Lemma 41.2.4.label Let $A$ be a $C^{*}$-algebra, $S \subset A$ be an operator system, then $(\ol{S})_{+} = \ol{S_+}$.
Proof. Let $x \in (\ol{S})_{+}$, then there exists $\seq{x_n}\subset S$ such that $x_{n} \to x$ as $n \to \infty$. In which case,
\[x = \frac{x + x^{*}}{2}= \limv{n}\frac{x_{n} + x_{n}^{*}}{2}\in \ol{S_{sa}}\]
Assume without loss of generality that $\seq{x_n}\subset S_{sa}$, so that $\sigma_{A}(x_{n}) \subset \real$ for all $n \in \natp$. By upper semicontinuity of the spectrum, $\lim_{n \to \infty}\inf(\sigma_{A}(x_{n})) \ge 0$. Therefore
\[x = \limv{n}x_{n} = \limv{n}[x_{n} + \inf(\sigma_{A}(x_{n})) \wedge 0] \in \ol{S_+}\]
by Corollary 38.6.3.$\square$
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