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+*.log
+*.aux
+*.bbl
+*.toc
+*.gz
+*.out
+*.toc
\ No newline at end of file
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diff --git a/latex_template/template.tex b/latex_template/template.tex
index 970913a..eab517b 100644
--- a/latex_template/template.tex
+++ b/latex_template/template.tex
@@ -25,46 +25,45 @@ \section{Random Examples}
 \nt{We will do topology in Normed Linear Space  (Mainly $\bbR^n$ and occasionally $\bbC^n$)using the language of Metric Space}
 \clm{Topology}{}{Topology is cool}
 \ex{Open Set and Close Set}{
-	\begin{tabular}{rl}
-		Open Set:   & $\bullet$ $\phi$                                              \\
-		            & $\bullet$ $\bigcup\limits_{x\in X}B_r(x)$ (Any $r>0$ will do) \\[3mm]
-		            & $\bullet$ $B_r(x)$ is open                                    \\
-		Closed Set: & $\bullet$ $X,\ \phi$                                          \\
-		            & $\bullet$ $\overline{B_r(x)}$                                 \\
-		            & $x-$axis $\cup$ $y-$axis
-	\end{tabular}}
+    \begin{tabular}{rl}
+        Open Set:   & $\bullet$ $\phi$                                              \\
+                    & $\bullet$ $\bigcup\limits_{x\in X}B_r(x)$ (Any $r>0$ will do) \\[3mm]
+                    & $\bullet$ $B_r(x)$ is open                                    \\
+        Closed Set: & $\bullet$ $X,\ \phi$                                          \\
+                    & $\bullet$ $\overline{B_r(x)}$                                 \\
+                    & $x-$axis $\cup$ $y-$axis
+    \end{tabular}}
 \thm{}{If $x\in$ open set $V$ then $\exists$ $\delta>0$ such that $B_{\delta}(x)\subset V$}
 \begin{myproof}By openness of $V$, $x\in B_r(u)\subset V$
-	\begin{center}
-		\begin{tikzpicture}
-			\draw[red] (0,0) circle [x radius=3.5cm, y radius=2cm] ;
-			\draw (3,1.6) node[red]{$V$};
-			\draw [blue] (1,0) circle (1.45cm) ;
-			\filldraw[blue] (1,0) circle (1pt) node[anchor=north]{$u$};
-			\draw (2.9,0.4) node[blue]{$B_r(u)$};
-			\draw [green!40!black] (1.7,0) circle (0.5cm) node [yshift=0.7cm]{$B_{\delta}(x)$} ;
-			\filldraw[green!40!black] (1.7,0) circle (1pt) node[anchor=west]{$x$};
-		\end{tikzpicture}
-	\end{center}
-
-	Given $x\in B_r(u)\subset V$, we want $\delta>0$ such that $x\in B_{\delta} (x)\subset B_r(u)\subset V$. Let $d=d(u,x)$. Choose $\delta $ such that $d+\delta<r$ (e.g. $\delta<\frac{r-d}{2}$)
-
-	If $y\in B_{\delta}(x)$ we will be done by showing that $d(u,y)<r$ but $$d(u,y)\leq d(u,x)+d(x,y)<d+\delta<r$$
+    \begin{center}
+        \begin{tikzpicture}
+            \draw[red] (0,0) circle [x radius=3.5cm, y radius=2cm] ;
+            \draw (3,1.6) node[red]{$V$};
+            \draw [blue] (1,0) circle (1.45cm) ;
+            \filldraw[blue] (1,0) circle (1pt) node[anchor=north]{$u$};
+            \draw (2.9,0.4) node[blue]{$B_r(u)$};
+            \draw [green!40!black] (1.7,0) circle (0.5cm) node [yshift=0.7cm]{$B_{\delta}(x)$} ;
+            \filldraw[green!40!black] (1.7,0) circle (1pt) node[anchor=west]{$x$};
+        \end{tikzpicture}
+    \end{center}
+
+    Given $x\in B_r(u)\subset V$, we want $\delta>0$ such that $x\in B_{\delta} (x)\subset B_r(u)\subset V$. Let $d=d(u,x)$. Choose $\delta $ such that $d+\delta<r$ (e.g. $\delta<\frac{r-d}{2}$)
+
+    If $y\in B_{\delta}(x)$ we will be done by showing that $d(u,y)<r$ but $$d(u,y)\leq d(u,x)+d(x,y)<d+\delta<r$$
 \end{myproof}
 
 \cor{}{By the result of the proof, we can then show...}
-\mlenma{}{Suppose $\vec{v_1}, \dots, \vec{v_n} \in \RR[n]$ is subspace of $\RR^n$.}
+\mlemma{}{Suppose $\vec{v_1}, \dots, \vec{v_n} \in \RR[n]$ is subspace of $\RR^n$.}
 \mprop{}{$1 + 1 = 2$.}
 
 \section{Random}
 \dfn{Normed Linear Space and Norm $\boldsymbol{\|\cdot\|}$}{Let $V$ be a vector space over $\bbR$ (or $\bbC$). A norm on $V$ is function $\|\cdot\|\ V\to \bbR_{\geq 0}$ satisfying \begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
-		\item \label{n:1}$\|x\|=0 \iff x=0$ $\forall$ $x\in V$
-		\item \label{n:2}	$\|\lambda x\|=|\lambda|\|x\|$ $\forall$ $\lambda\in\bbR$(or $\bbC$), $x\in V$
-		\item \label{n:3} $\|x+y\| \leq \|x\|+\|y\|$ $\forall$ $x,y\in V$ (Triangle Inequality/Subadditivity)
-	\end{enumerate}And $V$ is called a normed linear space.
-
-	$\bullet $ Same definition works with $V$ a vector space over $\bbC$ (again $\|\cdot\|\to\bbR_{\geq 0}$) where \ref{n:2} becomes $\|\lambda x\|=|\lambda|\|x\|$ $\forall$ $\lambda\in\bbC$, $x\in V$, where for $\lambda=a+ib$, $|\lambda|=\sqrt{a^2+b^2}$ }
+        \item \label{n:1}$\|x\|=0 \iff x=0$ $\forall$ $x\in V$
+        \item \label{n:2}	$\|\lambda x\|=|\lambda|\|x\|$ $\forall$ $\lambda\in\bbR$(or $\bbC$), $x\in V$
+        \item \label{n:3} $\|x+y\| \leq \|x\|+\|y\|$ $\forall$ $x,y\in V$ (Triangle Inequality/Subadditivity)
+    \end{enumerate}And $V$ is called a normed linear space.
 
+    $\bullet $ Same definition works with $V$ a vector space over $\bbC$ (again $\|\cdot\|\to\bbR_{\geq 0}$) where \ref{n:2} becomes $\|\lambda x\|=|\lambda|\|x\|$ $\forall$ $\lambda\in\bbC$, $x\in V$, where for $\lambda=a+ib$, $|\lambda|=\sqrt{a^2+b^2}$ }
 
 \ex{$\bs{p-}$Norm}{\label{pnorm}$V={\bbR}^m$, $p\in\bbR_{\geq 0}$. Define for $x=(x_1,x_2,\cdots,x_m)\in\bbR^m$ $$\|x\|_p=\Big(|x_1|^p+|x_2|^p+\cdots+|x_m|^p\Big)^{\frac1p}$$(In school $p=2$)}
 \textbf{Special Case $\bs{p=1}$}: $\|x\|_1=|x_1|+|x_2|+\cdots+|x_m|$ is clearly a norm by usual triangle inequality. \par
@@ -73,56 +72,56 @@ \section{Random}
 Now exercise
 \qs{}{\label{exs1}Prove that triangle inequality is true if $p\geq 1$ for $p-$norms. (What goes wrong for $p<1$ ?)}
 \sol{\textbf{For Property \ref{n:3} for norm-2}	\subsubsection*{\textbf{When field is $\bbR:$}} We have to show\begin{align*}
-		         & \sum_i(x_i+y_i)^2\leq \left(\sqrt{\sum_ix_i^2} +\sqrt{\sum_iy_i^2}\right)^2                                       \\
-		\implies & \sum_i (x_i^2+2x_iy_i+y_i^2)\leq \sum_ix_i^2+2\sqrt{\left[\sum_ix_i^2\right]\left[\sum_iy_i^2\right]}+\sum_iy_i^2 \\
-		\implies & \left[\sum_ix_iy_i\right]^2\leq \left[\sum_ix_i^2\right]\left[\sum_iy_i^2\right]
-	\end{align*}So in other words prove $\langle x,y\rangle^2 \leq \langle x,x\rangle\langle y,y\rangle$ where
-	$$\langle x,y\rangle =\sum\limits_i x_iy_i$$
-
-	\begin{note}
-		\begin{itemize}
-			\item $\|x\|^2=\langle x,x\rangle$
-			\item $\langle x,y\rangle=\langle y,x\rangle$
-			\item $\langle \cdot,\cdot\rangle$ is $\bbR-$linear in each slot i.e. \begin{align*}
-				      \langle rx+x',y\rangle=r\langle x,y\rangle+\langle x',y\rangle	\text{ and similarly for second slot}
-			      \end{align*}Here in $\langle x,y\rangle$ $x$ is in first slot and $y$ is in second slot.
-		\end{itemize}
-	\end{note}Now the statement is just the Cauchy-Schwartz Inequality. For proof $$\langle x,y\rangle^2\leq \langle x,x\rangle\langle y,y\rangle $$ expand everything of $\langle x-\lambda y,x-\lambda y\rangle$ which is going to give a quadratic equation in variable $\lambda $ \begin{align*}
-		\langle x-\lambda y,x-\lambda y\rangle & =\langle x,x-\lambda y\rangle-\lambda\langle y,x-\lambda y\rangle                                       \\
-		                                       & =\langle x ,x\rangle -\lambda\langle x,y\rangle -\lambda\langle y,x\rangle +\lambda^2\langle y,y\rangle \\
-		                                       & =\langle x,x\rangle -2\lambda\langle x,y\rangle+\lambda^2\langle y,y\rangle
-	\end{align*}Now unless $x=\lambda y$ we have $\langle x-\lambda y,x-\lambda y\rangle>0$ Hence the quadratic equation has no root therefore the discriminant is greater than zero.
-
-	\subsubsection*{\textbf{When field is $\bbC:$}}Modify the definition by $$\langle x,y\rangle=\sum_i\overline{x_i}y_i$$Then we still have $\langle x,x\rangle\geq 0$}
+                 & \sum_i(x_i+y_i)^2\leq \left(\sqrt{\sum_ix_i^2} +\sqrt{\sum_iy_i^2}\right)^2                                       \\
+        \implies & \sum_i (x_i^2+2x_iy_i+y_i^2)\leq \sum_ix_i^2+2\sqrt{\left[\sum_ix_i^2\right]\left[\sum_iy_i^2\right]}+\sum_iy_i^2 \\
+        \implies & \left[\sum_ix_iy_i\right]^2\leq \left[\sum_ix_i^2\right]\left[\sum_iy_i^2\right]
+    \end{align*}So in other words prove $\langle x,y\rangle^2 \leq \langle x,x\rangle\langle y,y\rangle$ where
+    $$\langle x,y\rangle =\sum\limits_i x_iy_i$$
+
+    \begin{note}
+        \begin{itemize}
+            \item $\|x\|^2=\langle x,x\rangle$
+            \item $\langle x,y\rangle=\langle y,x\rangle$
+            \item $\langle \cdot,\cdot\rangle$ is $\bbR-$linear in each slot i.e. \begin{align*}
+                      \langle rx+x',y\rangle=r\langle x,y\rangle+\langle x',y\rangle	\text{ and similarly for second slot}
+                  \end{align*}Here in $\langle x,y\rangle$ $x$ is in first slot and $y$ is in second slot.
+        \end{itemize}
+    \end{note}Now the statement is just the Cauchy-Schwartz Inequality. For proof $$\langle x,y\rangle^2\leq \langle x,x\rangle\langle y,y\rangle $$ expand everything of $\langle x-\lambda y,x-\lambda y\rangle$ which is going to give a quadratic equation in variable $\lambda $ \begin{align*}
+        \langle x-\lambda y,x-\lambda y\rangle & =\langle x,x-\lambda y\rangle-\lambda\langle y,x-\lambda y\rangle                                       \\
+                                               & =\langle x ,x\rangle -\lambda\langle x,y\rangle -\lambda\langle y,x\rangle +\lambda^2\langle y,y\rangle \\
+                                               & =\langle x,x\rangle -2\lambda\langle x,y\rangle+\lambda^2\langle y,y\rangle
+    \end{align*}Now unless $x=\lambda y$ we have $\langle x-\lambda y,x-\lambda y\rangle>0$ Hence the quadratic equation has no root therefore the discriminant is greater than zero.
+
+    \subsubsection*{\textbf{When field is $\bbC:$}}Modify the definition by $$\langle x,y\rangle=\sum_i\overline{x_i}y_i$$Then we still have $\langle x,x\rangle\geq 0$}
 
 \section{Algorithms}
 \begin{algorithm}[H]
-\KwIn{This is some input}
-\KwOut{This is some output}
-\SetAlgoLined
-\SetNoFillComment
-\tcc{This is a comment}
-\vspace{3mm}
-some code here\;
-$x \leftarrow 0$\;
-$y \leftarrow 0$\;
-\uIf{$ x > 5$} {
-    x is greater than 5 \tcp*{This is also a comment}
-}
-\Else {
-    x is less than or equal to 5\;
-}
-\ForEach{y in 0..5} {
-    $y \leftarrow y + 1$\;
-}
-\For{$y$ in $0..5$} {
-    $y \leftarrow y - 1$\;
-}
-\While{$x > 5$} {
-    $x \leftarrow x - 1$\;
-}
-\Return Return something here\;
-\caption{what}
+    \KwIn{This is some input}
+    \KwOut{This is some output}
+    \SetAlgoLined
+    \SetNoFillComment
+    \tcc{This is a comment}
+    \vspace{3mm}
+    some code here\;
+    $x \leftarrow 0$\;
+    $y \leftarrow 0$\;
+    \uIf{$ x > 5$} {
+        x is greater than 5 \tcp*{This is also a comment}
+    }
+    \Else {
+        x is less than or equal to 5\;
+    }
+    \ForEach{y in 0..5} {
+        $y \leftarrow y + 1$\;
+    }
+    \For{$y$ in $0..5$} {
+        $y \leftarrow y - 1$\;
+    }
+    \While{$x > 5$} {
+        $x \leftarrow x - 1$\;
+    }
+    \Return Return something here\;
+    \caption{what}
 \end{algorithm}
 
 \end{document}