From cddd7a4d55144759363a2d330978702d6d29c048 Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Wed, 24 Jun 2026 14:18:27 -0400
Subject: [PATCH] Cleaned up citation stlye changes.

---
 src/fa/lc/convex.tex | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/src/fa/lc/convex.tex b/src/fa/lc/convex.tex
index cb29764..1909eba 100644
--- a/src/fa/lc/convex.tex
+++ b/src/fa/lc/convex.tex
@@ -29,7 +29,7 @@
 
 
 
-\begin{lemma}[{{\cite[II.1.1]{SchaeferWolff}}}]
+\begin{lemma}
 \label{lemma:convex-interior}
     Let $E$ be a TVS over $K \in \RC$, $A \subset E$ be convex, $x \in A^o$, and $y \in \ol{A}$, then
     \[
@@ -37,7 +37,7 @@
     \]
 
 \end{lemma}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[II.1.1]{SchaeferWolff}}}. ]
     Fix $t \in (0, 1)$. Using translation, assume without loss of generality that $tx + (1 - t)y = 0$. In which case, $x = \alpha y$ where $\alpha = (1 - t)/t$. By (TVS2), $\alpha A^o \in \cn^o(y)$. Since $y \in \ol{A}$, $\alpha A^o \cap A \ne \emptyset$, so there exists $z \in A^o$ such that $\alpha z \in A$.
 
     Let $\mu = \alpha/(\alpha - 1)$, then since $\alpha < 0$, $\mu \in (0, 1)$ and
@@ -74,11 +74,11 @@
     converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$.
 \end{proof}
 
-\begin{proposition}[{{\cite[II.1.3]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:convex-interior-closure}
     Let $E$ be a TVS over $K \in \RC$ and $A \subset E$ be convex. If $A^o \ne \emptyset$, then $\ol{A} = \ol{A^o}$.
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[II.1.3]{SchaeferWolff}}}. ]
     Since $A^o \subset A$, $\ol{A^o} \subset \ol{A}$. Let $x \in A^o$, then for any $y \in \ol{A}$,
     \[
     y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}} \subset \ol{A^o}