diff --git a/blueprint/src/chapter/main.tex b/blueprint/src/chapter/main.tex
index 9236ae71..a810ef1b 100644
--- a/blueprint/src/chapter/main.tex
+++ b/blueprint/src/chapter/main.tex
@@ -750,7 +750,7 @@ \chapter{T. \ref{thm main 1} from finitary P.
 where $\sigma_1$ is the smallest integer such that $D^{\sigma_1-2}R_2>\frac 1{4D}$ and $\sigma_2$
 is the largest integer so that $D^{\sigma_2+2}R_1<\frac 12$. Here we restricted the summation index $s$
 by omitting the summands with $s<\sigma_1-2$
-or $s>s_2+2$ because for these summands the function $K_s$ vanishes on the domain of integration, and we have ommitted the restriction in the integral
+or $s>s_2+2$ because for these summands the function $K_s$ vanishes on the domain of integration, and we have omitted the restriction in the integral
 in the  summands in \eqref{middles} because in theses summands the support of $K_s$ is contained in
 the set described by this restriction.
 
@@ -1063,7 +1063,7 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
         \mu(B(y,2^{j}D^k))\le A^j \mu(B(y,D^k))\, .
     \end{equation}
     As $X$ is the union of the balls $B(y,2^{j}D^k)$ and $\mu$ is not zero, at least one of
-    the balls $B(y,2^{j}D^k)$ has positive measure und thus $B(y,D^k)$ has positive measure.
+    the balls $B(y,2^{j}D^k)$ has positive measure and thus $B(y,D^k)$ has positive measure.
 
     Applying \eqref{jballs} for $j'$ the smallest integer larger than $\ln_2(8D^{2S})$, using $-S\le k\le S$
     and $y\in B(o,4D^S)$ and the triangle inequality, we have
@@ -1299,7 +1299,7 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 
 Assume now the case $x\notin X_k$.
 By \eqref{definei3}, we have
-    $x\in I_2(y,k)$. Morover, for any $u<y$ in
+    $x\in I_2(y,k)$. Moreover, for any $u<y$ in
     $Y_k$ we have $x\not\in I_3(u,k)$.
     Let $u<y$. By transitivity of the order in $Y_k$, we conclude $x\not \in I_2(y,k)$.
 By \eqref{defineij} and the disjointness property of Lemma \ref{firstgridlemma}, we have
@@ -2472,7 +2472,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 Using $D = 2^{100a^2}$ and $a \ge 4$ and $\kappa Z \ge 1$ proves the lemma.
 \end{proof}
 
-\section{Auxilliary lemmas}
+\section{Auxiliary lemmas}
 
 Before proving Lemma \ref{subsecflemma} and Lemma \ref{subsecalemma}, we collect some useful properties of $\lesssim$.
 
@@ -3071,8 +3071,8 @@ \section{The density  arguments}\label{sec TT* T*T}
 \begin{proof}
 Let $\fp,\fp'$ and $x$ be given.
 Assume without loss of generality that $\ps(\fp)\le \ps(\fp')$.
-As  we have $x\in E(\fp)\subset \sc(\fp)$  and $x\in E(\fp')\subset \sc(\fp')$ by Definition \eqref{defineep}, we conclude
-withfor $i=1,2$ that
+As we have $x\in E(\fp)\subset \sc(\fp)$  and $x\in E(\fp')\subset \sc(\fp')$ by Definition \eqref{defineep}, we conclude
+for $i=1,2$ that
 $\tQ(x)\in\fc(\fp)$ and $\tQ(x)\in\fc(\fp')$. By \eqref{eq freq dyadic} we have $\fc(\fp')\subset \fc(\fp)$. By Definition
 \eqref{straightorder}, we conclude $\fp\le \fp'$. As $\mathfrak{A}$ is an antichain, we conclude $\fp=\fp'$.
 This proves the lemma.
@@ -3963,7 +3963,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 We turn to \eqref{eqanti2}.
 Consider the element $\fp'\in \mathfrak{A}'$ as above
 with $L\subset \sc(\fp')$ and $s(L)<s(\fp')$.
-As $L\subset L'$, we conclude twith the dyadic property that $L'\subset \sc(\fp')$.
+As $L\subset L'$, we conclude with the dyadic property that $L'\subset \sc(\fp')$.
 By maximality of $L$, we have
 $L'\not\in \mathcal{L}$.
 This together with the existence of the given $\fp'\in \mathfrak{A}$
@@ -5971,7 +5971,7 @@ \chapter{Proof of P. \ref{prop hlm}, Besicovitch covering and Hardy Littlewood}
     =2^{2a} q(q-q')^{-1}\|h\|_q\, .
 \end{equation}
 This proves \eqref{eq hlm} in general.
-and completes the proof of Propsoition \ref{prop hlm}.
+and completes the proof of Proposition \ref{prop hlm}.
 
 \chapter{Proof of Theorem \ref{classical}}
 
@@ -6026,7 +6026,7 @@ \section{The classical Carleson theorem}
 \begin{proof}
 Let $F$ be any set in $\mathbb{C}$. We show that
 $f_0^{-1}(F)$ is measurable. As
-$Z:=\{2\pi k/K, k\in \mathbb{Z}\}$ is countable, it suffcies to show
+$Z:=\{2\pi k/K, k\in \mathbb{Z}\}$ is countable, it suffices to show
 that $f_0^{-1}(F)\setminus Z$ is measurable. It then
 suffices to show that $f_0^{-1}(F)\setminus Z$ is open.
 Let $x\in f_0^{-1}(F)\setminus Z$. There is a $k$
@@ -6277,7 +6277,7 @@ \section{Piecewise constant functions.}
 
 
 
-We first compute the partial Fourier sumes for constant functions.
+We first compute the partial Fourier sums for constant functions.
 
 
 \begin{lemma}\label{constant}
@@ -6789,7 +6789,7 @@ \section{The truncated Hilbert transform}
 \begin{equation}\label{eqhil15}
     2^{10}\left(1+\frac{r}{|1-e^{ix}|^2}\right)\le 2^{3}r^{-1}\, .
 \end{equation}
-    For thos $x$, the left-hand side  of \eqref{eqhil17} becomes
+    For those $x$, the left-hand side  of \eqref{eqhil17} becomes
     \begin{equation}
     |\frac{1}{1-e^{-ix}}-\frac{1}{1-e^{-ix'}}|\, .
 \end{equation}