Approximate Identity: Projections

Question

Problem

Given a C*-algebra $\mathcal{A}$.

Denote the positive open unit ball by: $\mathcal{B}^+$

Then it has an approximate identity: $$A\in\mathcal{A}:\quad\|A-AE\|,\|A-EA\|\stackrel{E\to1}{\to}0\quad(E\in\mathcal{B}^+)$$

Denote the self adjoint projections by: $\mathcal{P}$

Do they form an approximate identity, too: $$A\in\mathcal{A}:\quad\|A-AP\|,\|A-PA\|\stackrel{P\to1}{\to}0\quad(P\in\mathcal{P})$$

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Answer 1

No, they don't!

First, consider the operator algebra generated by the right shift: $$\mathcal{A}_R:=\overline{\langle\{R\}\rangle}\subseteq\mathcal{B}(\ell^2(\mathbb{N}))$$

Then the collection of selfadjoint projections is even empty!

Next, consider a proper selfadjoint projections adjoined: $$\mathcal{A}_{RP}:=\overline{\langle\{R,P\neq1\}\rangle}\subseteq\mathcal{B}(\ell^2(\mathbb{N}))$$

Then the collection still does not form an approximate identity: $$\mathcal{P}_{RP}=\{P\}:\quad\|R-RP\|=1$$

Finally, that won't change until adjoining the identity as: $$P\in\mathcal{P}:\quad\|R-RP\|\equiv1\quad(P\neq1)$$ (That is because the right shift is isometric!)

Remark: From operator theory one knows they form only strong approximate identities...