Is $\Psi^0(\mathbb{R})$ (pseudodifferential operators with symbols obeying $ |\partial^\alpha_x \partial^\beta_\xi a(x,\xi)| \leq C_{\alpha,\beta} (1+|\xi|)^{-|\beta|} $ ) a $C^*$-algebra?

In other words, is $\Psi^0(\mathbb{R})$ is closed in $\mathcal{L}(L^2(\mathbb{R}))$ in the operator norm topology?

If not, then is there any nice characterization by the $C^*$-algebra generated by $\Psi^0$? Alternatively, what is the strongest (or just a reasonable) topology on $\mathcal{L}(L^2(\mathbb{R}))$ such that $\Psi^0$ is a closed subspace?

Is $\Psi^0(\mathbb{R})$ (pseudodifferential operators with symbols obeying $ |\partial^\alpha_x \partial^\beta_\xi a(x,\xi)| \leq C_{\alpha,\beta} (1+|\xi|)^{-|\beta|} $ ) a $C^*$-algebra?

In other words, is $\Psi^0(\mathbb{R})$ is closed in $\mathcal{L}(L^2(\mathbb{R}))$ in the operator norm topology?

If not, then is there any nice characterization by the $C^*$-algebra generated by $\Psi^0$? Alternatively, what is the strongest (or just a reasonable) topology on $\mathcal{L}(L^2(\mathbb{R}))$ such that $\Psi^0$ is a closed subspace?

Edit: Per Yemon Choi's comments below, the above question seems somewhat hopeless. As described here, $\Psi^0(\mathbb{R})$ is a Fréchet $*$-algebra with a topology stronger than the operator topology. I assume that this is the topology given by the seminorms on symbols: $$ \Vert a \Vert_{\alpha,\beta} = \sup_{x,\xi \in \mathbb{R}} (1+|\xi|)^{|\beta|} |\partial^\alpha_x \partial^\beta_\xi a(x,\xi)|. $$

So, in addition to the above question, I am adding the following question, to make it so that there might be an answer:

Is there a reasonable description of the smallest $C^*$-algebra containing $\Psi^0$?