diff --git a/src/fa/tvs/equicontinuous.tex b/src/fa/tvs/equicontinuous.tex
index 96e8db4..c8d3be3 100644
--- a/src/fa/tvs/equicontinuous.tex
+++ b/src/fa/tvs/equicontinuous.tex
@@ -29,7 +29,7 @@
     Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$. Suppose that one of the following holds:
     \begin{enumerate}
         \item[(B)] $E$ is a Baire space. 
-        \item[(B')] $E$ and $F$ are locally convex with $E$ being barreled.
+        \item[(B')] $E$ is barreled and $F$ is locally convex.
     \end{enumerate}
 
     and that
@@ -57,6 +57,49 @@
     (E1) $\Rightarrow$ (C1) + (C2): By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} and \autoref{proposition:equicontinuous-linear-closure}.
 \end{proof}
 
+\begin{lemma}
+\label{lemma:equicontinuous-bilinear}
+    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg \subset L^2(E, F; G)$ be continuous bilinear maps, then the following are equivalent:
+    \begin{enumerate}
+        \item $\alg$ is equicontinuous.
+        \item $\alg$ is equicontinuous at $0$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (2) $\Rightarrow$ (1): For each $(x_0, y_0), (x, y) \in E \times F$ and $\lambda \in \alg$, 
+    \[
+    \lambda(x, y) - \lambda(x_0, y_0) = \lambda(x - x_0, y - y_0) + \lambda(x - x_0, y_0) + \lambda(x_0, y - y_0)
+    \]
+
+    For each $U \in \cn_G(0)$ circled, there exists circled neighbourhoods $V \in \cn_E(0)$ and $W \in \cn_F(0)$ such that $\lambda(V \times W) \subset U$ for all $\lambda \in \alg$. In which case, there exists $\mu > 0$ such that $y_0 \in \mu W$ and $x_0 \in \mu V$. Thus if $(x, y) - (x_0, y_0) \in \mu^{-1}(V \times W)$, then for every $\lambda \in \alg$,
+    \[
+    \lambda(x - x_0, y_0) \in \lambda(\mu^{-1} V \times \mu W) = \lambda(V \times W) \subset U
+    \]
+
+    and $\lambda(x_0, y - y_0) \in U$ as well. Therefore $\alg$ is equicontinuous at $(x_0, y_0)$. 
+\end{proof}
+
+
+
+\begin{theorem}
+\label{theorem:separate-joint-bilinear}
+    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds:
+    \begin{enumerate}
+        \item[(B)] $E$ is Baire.
+        \item[(B')] $E$ is barreled and $G$ is locally convex. 
+    \end{enumerate}
+
+    and that
+    \begin{enumerate}
+        \item[(M)] $E$ and $F$ are both metrisable. 
+        \item[(E)] For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg} \subset L(F; G)$ is equicontinuous. 
+    \end{enumerate}
+
+    then $\alg$ is equicontinuous. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[III.5.1]{SchaeferWolff}}}. ]
+    Let $\seq{(x_n, y_n)} \subset E \times F$ and $\seq{\lambda_n} \subset \alg$ such that $(x_n, y_n) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and \autoref{proposition:equicontinuous-net}. By (B) or (B') and the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_n(x_n, y_n) \to 0$ as $n \to \infty$ by \autoref{proposition:equicontinuous-net}. By (M) and \autoref{proposition:equicontinuous-net}, $\alg$ is equicontinuous at $0$, and hence equicontinuous by \autoref{lemma:equicontinuous-bilinear}. 
+\end{proof}
 
 
 
diff --git a/src/topology/uniform/equicontinuous.tex b/src/topology/uniform/equicontinuous.tex
index a33a5fe..a4529c1 100644
--- a/src/topology/uniform/equicontinuous.tex
+++ b/src/topology/uniform/equicontinuous.tex
@@ -8,6 +8,22 @@
     The set $\cf \subset C(X; Y)$ is \textbf{equicontinuous} if it is equicontinuous at every point in $x$. 
 \end{definition}
 
+\begin{proposition}
+\label{proposition:equicontinuous-net}
+    Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, $\cf \subset Y^X$, and $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\cf$ is equicontinuous at $x$. 
+        \item For $\angles{x_\alpha}_{\alpha \in A} \subset X$ with $x_\alpha \to x$, $\angles{f_\alpha}_{\alpha \in A} \subset \cf$, and $U \in \fU$, there exists $\alpha_0 \in A$ such that $(f_\alpha(x_\alpha), f_\alpha(x)) \in U$ for all $\alpha \ge \alpha_0$.  
+        \item For any upward-directed set $A$ with $|A| \le |\cn_X(x)|$, $\angles{x_\alpha}_{\alpha \in A} \subset X$ with $x_\alpha \to x$, $\angles{f_\alpha}_{\alpha \in A} \subset \cf$, and $U \in \fU$, there exists $\alpha_0 \in A$ such that $(f_\alpha(x_\alpha), f_\alpha(x)) \in U$ for all $\alpha \ge \alpha_0$.  
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Since $\cf$ is equicontinuous at $x$, there exists $V \in \cn_X(x)$ such that $(f_\alpha(y), f_\alpha(x)) \in U$ for all $y \in V$ and $\alpha \in A$. Given that $x_\alpha \to x$, there exists $\alpha_0 \in A$ such that $x_\alpha \in V$ for all $\alpha \ge \alpha_0$, so $(f_\alpha(x_\alpha), f_\alpha(x)) \in U$ for all $\alpha \ge \alpha_0$.  
+
+    $\neg (1) \Rightarrow \neg (3)$: Direct $\cn_X(x)$ under reverse inclusion. If $\cf$ is not equicontinuous at $x$, then there exists $U \in \fU$ such that for every $V \in \cn_X(x)$, there exists $f_V \in \cf$ and $x_V \in V$ with $(f_V(x_V), f_V(x)) \not\in U$. In which case, $x_V \to x$ but $(f_V(x_V), f_V(x)) \not\in U$ for all $V \in \cn_X(x)$. 
+\end{proof}
+
+
 \begin{definition}[Uniformly Equicontinuous]
 \label{definition:uniformly-equicontinuous}
     Let $(X, \fU)$ and $(Y, \fV)$ be uniform spaces, and $\cf \subset UC(X; Y)$, then $\cf$ is \textbf{uniformly equicontinuous} if for every $V \in \fV$, there exists $U \in \fU$ such that $(f \times f)(V) \subset \fU$ for all $f \in \cf$.