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Let $A$ be a $B^{*}$ algebra and let $B$ be a sub $B^{*}$ algebra.From the fact that $d({\sigma}_{B}(x))$ is a subset of $d({\sigma}_{A}(x))$ where $d$ is the boundary., deduce that ${\sigma}_{B}(x)$/${\sigma}_{A}(x)$ is open. I was thinking if the boundary of set difference is set difference of boundary, then from the above, the boundary is empty but having tried to prove that I don't think that is true

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  • $\begingroup$ Use that fact that $\sigma_B(x)$ is the union of its boundary and of its interior. $\endgroup$
    – Etienne
    Feb 24, 2014 at 20:33
  • $\begingroup$ @Etienne From your hint, I thought maybe try 'de morgans' law but couldn't come up with anything useful $\endgroup$
    – user108605
    Feb 24, 2014 at 20:44

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Denote by $int(\sigma_B(x))$ the interior of $\sigma_B(x)$. Since $\sigma_B(x)$ is a closed set, we have $\sigma_B(x)=d(\sigma_B(x))\cup int(\sigma_B(x))$; and since $d(\sigma_B(x))\subset d(\sigma_A(x))\subset\sigma_A(x)$, it follows that $\sigma_B(x)\setminus \sigma_A(x)=\sigma_B(x)\cap \sigma_A(x)^c$ is equal to $int(\sigma_B(x))\cap \sigma_A(x)^c$. Hence $\sigma_B(x)\setminus \sigma_A(x)$ is indeed an open set.

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