diff --git a/chapters/sections/open_closed_sets.tex b/chapters/sections/open_closed_sets.tex
new file mode 100644
index 0000000..da783d1
--- /dev/null
+++ b/chapters/sections/open_closed_sets.tex
@@ -0,0 +1,338 @@
+\documentclass[../../script.tex] {subfiles}
+%! TEX root = ../../script.tex
+
+\begin{document}
+\section{Open and Closed Sets}
+
+\begin{defi}[Inner points and Boundary points]
+    Let $\metric$ be a metric space, $A \subset X$ and $x \in X$. 
+    \begin{enumerate}[(i)]
+        \item $x$ is said to be an inner point of $A$, if \[ \exists \epsilon > 0: ~~ \oball(x) \subset A \]
+        \item $x$ is said to be a boundary point of $A$ if 
+        \[
+            \forall \epsilon > 0: ~~\underbrace{\oball(x) \cap A \ne \emptyset}_{\substack{\oball(x) \text{ contains} \\ \text{points from } A}}
+            \wedge \underbrace{\oball(x) 
+            \cap (X \setminus A) \ne \emptyset}_{\substack{\oball(x) \text{ contains points}\\ \text{from outside of } A}}
+        \]
+        \item The set 
+        \[
+            \set[x \text{ is inner point of } A]{x \in X}
+        \]
+        is called the interior of $A$, and is denoted as $\interior{A}$.
+        \item The set 
+        \[
+            \set[x \text{ is boundary point of} A]{x \in X}
+        \]
+        is called the boundary of $A$, and is denoted as $\partial A$.
+        \item $A \cup \boundary{A}$ is said to be the closure of $A$, and is denoted as $\closure{A}$.
+    \end{enumerate}
+        
+        \begin{center}
+            \begin{tikzpicture}
+                \draw [->, thick] (-6, -3.5) -- (-6, 3);
+                \draw [->, thick] (-6, -3.5) -- (2, -3.5);
+                \draw [thick, fill=lightgray]  plot[smooth, tension=.7] coordinates {(-4,2.5) (-3,3) (-2,2.8) (-0.8,2.5) (-0.5,1.5) (0.5,0) (0,-2)(-1.5,-2.5) (-4,-2) (-3.5,-0.5) (-5,1) (-4,2.5)};
+            
+                \node[below right] at (-6, 3) (X) {$X$};
+                \node[below right] at (-4.5, 2) (A) {$A$};
+    
+                \draw[fill] (-2, -1) circle [radius=1pt];
+                \node[below right] at (-2, -1) (C) {$x$};
+                \draw[dashed] (-2, -1) circle [radius=20pt];
+                \node[below] at (-2, -1.75) (C) {$\oball(x)$};
+    
+                \draw[fill] (-0.63, 2) circle [radius=1pt];
+                \node[right] at (-0.63, 2) (C) {$x$};
+            \end{tikzpicture}
+        \end{center}
+\end{defi}
+
+\begin{eg}
+    Consider $X = \realn^2$. Then 
+    \begin{align*}
+        A &= \set[0 \le y < 1]{(x, y) \in \realn} \\
+        \mathring{A} &= \set[0 \le y < 1]{(x, y) \in \realn^2} \\
+        \boundary{A} &= \set[y = 1 \vee y = 0]{(x, y) \in \realn^2} \\
+        \closure{A} &= \set[0 \le y \le 1]{(x, y) \in \realn^2}
+    \end{align*}
+\end{eg}
+
+\begin{rem}
+    \begin{enumerate}[(i)]
+        \item $\interior{A} \subset A$
+        \item Boundary points of $A$ can be elements of $A$ or not.
+        \item $A \subset \interior{A} \cup \boundary{A}$, $~~\interior{A} \cap \boundary{A} = \varnothing$
+        \item $\boundary{A} = \boundary{X \setminus A}$
+    \end{enumerate}
+\end{rem}
+
+\begin{thm}
+    Let $\metric$ be a metric space, $A \subset X$ and $x$ an interior point or boundary point of $A$. Then 
+    \[
+        \exists \anyseqdef{A}: ~~x_n \conv{} x
+    \]
+\end{thm}
+\begin{proof}
+    If $x \in A$ then this is trivial, so let $x \notin A$. Then 
+    \begin{equation}
+        \forall n \in \natn ~\exists x_n \in \left(\oball[\frac{1}{n}](x) \cap A \ne \varnothing\right)
+    \end{equation}
+    We need to show that $\seq{x}$ converges to $x$.
+    \begin{equation}
+        \forall \epsilon > 0 ~\epsilon N \in \natn: ~~\frac{1}{N} < \epsilon
+    \end{equation}
+    For $n \ge N$ we have 
+    \begin{equation}
+        \frac{1}{n} \le \frac{1}{N} < \epsilon
+    \end{equation}
+    and thus 
+    \begin{equation}
+        d(x_n, x) < \frac{1}{n} < \epsilon
+    \end{equation}
+\end{proof}
+
+\begin{defi}[Open and Closed sets]
+    Let $\metric$ be a metric space. $A \subset X$ is said to be 
+    \begin{enumerate}[(i)]
+        \item open, if every point in $A$ is an interior point
+        \item closed, if $A$ contains all its boundary point 
+        \item neighbourhood of $x \in A$, if $x$ is an interiot point of $A$
+    \end{enumerate}
+\end{defi}
+
+\begin{thm}
+    Let $\metric$ be a metric space and $A \subset X$.
+    \[
+        A \text{ open} \iff X \setminus A \text{ closed}
+    \]
+\end{thm}
+\begin{proof}
+    \begin{subequations}
+    \begin{align}
+        A \text{ open} &\iff \forall x \in A: ~~x \in \interior{A} \\
+        &\iff \forall x \in A: ~~x \in \boundary{A} \\
+        &\iff X \setminus A \text{ contains all boundary point of } A \\
+        &\iff X \setminus A \text{ contains all boundary points of } X \setminus A \\
+        &\iff X \setminus A \text{ closed}
+    \end{align}
+    \end{subequations}
+\end{proof}
+
+\begin{rem}
+    That doesn't mean $A$ has to be either open and closed.
+\end{rem}
+
+\begin{eg}
+    Let $\metric$ be a metric space, $x \in X$ and $r > 0$. Then 
+    \begin{align*}
+        \Oball(x) &= \set[d(x, y) < r]{y \in X} \text{ is open} \\
+        \Cball(x) &= \set[d(x, y) < r]{y \in X} \text{ is closed}
+    \end{align*}
+\end{eg}
+
+\begin{rem}
+    Consider the special case $a, b \in \realn$ with $a < b$
+    \begin{align*}
+        (a, b) &= \oball[\frac{b - a}{2}]\left(\frac{a + b}{2}\right) \text{ open} \\
+        [a, b] &= \cball[\frac{b - a}{2}]\left(\frac{a + b}{2}\right) \text{ closed}
+    \end{align*}
+\end{rem}
+
+\begin{thm}
+    Let $\metric$ be a metric space and $A \subset X$. 
+    \[
+        A \text{ closed} \iff \forall \anyseqdef{A} \text{ convergent}: ~~\limn{x_n} \in A
+    \]
+\end{thm}
+\begin{proof}
+    Assume $A$ is closed. Let $\anyseqdef{A}$ be convergent to $x$. then
+    \begin{equation}
+        \forall \epsilon > 0 ~\exists N \in \natn: ~~x_n \in \oball(x) ~~\forall n \ge N
+    \end{equation}
+    This means that every $\epsilon$-ball around $x$ contains at least one point from $A$.
+    I.e. $x$ is always a point (or a boundary point) of $A$. From $A$ closed follows $x \in A$.
+
+    Now assume $x \in \boundary{A}$. Then 
+    \begin{equation}
+        \exists \anyseqdef{A}: ~~\seq{x} \conv{} x
+    \end{equation}
+    According to the prerequisites, $x \in A$.
+\end{proof}
+
+\begin{thm}
+    Let $\metric$ be a metric space, and $\tau$ the set of all open subsets. Then 
+    \begin{enumerate}[(i)]
+        \item $\varnothing \in \tau$, $~~X \in \tau$
+        \item The union of any number of sets from $\tau$ is an open set
+        \[
+            \bigcup_{t \in \tau} t \in \tau
+        \]
+        \item The intersection of finitely many sets from $\tau$ is an open set 
+        \[
+            \bigcap_{t \in \tau} t \in \tau
+        \]
+    \end{enumerate}
+\end{thm}
+\begin{proof}
+    \reader
+\end{proof}
+
+\begin{rem}
+    \begin{enumerate}[(i)]
+        \item $\tau$ is said to be the topology induced by $d$
+        \item \begin{itemize}
+            \item $\varnothing$, $X$ are also closed 
+            \item The intersection of any number of closed sets is closed 
+            \item The union of finitely many closed sets is closed 
+        \end{itemize}
+        \item Infinitely many intersections of open sets are not open in general.
+    \end{enumerate}
+\end{rem}
+
+\begin{thm}
+    Let $\metric$ be a metric space and $A \subset X$. Then 
+    \[
+        \interior{A} \text{ open} \implies \boundary{A}, \closure{A} \text{ closed}
+    \]
+\end{thm}
+\begin{proof}
+    Let $\interior{A}$ be open and $x \in \interior{A} \subset A$. This means 
+    \begin{equation}
+        \exists \epsilon > 0: ~~\oball(x) \subset A
+    \end{equation}
+    We have to show that $\oball(x) \subset \interior{A}$. Let $y \in \oball(x)$. 
+    Since $\oball(x)$ is open
+    \begin{equation}
+        \exists \delta > 0: ~~\oball[\delta](y) \subset \oball(x) \subset A
+    \end{equation}
+    This means that $y \in \oball(x)$ is interior point $A$. I.e. $\subset(x) \subset \interior{A}$, and thus $x$ is interior point of $\interior{A}$.
+
+    Let $B = X \setminus A$. Then $\boundary{A} = \boundary{B}$
+    \begin{equation}
+        X = A \cup B = \interior{A} \cup \boundary{A} \cup \interior{B} \cup \boundary{B} = \interior{A} \cup \boundary{A} \cup \interior{B}
+    \end{equation}
+    Then 
+    \begin{subequations}
+    \begin{align}
+        A \text{ and } B \text{ are disjoint } &\implies \interior{A}, \interior{B} \text{ disjoint} \\
+        &\implies \boundary{A} \text{ disjoint to } \interior{A}, \interior{B}
+    \end{align}
+    \end{subequations}
+    This results in 
+    \begin{equation}
+        \boundary{A} = X \setminus (\underbrace{\interior{A} \cup \interior{B}}_{\text{open}}) \implies \boundary{A} \text{ closed}
+    \end{equation}
+    and 
+    \begin{equation}
+        \closure{A} = A \cup \boundary{A} = \interior{A} \cup \boundary{A} = X \setminus \interior{B} \text{ closed}
+    \end{equation}
+\end{proof}
+
+\begin{thm}
+    Let $\metric$ be a metric space and $A \subset X$
+    \begin{align*}
+        \bigcup_{\substack{O \text{ open} \\ O \subset A}} O = \interior{A} && \text{and} && \bigcap_{\substack{C \text{ closed} \\ A \subset C}} C = \closure{A}
+    \end{align*}
+\end{thm}
+\begin{proof}
+    Let $\interior{A}$ is open and $\interior{A} \subset A$
+    \begin{equation}
+        \implies \bigcup_{O \subset A \text{ open}} \supset \interior{A}
+    \end{equation}
+    Now let $O \subset A$ be open and $x \in O$, i.e.
+    \begin{equation}
+        \exists \epsilon > 0: ~~\oball(x) \subset O \subset A \implies x \in \interior{A}
+    \end{equation}
+    This implies that $O \subset \interior{A}$. Since this holds for all open $O \subset A$, this statement is proven.
+    The other statement follows from the complement.
+\end{proof}
+
+\begin{thm}
+    Let $\metric$ be a complete space and $A \subset X$ be closed. Then $(A, d_A)$ is complete.
+\end{thm}
+\begin{proof}
+    \reader
+\end{proof}
+
+\begin{rem}
+    Topological terms (open, closed, continuous, compact) don't just depend on $A$, but also on $X$.
+\end{rem}
+
+\begin{defi}
+    Let $\metric$ be a metric space and $x \in X$.
+    \begin{enumerate}[(i)]
+        \item $x$ is said to be an isolated point if $\exists \epsilon > 0$ such that $\oball(x) = \set{x}$.
+        \item $x$ is said to be a limit point if it's not an isolated point.
+    \end{enumerate}
+\end{defi}
+
+\begin{defi}[Punctured neighbourhood, Punctured ball]
+    $\dot{U} \subset X$ is said to be a punctured neighbourhood, if there is a neighbourhood $U$ of $x$ with $\dot{U} = U \setminus \set{x}$
+
+    A punctured ball is $\dot{\oball}(x) = \oball \setminus \set{x}$.
+\end{defi}
+
+\begin{defi}[Limit of mappings]
+    Let $(X, d_X), (Y, d_Y)$ and $x$ a limit point of $X$. Let $\dot{U}$ be a punctured neighbourhood of $x$ and $f: \dot{U} \rightarrow Y$.
+    Then $f$ converges to $y \in Y$ in $x$ ($y$ is said to be the limit of $f$ in $x$), if 
+    \[
+        \forall \epsilon > 0 ~\exists \delta > 0: ~~f(\tilde{x}) \in \oball(y) ~~[d(f(\tilde{x}), y) < \epsilon]
+    \]
+    if $\tilde{x} \in \dot{\oball}(x)$ [$d(\tilde{x}, x) < \delta$]
+\end{defi}
+
+\begin{eg}
+    Let $f, g: \realn^2 \setminus \set{0} \rightarrow \realn$.
+    \begin{align*}
+        f(x) := \norm{x}^2 && g(x) := \frac{1}{\norm{x}}
+    \end{align*}
+    Then $\limes{x}{0} f(x) = 0$, because for $\epsilon > 0$ and $\delta = \sqrt{\epsilon}$ we have 
+    \[
+        d(\tilde{x}, 0) = \norm{\tilde{x} - 0} = \tilde{x} < \delta \implies d(f(\tilde{x}), 0) = \abs{\norm{\tilde{x}}^2 - 0} = \norm{\tilde{x}}^2 < \epsilon = \delta^2
+    \]
+\end{eg}
+
+\begin{thm}
+    \[ 
+        f \text{ converges to } y \in Y \text{ in } x \iff \forall \anyseqdef{X}: ~~f(x_n) \conv{x_n \rightarrow x} y
+    \]
+\end{thm}
+\begin{proof}
+    Let $\anyseqdef{X}$ with $x_n \conv{} x$. Let $\epsilon > 0$, then 
+    \begin{equation}
+        \exists \delta > 0: ~~f(\tilde{x}) \in \oball(y) \text{ if } \tilde{x} \in \oball[\delta](x)
+    \end{equation}
+    Furthermore
+    \begin{equation}
+        \exists N \in \natn: ~~x_n \in \oball[\delta](x) ~~\forall n \ge N
+    \end{equation}
+    Then 
+    \begin{equation}
+        f(x_n) \in \oball(y) ~~\forall n \ge N
+    \end{equation}
+    To prove the other direction, assume $f$ doesn't converge to $y$ in $y$. This means 
+    \begin{equation}
+        \exists \epsilon > 0: ~~\exists \tilde{x} \in \oball[\delta](x) \text{ but } f(\tilde{x}) \notin \oball(y) ~~\forall \delta > 0
+    \end{equation}
+    Therefore 
+    \begin{equation}
+        \forall n \in \natn: ~~\exists x_n \in \oball[\frac{1}{n}](x)
+    \end{equation}
+    We know that $x_n \conv{} x$ since $d(x_n, x) < \frac{1}{n}$, but $f(x_n)$ doesn't converge to $y$ since $d(f(x_n), y) \ge \epsilon$.
+\end{proof}
+
+\begin{cor}
+    Let $\metric$ be a metric space, $x \in X$ a limit point and $\dot{U}$ a punctured neighbourhood of $x$. Let $f, g: \dot{U} \rightarrow \field$ with 
+    \begin{align*}
+        \limes{\tilde{x}}{x} f(\tilde{x}) = y_1 && \limes{\tilde{x}}{x} g(\tilde{x}) = y_2
+    \end{align*}
+    Then 
+    \begin{align*}
+        \limes{\tilde{x}}{x} (f + g)(\tilde{x}) = y_1 + y_2 && \limes{\tilde{x}}{x} (f \cdot g)(\tilde{x}) = y_1 \cdot y_2 \\
+        \limes{\tilde{x}}{x} \left(\frac{f}{g}\right)(\tilde{x}) = \frac{y_1}{y_2}
+    \end{align*}
+\end{cor}
+\begin{hproof}
+    Draw parallels back to number sequences
+\end{hproof}
+\end{document}
\ No newline at end of file
diff --git a/chapters/topo_of_metr_spaces.tex b/chapters/topo_of_metr_spaces.tex
index b6e1af2..d7ee84d 100644
--- a/chapters/topo_of_metr_spaces.tex
+++ b/chapters/topo_of_metr_spaces.tex
@@ -6,4 +6,5 @@
 
     \subfile{sections/metr_and_normed.tex}
     \subfile{sections/seq_ser_limits.tex}
+    \subfile{sections/open_closed_sets.tex}
 \end{document}
\ No newline at end of file
diff --git a/script.pdf b/script.pdf
index 9b4b55b..b3b9c91 100644
Binary files a/script.pdf and b/script.pdf differ
diff --git a/script.tex b/script.tex
index d404560..8f54bc8 100644
--- a/script.tex
+++ b/script.tex
@@ -59,7 +59,6 @@
 \newcommand{\series}[2][\infty]{\sum_{#2 = 1}^{#1}}
 \newcommand{\finite}{\text{ finite}}
 \newcommand{\conj}[1]{\overline{#1}}
-\newcommand{\closure}[1]{\overline{#1}}
 \newcommand{\must}[1][=]{\stackrel{!}{#1}}
 
 \newcommand{\metric}[1][X]{(#1, d)}
@@ -72,6 +71,10 @@
 \newcommand{\Oball}{\oball[r]}
 \newcommand{\Cball}{\cball[r]}
 
+\newcommand{\interior}[1]{\mathring{#1}}
+\newcommand{\boundary}[1]{\partial #1}
+\newcommand{\closure}[1]{\bar{#1}}
+
 \newcommand{\conv}[1]{\xrightarrow{\makebox[2em][c]{$\scriptstyle#1$}}}
 \newcommand{\convinf}{\conv{n \rightarrow \infty}}