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 $\lim\limits_{n\to\infty}y_n=y$.
Questions:
Is it true that we can pick this $\{y_n\}$ in such a way that $y_n\leq y_{n+1}$ for any $n$ with respect to the partial order $\leq$ on $A^{LS}$ defined by $(A^{LS})^+$?
More generally. Suppose we are given a unital separable $C^*$-algebra $A$ with a unital dense $*$-subalgebra $\widetilde{A}\subset A$. Is it true that any $y\in A^+$ can be approximated by a monotone (in the sence of the partial order defined by positive elements) sequence of elements from $\left(\widetilde{A}\right)^+$? Is it possible at least for any $y\in A^+$ and any open neighbourhood of $y$ find an element $z\in \left(\widetilde{A}\right)^+$ such that $z$ belongs to that neighbourhood and $z\leq y$?
Thanks in advance!
This only answers question 2. I've been trying question 1 for days but with no progress. I am convinced that question 1 also fails, but I cannot construct a counter-example. I believe the easiest way to obtain a counter-example is by looking at the UHF algebra $\mathbb{M}_{2^\infty}$, But I cannot find a counter example.
Remark: If $A$ is a $C^*$-algebra and $D$ is a dense subalgebra, we can approximate any positive element of $A$ by positive elements of $D$ (take $a^*a\in A_+$, approximate $a$ by a sequence $d_n$ in $D$, then $d_n^*d_n$ is a sequence of positive elements of $D$ that approximates $a^*a$).
The general question, i.e. question 2 fails. Consider the $C^*$-algebra $C[0,1]$. Denote by $D$ the $*$-subalgebra of the polynomials. By Weierstrass' theorem, $D$ is dense in $C[0,1]$. Now let $f(t)$ be any continuous, not constantly $0$ map on $[0,1]$ such that $f(t)=0$ on some interval $[\alpha,\beta]\subset[0,1]$ (for example set $f(t)=0$ on $[0,1/2]$, $f(1)=1$ and extend linearly). Assume that there exists an increasing sequence of positive polynomials $(p_n(t))$ such that $p_n\to f$ uniformly (or even pointwise). Then $p_n(t)\leq p_{n+1}(t)$ for all $n$ and all $t\in[0,1]$, so $f(t)=\sup_{n}p_n(t)$. But then this means that $p_n(t)=0$ on an interval $[\alpha,\beta]$ for all $n$, so each $p_n$ has infinitely many roots, and, by the fundamental theorem of algebra, each $p_n$ is constantly zero. This is a contradiction. This example also gives a negative answer to the second part of question 2.