chore:Clean up Blueprint

blueprint/src/BlueprintFiles/NormalityandSigmaWeak.tex

@@ -3,11 +3,11 @@ \chapter{Normality and Ultraweak Continuity for Positive Functionals}
 In what follows, let $M$ be a (nonzero) $W^{*}$--algebra. Let $\mathcal{P}(M)$
 denote the projection lattice of $M$. Let $\varphi$ be a positive linear functional $\varphi$ on $M$.
 We say $\varphi$ is \textbf{normal} if whenever $(p_{\alpha})$ is an increasing net of projections in $M$ with supremum $p$,
-we have $\varphi(p_{\alpha})\to \varphi(p)$. In this section we show that this property is equivalent to 
+we have $\varphi(p_{\alpha})\to \varphi(p)$. In this section we show that this property is equivalent to
 $\sigma(M, M_{*})$--continuity. We say that a linear functional $\varphi$ on $M$
 is \textbf{positive} if $\varphi(x)\ge 0$ whenever $x\ge 0$.
 
-\section{Ultraweakly Continuous Implies Normal} 
+\section{Ultraweakly Continuous Implies Normal}
 
 In what follows, let $T$ denote the set of all $\sigma$--continuous positive
 linear functionals on $M$, and $E$ the linear space of all finite linear combinations
@@ -17,7 +17,7 @@ \section{Ultraweakly Continuous Implies Normal}
   \label{lem:homeo_of_cts_bij_cpt_haus}
   \lean{isHomeomorph_iff_continuous_bijective}
   \mathlibok
-  Every continuous bijection from a compact space to a Hausdorff space is a 
+  Every continuous bijection from a compact space to a Hausdorff space is a
   homeomorphism.
 \end{lemma}
 
@@ -42,27 +42,30 @@ \section{Ultraweakly Continuous Implies Normal}
 \begin{theorem}
   \label{lem:sigma_cont_of_normal_Sak_1_7_4} (Sak 1.7.4)
   \uses{thm:Banach_Alaoglu,lem:homeo_of_cts_bij_cpt_haus,lem:uw_pos_sep_pts,lem:selfadjoint_le_norm,def:sigma_top,thm:comm_cstar_cts_fcns_Sak_1_2_1}
+  \leanok
+  \lean{Ultraweak.continuous_of_forall_posCLM,Ultraweak.continuousOn_of_forall_posCLM,Ultraweak.continuousAt_of_forall_posCLM,Ultraweak.continuousWithinAt_of_forall_posCLM,Ultraweak.DirectedOn.isLUB_star_right_conjugate}
   Every uniformly bounded, increasing net in $M^{s}$ converges
   to its least upper bound in the $\sigma$--topology. Further, if
   $x=\sup_{\lambda}x_{\lambda}$ then $a^{*}xa=\sup_{\lambda}a^{*}x_{\lambda}a$.
 \end{theorem}
 
 \begin{proof}
+  \leanok
   Letting $T$ denote the set of $\sigma$--continuous positive functionals on $M$ and $E$ its linear span, by
   Lemma \ref{lem:uw_pos_sep_pts}, the subspace $E$ separates the points of $M$ and so the $\sigma(M,E)$
   is Hausdorff. By the Banach-Alaoglu Theorem (\ref{thm:Banach_Alaoglu}) the closed unit ball $B$ of $M$ is
   compact in the $\sigma(M,M_{*})$--topology. Since $\sigma(M,E)$ is weaker than $\sigma(M,M_{*})$, the identity
-  map on $B$ is $\sigma(M,M_{*})-\sigma(M,E)$--continuous, and therefore a homeomorphism, 
+  map on $B$ is $\sigma(M,M_{*})-\sigma(M,E)$--continuous, and therefore a homeomorphism,
   by Lemma \ref{lem:homeo_of_cts_bij_cpt_haus}. So, to prove that a uniformly bounded net $(x_{t})$ is $\sigma$--
   Cauchy, it's enough to check that for all $\varepsilon>0$ and $\varphi\in T$ that there is a $\Lambda$ such that
   $\varphi(x_{\lambda}-x_{\mu})\le \varepsilon$ for $\lambda \ge \mu \ge \Lambda$.
-  
+
   If $(x_{\lambda})$ is a uniformly bounded increasing net in $M^{s}$, and $\varphi\in T$, then $(\varphi(x_{\lambda}))$ is
   a uniformly bounded increasing net of real numbers and therefore is Cauchy. By the above paragraph, $(x_{\lambda})$ is
   $\sigma$--Cauchy and by the $\sigma$--compactness of $B$ and the fact that $(x_{\lambda})$ is increasing there exists $x\in B$
-  so that $x_{\lambda} \to x$ in the $\sigma$--topology. Since for any $\varphi\in T$ and any $\lambda$ we have 
+  so that $x_{\lambda} \to x$ in the $\sigma$--topology. Since for any $\varphi\in T$ and any $\lambda$ we have
   $0 \le \varphi(x-x_{\lambda})$, by Lemma \ref{lem:non_pos_elem_neg_for_some_state_Sak_1_7_2} we have $0\le x - x_{\lambda}$.
-  Therefore $\sup(x_{\lambda})\le x$. For a given $\varphi\in T$ and $\varepsilon = \varphi(x-\sup(x_{\lambda}))$ one find 
+  Therefore $\sup(x_{\lambda})\le x$. For a given $\varphi\in T$ and $\varepsilon = \varphi(x-\sup(x_{\lambda}))$ one find
   $x_{\lambda}$ so that $\varphi(x-x_{\lambda})<\varphi (x - \sup(x_{\lambda}))$, which is a contradiction, therefore $x=\sup(x_{\lambda})$.
 
   By Lemma \ref{lem:star_conj_pos} and the fact that $(u^{-1})^{*}=(u^{*})^{-1}$, then $u^{*}x_{\lambda}u\le u^{*}xu$
@@ -71,13 +74,13 @@ \section{Ultraweakly Continuous Implies Normal}
   If $a\in M$ is not invertible, one can shift the spectrum horizontally to avoid zero, i.e. there is a $c>0$ so that $c1+a$ is invertible. Indeed
   observe that $a-t1=(a+c1)-(t+c)1$ and therefore $t\in \sigma(a)\iff t+c \in \sigma(a+c1)$. In particular if we let $t=-c$ here we find that
   $0\in \sigma(a)+c \iff 0 \in \sigma(a+c1)$.
-  By the invertible case above, 
+  By the invertible case above,
   \begin{align}
     c^2\varphi(x_{\lambda})+c\varphi(a^{*}x_{\lambda}+x_{\lambda}a)+\varphi(a^{*}x_{\lambda}a)&=\varphi((c1+a)^{*}x_{\lambda}(c1+a))\\\nonumber
     &\to \varphi((c1+a)^{*}x(c1+a))
   \end{align}
-  for all $\varphi\in T$. 
-  It suffices to show that $\varphi(a^{*}x_{\lambda}+x_{\lambda}a)\to \varphi(a^{*}x+xa)$ for all $\varphi\in T$, 
+  for all $\varphi\in T$.
+  It suffices to show that $\varphi(a^{*}x_{\lambda}+x_{\lambda}a)\to \varphi(a^{*}x+xa)$ for all $\varphi\in T$,
   because then $\varphi(a^{*}x_{\lambda}a)\to \varphi(a^{*}xa)$ by the above. We can use Cauchy-Schwartz for this as follows. If $\alpha\le \beta$,
   we have $x_{\beta}-x_{\alpha}\ge 0$, and since the net $(x_{\lambda})$ is uniformly bounded there exists $M>0$ such that (using \ref{lem:selfadjoint_le_norm})
   \begin{align}
@@ -95,7 +98,7 @@ \section{Normal Implies Ultraweakly Continuous}
 \end{proof}
 
 If $P:\mathcal{P}(M)\to \text{Prop}$ is a predicate, the usual ordering ``$\le$'' on projections
-induces an order on the set $\{p\in \mathcal{P}(M)|P(p)\}$. In what follows there will be no confusion if 
+induces an order on the set $\{p\in \mathcal{P}(M)|P(p)\}$. In what follows there will be no confusion if
 we also denote this induced order by ``$\le$''.
 
 \begin{lemma}
@@ -117,40 +120,40 @@ \section{Normal Implies Ultraweakly Continuous}
 For all positive linear functionals $\varphi,\psi$ with
 $\varphi$ normal and $\psi$ $\sigma(M,M_{*})$--continuous and every nonzero $p\in \mathcal{P}(M)$
 such that $\varphi(p)<\psi(p)$, there exists a nonzero $p_1 \in \mathcal{P}(M)$ such that $p_1\le p$ and for all nonzero
-$q\in \mathcal{P}(M)$ with $q\le p_1$, we have $\varphi(q)<\psi(q)$. 
+$q\in \mathcal{P}(M)$ with $q\le p_1$, we have $\varphi(q)<\psi(q)$.
 \end{lemma}
 
 \begin{proof}
   We proceed by contradiction. Suppose the conclusion does not hold. Then, for every nonzero subprojection
   $p_1$ of $p$ there is a nonzero subprojection $q\le p_1$ such that $\varphi(q)\ge \psi(q)$. In particular,
   (letting $p_1=q$) there is a $q\le p$ such that $\varphi(q)\ge \psi(q)$. If $(q_\alpha)$ is a chain
-  of such nonzero projections, $q_{\alpha}\to \sup q_{\alpha}$ in the $\sigma$--topology by Theorem \ref{lem:sigma_cont_of_normal_Sak_1_7_4}, 
+  of such nonzero projections, $q_{\alpha}\to \sup q_{\alpha}$ in the $\sigma$--topology by Theorem \ref{lem:sigma_cont_of_normal_Sak_1_7_4},
   and by Lemma \ref{prop:proj_compl_lat_wstar_Sak_1_10_2} we know this supremum is a subprojection of $p_1$. Since $\varphi$
   is positive and normal, and $\psi$ is positive and $\sigma$--continuous, we have $\varphi(\sup q_{\alpha})\ge \psi(\sup q_{\alpha})$.
-  Therefore, by Zorn's Lemma, there is a maximal $q_{0}\le p$ such that $\varphi(q_0)\ge \psi(q_0)$. We claim $q_{0}=p$. If not, 
+  Therefore, by Zorn's Lemma, there is a maximal $q_{0}\le p$ such that $\varphi(q_0)\ge \psi(q_0)$. We claim $q_{0}=p$. If not,
   $p-q_0$ is a nonzero subprojection of $p$ by Corollary \ref{cor:proj_sub_of_subproj}, and there exists a nonzero projection $q_{1}\le p-q_{0}$
-  such that $\varphi(q_{1})\ge \psi(q_{1})$. But then $q_{0}$ is a proper subprojection of $q_0+q_1$ and by linearity and positivity 
-  $\varphi(q_0+q_1)\ge \psi(q_0+q_1)$ contradicting the maximality of $q_0$. Thus $p=q_0$ and so $\varphi(p) \ge \psi(p)$, contradicting 
-  the hypothesis $\varphi(p)<\psi(p)$.  
+  such that $\varphi(q_{1})\ge \psi(q_{1})$. But then $q_{0}$ is a proper subprojection of $q_0+q_1$ and by linearity and positivity
+  $\varphi(q_0+q_1)\ge \psi(q_0+q_1)$ contradicting the maximality of $q_0$. Thus $p=q_0$ and so $\varphi(p) \ge \psi(p)$, contradicting
+  the hypothesis $\varphi(p)<\psi(p)$.
 \end{proof}
 
-Note that the following likely contains the most crucial use of the normality of $\varphi$ for the later proof 
+Note that the following likely contains the most crucial use of the normality of $\varphi$ for the later proof
 of Theorem \ref{thm:sigma_cts_of_normal_Sak_1_13_2}.
 
 \begin{lemma}
   \label{lem:zorn_base}
   \uses{cor:proj_sub_of_subproj,lem:selfadjoint_le_norm,lem:sigma_cont_of_normal_Sak_1_7_4,prop:proj_compl_lat_wstar_Sak_1_10_2,def:sigma_top}
-  Let $\varphi$ be a normal positive linear functional on $M$ and consider the predicate $P:\mathcal{P}(M)\to \text{Prop}$ defined, for $p\in \mathcal{P}(M)$, by 
+  Let $\varphi$ be a normal positive linear functional on $M$ and consider the predicate $P:\mathcal{P}(M)\to \text{Prop}$ defined, for $p\in \mathcal{P}(M)$, by
   ``$M\ni x \mapsto \varphi(xp)$ is $\sigma(M,M_{*})$--continuous''. If $(p_{\alpha})$ is a chain of projections in $M$ such that $P(p_{\alpha})$ is true for each $\alpha$,
-  then $P(\sup(p_{\alpha}))$ is true. Hence by Zorn's Lemma there is a maximal $p_0\in \mathcal{P}(M)$ such that $P(p_0)$ is true.  
+  then $P(\sup(p_{\alpha}))$ is true. Hence by Zorn's Lemma there is a maximal $p_0\in \mathcal{P}(M)$ such that $P(p_0)$ is true.
 \end{lemma}
 
 \begin{proof}
-  Let $x$ be on the unit sphere and let $p$ be the supremum of the $p_{\alpha}$. By Theorem \ref{lem:sigma_cont_of_normal_Sak_1_7_4} and Lemma \ref{prop:proj_compl_lat_wstar_Sak_1_10_2} 
-  we know $p_{\alpha}$ converges in $\sigma$--topology to $p$, and $p$ is a projection. 
+  Let $x$ be on the unit sphere and let $p$ be the supremum of the $p_{\alpha}$. By Theorem \ref{lem:sigma_cont_of_normal_Sak_1_7_4} and Lemma \ref{prop:proj_compl_lat_wstar_Sak_1_10_2}
+  we know $p_{\alpha}$ converges in $\sigma$--topology to $p$, and $p$ is a projection.
   By Corollary \ref{cor:proj_sub_of_subproj} we know $p-p_\alpha$ is a projection. By the $C^{*}$--property
   of the norm ($\|x^{*}x\|=\|x\|^{2}$), Lemma \ref{lem:selfadjoint_le_norm} and the positivity of $\varphi$ we have $\varphi(x^{*}x)\le \varphi(1)$.
-  Now these facts together with Cauchy-Schwartz and the monotonicity of square roots, 
+  Now these facts together with Cauchy-Schwartz and the monotonicity of square roots,
   \begin{align}
     |\varphi(x(p-p_{\alpha}))|&\le \varphi(x^{*}x)^{1/2}\varphi(p-p_{\alpha})^{1/2}\\ \nonumber
     &\le \varphi(1)^{1/2}\varphi(p-p_{\alpha})^{1/2}.
@@ -182,11 +185,11 @@ \section{Normal Implies Ultraweakly Continuous}
   a maximal $p_0\in \mathcal{P}(M)$ such that $M\ni x \mapsto \varphi(xp_0)$ is $\sigma(M,M_{*})$--continuous. Assume for the purposes of finding
   a contradiction that $p_0\ne 1$. By Lemma \ref{lem:exists_uw_ge_normal}
   there is a $\sigma(M,M_{*})$--continuous positive functional $\psi$ on $M$ such that $\varphi(1-p_0)<\psi(1-p_0)$. By Lemma \ref{lem:msr_th_lemma}
-  there is a nonzero subprojection $p\le 1-p_0$ in $M$ such that $\varphi(q)< \psi(q)$ for every nonzero $q\le p$ in $M$. Let $x\in pMp$ be on the unit sphere. 
+  there is a nonzero subprojection $p\le 1-p_0$ in $M$ such that $\varphi(q)< \psi(q)$ for every nonzero $q\le p$ in $M$. Let $x\in pMp$ be on the unit sphere.
   Then $x^{*}x$ is positive and hence normal, so the $C^{*}$--subalgebra of $pMp$ generated by $x^{*}x$ and $p$ is commutative, and is hence contained in a maximal abelian
-  $*$--subalgebra $A$ of $pMp$. Now $A$ is a $W^{*}$--subalgebra of $pMp$ by Lemma \ref{lem:wstar_of_masa} and hence is a maximal commutative $C^{*}$--subalgebra of $pMp$. Via the Gelfand Transform, 
-  $A$ is star isomorphic to $C(K)$, where $K$ is Stonean by Lemmas \ref{lem:cut_down_of_wstar} and \ref{lem:spec_masa_wstar_stonean_Sak_1_7_5}. 
-  By Lemma \ref{lem:fin_lin_approx_of_stonean_Sak_1_3_1} it follows that $\varphi(a)\le \psi(a)$ for every $a\ge 0$ in $A$, which holds a fortiori for $a\ge 0$ in $C^{*}(x^{*}x,p)$. 
+  $*$--subalgebra $A$ of $pMp$. Now $A$ is a $W^{*}$--subalgebra of $pMp$ by Lemma \ref{lem:wstar_of_masa} and hence is a maximal commutative $C^{*}$--subalgebra of $pMp$. Via the Gelfand Transform,
+  $A$ is star isomorphic to $C(K)$, where $K$ is Stonean by Lemmas \ref{lem:cut_down_of_wstar} and \ref{lem:spec_masa_wstar_stonean_Sak_1_7_5}.
+  By Lemma \ref{lem:fin_lin_approx_of_stonean_Sak_1_3_1} it follows that $\varphi(a)\le \psi(a)$ for every $a\ge 0$ in $A$, which holds a fortiori for $a\ge 0$ in $C^{*}(x^{*}x,p)$.
   In particular, $\varphi(px^{*}xp)\le \psi(px^{*}xp)$. Therefore,
   \begin{align}
     |\varphi(x(p_0+p))|&\le |\varphi(xp_0)|+|\varphi(xp)|\\ \nonumber
@@ -195,8 +198,8 @@ \section{Normal Implies Ultraweakly Continuous}
   \end{align}
   Since $x\mapsto \varphi(xp_0)$ is $\sigma$--continuous, it is $s$-continuous by Lemma \ref{lem:bdd_sigma_cts_iff_bdd_s_cts_Sak_1_8_10}. The seminorm
   $x \mapsto \psi(px^{*}xp)^{1/2}$ is a defining seminorm for the $s$--topology on $M$. It follows that $x\mapsto \varphi(x(p_0+p))$ is $s$-continuous
-  and therefore $\sigma$--continuous, by Corollary \ref{cor:sigma_eq_s_on_ball}. This contradicts the maximality of $p_0$, and therefore $p_0=1$ and the result follows. 
+  and therefore $\sigma$--continuous, by Corollary \ref{cor:sigma_eq_s_on_ball}. This contradicts the maximality of $p_0$, and therefore $p_0=1$ and the result follows.
 \end{proof}
 
 The above result gives us that the $\sigma$-weak continuity of a positive functional is determined entirely by the order structure on $M$,
-and thus is independent of the choice of predual. 
+and thus is independent of the choice of predual.

blueprint/src/BlueprintFiles/StoneanMaxCom.tex

@@ -8,13 +8,15 @@ \chapter{Stonean Spaces and Maximal Commutative Subalgebras}
 If $K$ is Stonean, there are many such connected components:
 
 \begin{lemma}
-  \label{lem:fin_lin_approx_of_stonean_Sak_1_3_1} (Sak  1.3.1)
-  \uses{cor:spec_sa_homeo_spec_cstar}
-  Let $K$ be a Stonean space. Then every positive self-adjoint element $a$ in $C(K)$ can be uniformly approximated by 
-  finite linear combinations of projections in $C(K)$ having nonnegative coefficients. 
+  \label{lem:fin_lin_approx_of_stonean_Sak_1_3_1} (Sak 1.3.1)
+  \lean{IsSelfAdjoint.mem_span_isStarProjection_of_finite_quasispectrum,instFiniteSpectrumContinuousMapOfTotallySeparatedSpaceOfCompactSpace}
+  \leanok
+  Let $K$ be a Stonean space. Then every positive self-adjoint element $a$ in $C(K)$ can be uniformly approximated by
+  finite linear combinations of projections in $C(K)$ having nonnegative coefficients.
 \end{lemma}
 
 \begin{proof}
+  \leanok
 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 $|\lambda_{i+1}-\lambda_{i}|<\varepsilon$.
@@ -29,13 +31,13 @@ \chapter{Stonean Spaces and Maximal Commutative Subalgebras}
 
 In Sakai, the pointwise order on $C(K)$ in Proposition \ref{prop:stonean_of_cts_fcns_incr_cond_complete_Sak_1_3_2}
 isn't specified beforehand in the book. Following $C^{*}$--algebra theory, $a\ge 0$ in $C(K)$
-iff the range of $a$ is contained in $\mathbb{R}_{\ge 0}$, and $b\le a$ iff $b-a \ge 0$. The spectrum 
+iff the range of $a$ is contained in $\mathbb{R}_{\ge 0}$, and $b\le a$ iff $b-a \ge 0$. The spectrum
 of a function in $C(K)$ is just its range. So $b\le a$ iff $b(t)\le a(t)$ for all $t\in K$. A net
  $(f_{\alpha})$ in $C(K)$ is \textbf{bounded} if there is a positive constant that bounds the norms of all $f_{\alpha}$.
 A net is said to be \textbf{increasing} if $\alpha \le \beta$ implies $f_{\alpha}\le f_{\beta}$ for all $\alpha,\beta$
 in the associated directed set. We ought to note, here, that Sakai opts not to use net language
 (although he sort of does when introducing subscripts) since his directed set is identical to his net
-in Proposition \ref{prop:stonean_of_cts_fcns_incr_cond_complete_Sak_1_3_2}. Recall that 
+in Proposition \ref{prop:stonean_of_cts_fcns_incr_cond_complete_Sak_1_3_2}. Recall that
 the \textbf{support} of $f\in C(K)$ is $\overline{\{x\in K | f(x)\ne 0\}}$.
 
 \begin{lemma}
@@ -45,7 +47,7 @@ \chapter{Stonean Spaces and Maximal Commutative Subalgebras}
 \end{lemma}
 
 \begin{proof}
-    The usual order is a partial order, and given $f,g\in C(K)$ supported on $U$ we have $f\vee g \in C(K)$, 
+    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 $f\vee g(t)=\max\{f(t),g(t)\}$.
 \end{proof}
 
@@ -54,20 +56,20 @@ \chapter{Stonean Spaces and Maximal Commutative Subalgebras}
     \uses{lem:cts_fns_ge_zero_le_one_directed,lem:selfadjoint_le_norm}
     Let $K$ be a compact Hausdorff space. Suppose every bounded increasing net
     of real valued, non-negative functions in $C(K)$ has a least upper bound in $C(K)$.
-    Then $K$ is Stonean. 
+    Then $K$ is Stonean.
 \end{proposition}
 
 \begin{proof}
     Suppose for the sake of a contradiction that $K$ isn't Stonean, i.e. that there exists an
     open set $U$ whose closure isn't open, i.e. that $\overline{U}^{c}$ isn't closed.
-    Consider the set of $f$ in $C(K)$ with support contained in $U$ and such that $0\le f\le 1$. 
-    By Lemma \ref{lem:selfadjoint_le_norm} and Lemma \ref{lem:cts_fns_ge_zero_le_one_directed}, 
-    this is a bounded, increasing net with respect to the order on $C(K)$. 
+    Consider the set of $f$ in $C(K)$ with support contained in $U$ and such that $0\le f\le 1$.
+    By Lemma \ref{lem:selfadjoint_le_norm} and Lemma \ref{lem:cts_fns_ge_zero_le_one_directed},
+    this 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 $\overline{U}^{c}\subset U^{c}$, 
+    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 $\overline{U}^{c}\subset U^{c}$,
     $f_c$ 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 $f_c(p)=0$ But $f_c(p)=1$ by the above, which is a contradiction.
 \end{proof}
@@ -80,7 +82,7 @@ \chapter{Stonean Spaces and Maximal Commutative Subalgebras}
 \end{lemma}
 
 \begin{proof}
-  Suppose $(a_{\lambda})$ is a uniformly bounded, increasing net of positive elements in $C$ with supremum $a$. 
+  Suppose $(a_{\lambda})$ is a uniformly bounded, increasing net of positive elements in $C$ with supremum $a$.
   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 \ref{prop:stonean_of_cts_fcns_incr_cond_complete_Sak_1_3_2} after applying the Gelfand transform.