Since a is self-adjoint, its spectrum is its range in \mathbb{R}_{\ge 0}, and is therefore contained in [0, \|a\|], where \|a\| = \sup_{t \in K} |a(t)|. Let \varepsilon > 0 and choose real numbers 0 < \lambda_1 < \lambda_2 < \ldots < \lambda_n < \|a\| + 1 so that |\lambda_{i+1} - \lambda_i| < \varepsilon. Now let G_1 = \overline{\{t \in K \mid a(t) < \lambda_1\}} and G_i = \overline{\{t \in K \setminus \bigcup_{j=1}^{i-1} G_j \mid a(t) < \lambda_i\}} for i > 1. These are pairwise disjoint, and the \chi_{G_i} are each continuous projections. Furthermore, for all t \in K,

\left| a(t) - \sum_{i=1}^{n} \lambda_i \chi_{G_i}(t) \right| < \varepsilon

and therefore \left\| a - \sum_{i=1}^{n} \lambda_i \chi_{G_i} \right\| < \varepsilon.

The usual order is a partial order, and given f, g \in C(K) supported on U we have f \vee g \in C(K), supported on U, where (f \vee g)(t) = \max\{f(t), g(t)\}.

Suppose for the sake of a contradiction that K is not Stonean, i.e. that there exists an open set U whose closure is not open, i.e. that \overline{U}^c is not closed. Consider the set of f in C(K) with support contained in U and such that 0 \le f \le 1. It is directed (Lemma 6.2), and it is bounded by Lemma 3.3; so it is a bounded, increasing net with respect to the order on C(K).

For any point t \in U, by Urysohn's Lemma (compact Hausdorff spaces are regular) there exists a continuous function g \in C(K) such that g(t) = 1 and g is zero on U^c. It follows that if f_s is the least upper bound of the above net, then 1 = g(t) \le f_s(t), and therefore (since we can play this game for any t \in U) it must be that f_s(t) = 1 for any t \in U. By continuity, f_s(t) = 1 for all t \in \overline{U}. Now, since \overline{U}^c \subset U^c, f_s is identically zero on U^c. Since \overline{U}^c is not closed, it must have a limit point p \in \overline{U}, and by continuity f_s(p) = 0. But f_s(p) = 1 by the above, which is a contradiction.

Suppose (a_\lambda) is a uniformly bounded, increasing net of positive elements in C with supremum a in M, which exists by Theorem 7.1.4. Given any unitary u \in C, we have u^* \sup_\lambda a_\lambda u = \sup_\lambda u^* a_\lambda u, and since every element in C is a linear combination of unitary elements from C, we have that \sup_\lambda a_\lambda \in C. The result follows from Proposition 6.3 after applying the Gelfand transform.

Place the positive element a in a maximal abelian *-subalgebra C of M. By Lemma 6.4 the spectrum space of C is Stonean, so Lemma 6.1 applies to the Gelfand transform of a. The second statement follows since every element of M is a linear combination of positive elements.