Typo fixes.

src/dg/derivative/higher.tex

@@ -35,26 +35,33 @@
         A(h, k) &= Df(x + h)(k) + Df(x)(k) \\
         &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
         &- [f(x + k) - f(x) - Df(x)(k)] \\
-        &= D^2f(x)(h, k) + r_1(h) \cdot Df(x)(k)
+        &= D^2f(x)(h, k) + r_1(h) \cdot Df(x)(k) \\
         &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
         &- [f(x + k) - f(x) - Df(x)(k)] \\
     \end{align*}
     Let $B_h: B_E(0, r) \to F$ be defined by
-    \[
+    \begin{align*}
-    B_h(k) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k)
+    B_h(k) &= f(x + h + k) - f(x + k) \\
-    \]
+    &- Df(x + h)(k) + Df(x)(k)
+    \end{align*}
+    
 
     then
-    \[
+    \begin{align*}
-    B_h(k) - B_h(0) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k) -f(x + h) + f(x)
+    B_h(k) - B_h(0) &= f(x + h + k) - f(x + k) \\
-    \]
+    &- Df(x + h)(k) + Df(x)(k) \\
+    &-f(x + h) + f(x)
+    \end{align*}
+    
 
     Now, there exists $r_2, r_3 \in \mathcal{R}_{B(E)}$ such that for any $k \in B(0, r)$,
     \begin{align*}
     DB_h(k) &= Df(x + h + k) - Df(x + k) - Df(x + h) + Df(x) \\
-    &= D^2f(x)(h + k) + Df(x) - D^2f(x)(h) - Df(x) - D^2f(x)(k) + r_2(k) + r_3(h) \\
+    &= D^2f(x)(h + k) + Df(x) - D^2f(x)(h) \\
+    &- Df(x) - D^2f(x)(k) + r_2(k) + r_3(h) \\
     &=r_2(k) + r_3(h)
     \end{align*}
+
     By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem},
     \[
     \norm{B_h(k) - B_h(0)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
@@ -68,9 +75,12 @@
     so $D^2f(x)(h, k) - D^2f(x)(k, h) = 0$.
 
     Now suppose that the proposition holds for $n$. Identify $L^n(E; F) = L^{2}(E; L^{n-2}(E; F))$, then for any $\seqf[n]{x_j} \subset E$,
-    \[
+    \begin{align*}
-    Df(x)(x_1, \cdots, x_n) = Df(x)(x_1, x_2)(x_3, \cdots, x_n) = Df(x)(x_2, x_1)(x_3, \cdots, x_n) = Df(x)(x_2, x_1, x_3, \cdots, x_n)
+    Df(x)(x_1, \cdots, x_n) &= Df(x)(x_1, x_2)(x_3, \cdots, x_n) \\
-    \]
+    &= Df(x)(x_2, x_1)(x_3, \cdots, x_n) \\
+    &= Df(x)(x_2, x_1, x_3, \cdots, x_n)
+    \end{align*}
+    
 
     Since any element $\sigma \in S_n$ that does not fix $x_1$ is the composition of the transposition $(12)$ and an element that fixes $x_1$, $Df(x)$ is symmetric.
 \end{proof}

src/topology/main/c0.tex

@@ -5,7 +5,7 @@
 \label{definition:vanish-at-infinity}
     Let $X$ be a topological space, $E$ be a TVS over $K \in \RC$, and $f \in C(X; E)$, then $f$ \textbf{vanishes at infinity} if for every $U \in \cn_E^o(0)$, $\bracs{f \not\in U}$ is compact.
 
-    The set $C_0(X; \complex)$ is the space of all functions that vanish at infinity. 
+    The set $C_0(X; E)$ is the space of all functions that vanish at infinity. 
 \end{definition}
 
 \begin{proposition}

src/topology/uniform/metric.tex

@@ -70,14 +70,15 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
     \begin{enumerate}
         \item[(FB1)] For any $J, J' \subset I$ finite and $r, r' > 0$,
         \[
-        \bigcap_{j \in J \cup J'}E(d_j, \min(r,r')E(d_j, r \wedge r') \subset \paren{\bigcap_{j \in J}E(d_j, r)} \cap \paren{\bigcap_{j \in J'}E(d_j, r')}
+        \bigcap_{j \in J \cup J'}E(d_j, r \wedge r') \subset \paren{\bigcap_{j \in J}E(d_j, r)} \cap \paren{\bigcap_{j \in J'}E(d_j, r')}
         \]
 
         \item[(UB1)] For each $i \in I$, $d(x, x) = 0$ for all $x \in X$. Thus for any $i \in I$ and $r > 0$, $E(d_i, r)$ contains the diagonal.
         \item[(UB2)] For each $J \subset I$ finite and $r > 0$,
-        \[
+        \begin{align*}
-        \paren{\bigcap_{j \in J}E(d_j, r/2)} \circ \paren{\bigcap_{j \in J}E(d_j, r)} \subset \bigcap_{j \in J}E(d_j, r/2) \circ E(d_j, r/2) \subset \bigcap_{j \in J}E(d_j, r)
+        \paren{\bigcap_{j \in J}E(d_j, r/2)} \circ \paren{\bigcap_{j \in J}E(d_j, r)} &\subset \bigcap_{j \in J}E(d_j, r/2) \circ E(d_j, r/2) \\
-        \]
+        &\subset \bigcap_{j \in J}E(d_j, r)
+        \end{align*}
         
         by the triangle inequality.
     \end{enumerate}