Bounding a positive element from below using a dense subset

Question

Let $A$ be a $C^*$-algebra. Let $a$ be a nonzero positive element of $A$. Suppose that $A$ equals the closed span of a subset $B$ of $A$, where $B$ is closed under the $*$ operation, linear combinations, and products. Is it possible to find $b \in B$ such that $b$ is positive, nonzero, and $a \geq b$ (i.e. $a-b$ is positive)?

Answer 1

No. Let $A=C[-1,1]$, and $a(t)=e^{-1/t^2}$ (of course, with $a(0)=0$). Take $B=\mathbb C[x]$, the polynomials. If $p\in B$ and $0\leq p\leq a$, so $p(0)=0$. But then $$ |p'(0)|=\lim_h\left|\frac{p(h)-p(0)}h\right|=\lim_h\left|\frac{p(h)}h\right|\leq\lim_h\frac{|a(h)|}{|h|}=0. $$ And then we can iterate this reasoning to get $p^{(k)}(0)=0$ for all $k$; thus $p=0$.

Answer 2

In general, it seems like no. Consider $A = \mathbb{C}$. For any $\lambda \in \mathbb{C} \setminus \{0\}$, $\langle \lambda \rangle = \mathbb{C}$. However, $\lambda$ can be chosen so that $B = \{\lambda\}$ does not even contain a positive element.