Very minor polish

ch-approximate.tex

@@ -762,7 +762,9 @@ \subsection{Differentiation}
 
 \begin{example}
 Let us construct a continuous nowhere differentiable function.
-Such functions are often called Weierstrass functions, although this
+Such functions are often called
+\emph{Weierstrass functions}\index{Weierstrass function},
+although this
 particular one is a different example than what Weierstrass gave.
 
 Define
@@ -2415,7 +2417,7 @@ \section{Equicontinuity and the Arzel{\`a}--Ascoli theorem}
 
 \begin{example}
 The sequence $f_n(x) := x^n$ of functions on $[0,1]$
-is uniformly bounded, but contains no sequence that converges
+is uniformly bounded, but contains no subsequence that converges
 uniformly,
 although the sequence converges pointwise to a discontinuous function.
 \end{example}
@@ -2504,7 +2506,7 @@ \section{Equicontinuity and the Arzel{\`a}--Ascoli theorem}
 is uniformly equicontinuous.  The definition is really interesting
 if $S$ is infinite.
 
-And just as for continuity, one can define equicontinuity at a point.
+Just as for continuity, one can define equicontinuity at a point.
 That is, $S$ is \emph{\myindex{equicontinuous}} at $x \in X$
 if for every $\epsilon > 0$, there is a $\delta > 0$
 such that if $y \in X$ with $d(x,y) < \delta$ we have

ch-several-vars-ders.tex

@@ -215,11 +215,12 @@ \subsection{Vector spaces}
 $S$ is closed under scalar multiplication, multiplying a vector in
 $S$ by a scalar gets us a vector in $S$.
 \end{enumerate}
-2) and 3)
+Items 2) and 3)
 make sure that the addition and scalar multiplication are in fact defined on
-$S$.  1) is required
-to fullfill \ref{vecspacedefn:addidentity}.  Existence
-of additive inverse $-v$ follows because $-v = (-1)v$ and 3) says that
+$S$.  Item 1) is required
+to fullfill item \ref{vecspacedefn:addidentity} from the deifnition of vector
+space.  Existence
+of additive inverse $-v$ follows because $-v = (-1)v$ and item 3) says that
 $-v \in S$ if $v \in S$.  All other properties are certain equalities
 which are already satisfied in $X$ and thus must be satisfied in a subset.
 \end{remark}