Monotone approximation of elements in AF-algebras

Question

Suppose that we are given an AF-algebra $A$ and a sequence of finite-dimensional subalgebras $\mathbb{C}=A_0\subset A_1\subset A_2\subset\ldots$ such that $A=\overline{\bigcup\limits_{n\geq 0}A_n}$. Let me denote this dense subalgebra of $A$ by $A^{LS}$, i.e. $A^{LS}= \bigcup\limits_{n\geq 0}A_n$.

Next, we define the positive elements as $$A^+=\left\{h^2\ \middle|\ h\in A,\ h^*=h \right\}$$ and $$(A^{LS})^+=\left\{h^2\ \middle|\ h\in A^{LS},\ h^*=h \right\}.$$

Then for any $y\in A^+$ we can find a sequence $\{y_n\}_{n\geq 1}\subset (A^{LS})^+$ such that $\lim\limits_{n\to\infty}y_n=y$.

Question: Can we pick this $\{y_n\}$ in such a way that $y_n\leq y$ for any $n$ in the sense of the partial order defined by the cone $A^+$?

Note: existence of the desired sequence is equivalent to existence of the sequence $\{y_n\}$ such that $y_n\to y$ and for any $n$ there exists $M$ such that for any $m\geq M$ we have $y_n\leq y_{m}$.

Answer 1

No, this is not possible in general. $A^{LS}$ might have trivial intersection with a non-zero hereditary $C^\ast$-subalgebra of $A$, and thus any non-zero positive element in such a hereditary $C^\ast$-subalgebra cannot be approximated from below by elements in $A^{LS}$.

For a concrete example, I will give a non-unital example. You can simply unitise this example to get a unital counter-example (but I leave this to the reader).

Consider $\mathbb C^n \subseteq \ell^2(\mathbb N)$ in the obvious way, and let $A = \mathcal K(\ell^2(\mathbb N))$ with finite dimensional $C^\ast$-subalgebras $A_n = \mathcal K(\mathbb C^n)$. Fix any unit vector $\xi \in \ell^2(\mathbb N) \setminus \bigcup_n \mathbb C^n$ and let $p$ be the projection onto the span of $\xi$. Then $pAp \cap \bigcup_n A_n = \{0\}$, and thus $0$ is the only positive element in $\bigcup A_n$ which is below $p$.