Consider a separable Hilbert space $\mathcal{H}$, we can first define $T(\mathcal{H})$ the trace class of $\mathcal{H}$, then $D(\mathcal{H})$ to denote the interesection of positive operator and trace class with trace no more than 1. 


For $A,B\in D(\mathcal{H})$, we say $A\leq B$ if $A-B$ is positive, or (A,B) satisfies the Lowner order.


Now suppose we have increasing sequence 
$$
A\_0,A\_1,\cdots,A\_n,\cdots
$$
with $(A\_i,A\_{i+1})$ satisfies the Lowner order, and $A\_i\in D(\mathcal{H})$.


Does this sequence has least upper bound? In other words, does this class has limit?



Completeness of Lowner order in separable Hilbert space
This isn't research level, but if the sequence is increasing and bounded then yes, it has a least upper bound. For each $v \in H$ the sequence $\langle A\_iv,v\rangle$ is bounded and increasing and therefore converges to its least upper bound in $\mathbb{R}$. By polarization it follows that for all $v,w \in H$ the sequence $\langle A\_iv,w\rangle$ converges. Thus the sequence $(A\_i)$ converges in the weak operator topology. (In fact, it also converges strong operator.)


(Also, it's "Loewner".)


