{12}|^2$$
The determinant condition on $A≤\mathbb1$ shows that the diagonal elements must be $≤1$, but the determinant condition on $A≥P_1,P_2$ together with $a_{ii}≥0$ gives ou then that they must be exactly $1$ and then $|a_{12}|^2=0$.
So actually $A=\mathbb1$ follows from $A≤\mathbb1$ and $A\in\mathcal A$.
Now on the other hand:
$$B=\begin{pmatrix}3&2.5\\2.5&4\end{pmatrix}$$
is incomparable with $\mathbb1$ (it has eigenvalues $\approx 6$ and $0.95$) but is larger than $P_1,P_2$, as can be seen by checking the conditions.
