I have a $C^*$-algebra $\mathcal{A}$, and would like to make use of the spectral order $\preceq$ coming from (the self-adjoint part of) its enveloping von Neumann algebra $\mathcal{A}^{**}$.

I am most interested in checking that the spectral join/meet of finite subsets of $\mathcal{A}^\text{sa}$ are contained in $\mathcal{A}^\text{sa}$. It suffices to check for pairs only.

Given self-adjoint $a, b \in \mathcal{A}^{**}$, the spectral join $a \vee b$ turns out to be the self-adjoint element of $\mathcal{A}^{**}$ whose spectral projections satisfy $$\chi_{(\lambda, \infty)}(a \vee b) = \chi_{(\lambda, \infty)}(a) \vee \chi_{(\lambda, \infty)}(b),$$ where $\chi_S$ denotes the indicator function of a subset $S \subset \mathbb{R}$, and the $\vee$ on the right is the standard join for projections of a von Neumann algebra.

If we restrict $a$ and $b$ to be elements of $\mathcal{A}$, then the two projections on the right above are open (in the sense described by Akemann in The general Stone-Weierstrass problem, 1969), and hence so is the one on the left (which is necessary for inclusion of $a \vee b$ in $\mathcal{A}$). However, I am not sure how to guarantee that $a \vee b \in \mathcal{A}$.

Would anyone have a reference or proof relevant to my objective? If it happens that $a \vee b$ is not an element of $\mathcal{A}$, might $a$ and $b$ still have a larger join in the smaller lattice $(\mathcal{A}, \preceq)$? I am fairly new to the operator algebra world, so I apologise if this is a standard result that I just haven't been able to track down in the literature.

I have a $C^*$-algebra $\mathcal{A}$, and would like to make use of the spectral order $\preceq$ coming from (the self-adjoint part of) its enveloping von Neumann algebra $\mathcal{A}^{**}$.

I am most interested in checking that the spectral join/meet of finite subsets of $\mathcal{A}^\text{sa}$ are contained in $\mathcal{A}^\text{sa}$. It suffices to check for pairs only.

Given self-adjoint $a, b \in \mathcal{A}^{**}$, the spectral join $a \vee b$ turns out to be the self-adjoint element of $\mathcal{A}^{**}$ whose spectral projections satisfy $$\chi_{(\lambda, \infty)}(a \vee b) = \chi_{(\lambda, \infty)}(a) \vee \chi_{(\lambda, \infty)}(b),$$ where $\chi_S$ denotes the indicator function of a subset $S \subset \mathbb{R}$, and the $\vee$ on the right is the standard join for projections of a von Neumann algebra.

If we restrict $a$ and $b$ to be elements of $\mathcal{A}$, then the two projections on the right above are open (in the sense described by Akemann in The general Stone-Weierstrass problem, 1969), and hence so is the one on the left (which is necessary for inclusion of $a \vee b$ in $\mathcal{A}$). However, I am not sure how to guarantee that $a \vee b \in \mathcal{A}$.

Would anyone have a reference or proof relevant to my objective? If it happens that $a \vee b$ is not an element of $\mathcal{A}$, might $a$ and $b$ still have a larger join in the smaller poset $(\mathcal{A}, \preceq|_{\mathcal{A}})$? I am fairly new to the operator algebra world, so I apologise if this is a standard result that I just haven't been able to track down in the literature.