diff --git a/docs/en/mathematics/functional-analysis/normed-spaces/compactness.md b/docs/en/mathematics/functional-analysis/normed-spaces/compactness.md
index e69de29..1cf5f59 100644
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+# Compactness
+
+> *Definition 1*: a metric space $X$ is **compact** if every sequence in $X$ has a convergent subsequence. A subset $M$ of $X$ is compact if every sequence in $M$ has a convergent subsequence whose limit is an element of $M$.
+
+A general property of compact sets is expressed in the following proposition.
+
+> *Proposition 1*: a compact subset $M$ of a metric space $(X,d)$ is closed and bounded. 
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+The converse of this proposition is generally false.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+However, for a finite dimensional normed space we have the following proposition.
+
+> *Proposition 2*: in a finite dimensional normed space $(X, \|\cdot\|)$ a subset $M \subset X$ is compact if and only if $M$ is closed and bounded.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+A source of interesting results is the following lemma.
+
+> *Lemma 1*: let $Y$ and $Z$ be subspaces of a normed space $(X, \|\cdot\|)$, suppose that $Y$ is closed and that $Y$ is a strict subset of $Z$. Then for every $\alpha \in (0,1)$ there exists a $z \in Z$, such that
+>
+> 1. $\|z\| = 1$,
+> 2. $\forall y \in Y: \|z - y\| \geq \alpha$.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+Lemma 1 gives the following remarkable proposition.
+
+> *Proposition 3*: if a normed space $(X, \|\cdot\|)$ has the property that the closed unit ball $M = \{x \in X | \|x\| \leq 1\}$ is compact, then $X$ is finite dimensional.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+Compact sets have several basic properties similar to those of finite sets and not shared by non-compact sets. Such as the following.
+
+> *Proposition 4*: let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and let $T: X \to Y$ be a continuous mapping. Let $M$ be a compact subset of $(X,d_X)$, then $T(M)$ is a compact subset of $(Y,d_Y)$. 
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+From this proposition we conclude that the following property carries over to metric spaces.
+
+> *Corollary 1*: let $M \subset X$ be a compact subset of a metric space $(X,d)$ over a field $F$, a continuous mapping $T: M \to F$ attains a maximum and minimum value.
+
+??? note "*Proof*:"
+
+    Will be added later.
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diff --git a/docs/en/mathematics/functional-analysis/normed-spaces/linear-operators.md b/docs/en/mathematics/functional-analysis/normed-spaces/linear-operators.md
index e69de29..38c5e5a 100644
--- a/docs/en/mathematics/functional-analysis/normed-spaces/linear-operators.md
+++ b/docs/en/mathematics/functional-analysis/normed-spaces/linear-operators.md
@@ -0,0 +1,2 @@
+# Linear operators
+
diff --git a/docs/en/mathematics/functional-analysis/normed-spaces/normed-spaces.md b/docs/en/mathematics/functional-analysis/normed-spaces/normed-spaces.md
index e69de29..a011836 100644
--- a/docs/en/mathematics/functional-analysis/normed-spaces/normed-spaces.md
+++ b/docs/en/mathematics/functional-analysis/normed-spaces/normed-spaces.md
@@ -0,0 +1,198 @@
+# Normed spaces
+
+> *Definition 1*: a vector space $X$ is a **normed space** if a norm $\| \cdot \|: X \to F$ is defined on $X$, satisfying
+>
+> 1. $\forall x \in X: \|x\| \geq 0$,
+> 2. $\|x\| = 0 \iff x = 0$,
+> 3. $\forall x \in X, \alpha \in F: \|\alpha x\| = |\alpha| \|x\|$,
+> 4. $\forall x, y \in X: \|x + y\| \leq \|x\| + \|y\|$.
+
+Also called a *normed vector space* or *normed linear space*. 
+
+> *Definition 2*: a norm on a vector space $X$ defines a metric $d$ on $X$ given by
+>
+> $$
+>   d(x,y) = \|x - y\|,
+> $$
+>
+> for all $x, y \in X$ and is called a **metric induced by the norm**.
+
+Furthermore, there is a category of normed spaces with interesting properties which is given in the following definition.
+
+> *Definition 3*: a **Banach space** is a complete normed space with its metric induced by the norm. 
+
+If we define the norm $\| \cdot \|$ of the Euclidean vector space $\mathbb{R}^n$ by
+
+$$
+    \|x\| = \sqrt{\sum_{j=1}^n |x(j)|^2},
+$$
+
+for all $x \in \mathbb{R}^n$, then it yields the metric
+
+$$
+    d(x,y) = \|x - y\| = \sqrt{\sum_{j=1}^n |x(j) - y(j)|^2},
+$$
+
+for all $x, y \in \mathbb{R}^n$ which imposes completeness. Therefore $(\mathbb{R}^n, \|\cdot\|)$ is a Banach space.
+
+This adaptation also works for $C$, $l^p$ and $l^\infty$, obviously. Obtaining that $\mathbb{R}^n$, $C$, $l^p$ and $l^\infty$ are all Banach spaces.
+
+> *Lemma 1*: a metric $d$ induced by a norm on a normed space $(X, \|\cdot\|)$ satisfies
+>
+> 1. $\forall x, y \in X, \alpha \in F: d(x + \alpha, y + \alpha) = d(x,y)$,
+> 2. $\forall x, y \in X, \alpha \in F: d(\alpha x, \alpha y) = |\alpha| d(x,y)$.
+
+??? note "*Proof*:"
+
+    We have
+
+    $$
+        d(x + \alpha, y + \alpha) = \|x + \alpha - (y + \alpha)\| = \|x - y\| = d(x,y),
+    $$
+
+    and
+
+    $$
+        d(\alpha x, \alpha y) = \|\alpha x - \alpha y\| = |\alpha| \|x - y\| = |\alpha| d(x,y).
+    $$
+
+By definition, a subspace $M$ of a normed space $X$ is a subspace of $X$ with its norm induced by the norm on $X$. 
+
+> *Definition 4*: let $M$ be a subspace of a normed space $X$, if $M$ is closed then $M$ is a **closed subspace** of $X$. 
+
+By definition, a subspace $M$ of a Banach space $X$ is a subspace of $X$ as a normed space. Hence, we do not require $M$ to be complete.
+
+> *Theorem 1*: a subspace $M$ of a Banach space $X$ is complete if and only if $M$ is a closed subspace of $X$.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+Convergence in normed spaces follows from the definition of convergence in metric spaces and the fact that the metric is induced by the norm. 
+
+## Convergent series
+
+> *Definition 5*: let $(x_k)_{k \in \mathbb{N}}$ be a sequence in a normed space $(X, \|\cdot\|)$. We define the sequence of partial sums $(s_n)_{n \in \mathbb{N}}$ by 
+>
+> $$
+>   s_n = \sum_{k=1}^n x_k,
+> $$
+>
+> if $s_n$ converges to $s \in X$, then
+>
+> $$
+>   \lim_{n \to \infty} \sum_{k=1}^n x_k, 
+> $$
+>
+> is convergent, and $s$ is the sum of the series, writing
+>
+> $$
+>   s = \lim_{n \to \infty} \sum_{k=1}^n x_k = \sum_{k=1}^\infty x_k = \lim_{n \to \infty } s_n.
+> $$
+>
+> If the series 
+>
+> $$
+>   \sum_{k=1}^\infty \|x_k\|,
+> $$
+>
+> is convergent in $F$, then the series is **absolutely convergent**.
+
+From the notion of absolute convergence the following theorem may be posed.
+
+> *Theorem 2*: absolute convergence of a series implies convergence if and only if $(X, \|\cdot\|)$ is complete.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## Schauder basis
+
+> *Definition 6*: let $(X, \|\cdot\|)$ be a normed space and let $(e_k)_{k \in \mathbb{N}}$ be a sequence of vectors in $X$, such that for every $x \in X$ there exists a unique sequence of scalars $(\alpha_k)_{k \in \mathbb{N}}$ such that
+>
+> $$
+>   \lim_{n \to \infty} \|x - \sum_{k=1}^n \alpha_k e_k\| = 0,
+> $$
+>
+> then $(e_k)_{k \in \mathbb{N}}$ is a **Schauder basis* of $(X, \|\cdot\|)$.
+
+The expansion of a $x \in X$ with respect to a Schauder basis $(e_k)_{k \in \mathbb{N}}$ is given by
+
+$$
+    x = \sum_{k=1}^\infty \alpha_k e_k.
+$$
+
+> *Lemma 2*: if a normed space has a Schauder basis then it is seperable.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## Completion
+
+> *Theorem 3*: for every normed space $(X, \|\cdot\|_X)$ there exists a Banach space $(Y, \|\cdot\|_Y)$ that contains a subspace $W$ that satisfies the following conditions
+>
+> 1. $W$ is a normed space isometric with $X$. 
+> 2. $W$ is dense in $Y$.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+The Banach space $(Y, \|\cdot\|_Y)$ is unique up to isometry.
+
+## Finite dimension
+
+> *Lemma 3*: let $\{x_k\}_{k=1}^n$ with $n \in \mathbb{N}$ be a linearly independent set of vectors in a normed space $(X, \|\cdot\|)$, then there exists a $c > 0$ such that
+>
+> $$
+>   \Big\| \sum_{k=1}^n \alpha_k x_k \Big\| \geq c \sum_{k=1}^n |\alpha_k|,
+> $$
+>
+> for all $\{\alpha_k\}_{k=1}^n \in F$.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+As a first application of this lemma, let us prove the following.
+
+> *Theorem 4*: every finite-dimensional subspace $M$ of a normed space $(X, \|\cdot\|)$ is complete. 
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+In particular, every finite dimensional normed space is complete.
+
+> *Proposition 1*: every finite-dimensional subspace $M$ of a normed space $(X, \|\cdot\|)$ is a closed subspace of $X$. 
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+Another interesting property of finite-dimensional vector space $X$ is that all norms on $X$ lead to the same topology for $X$. That is, the open subsets of $X$ are the same, regardless of the particular choice of a norm on $X$. The details are as follows.
+
+> *Definition 7*: a norm $\|\cdot\|_1$ on a vector space $X$ is **equivalent** to a norm $\|\cdot\|_2$ on $X$ if there exists $a,b>0$ such that
+>
+> $$
+>   \forall x \in X: a \|x\|_1 \leq \|x\|_2 \leq b \|x\|_1.
+> $$
+
+This concept is motivated by the following proposition.
+
+> *Proposition 2*: equivalent norms on $X$ define the same topology for $X$.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+Using lemma 3 we may now prove the following theorem.
+
+> *Theorem 5*: on a finite dimensional vector space $X$ any norm $\|\cdot\|_1$ is equivalent to any other norm $\|\cdot\|_2$. 
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+This theorem is of considerable importance. For instance, it implies that convergence or divergence of a sequence in a finite dimensional vector space does not depend on the particular choice of a norm on that space.
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