Added the sections convergence, completeness and completion to functional analysis.

config/en/mkdocs.yaml

@@ -102,7 +102,8 @@ nav:
         - 'Definition': mathematics/functional-analysis/metric-spaces/definition.md
         - 'Topological notions': mathematics/functional-analysis/metric-spaces/topological-notions.md
         - 'Convergence': mathematics/functional-analysis/metric-spaces/convergence.md
-        - 'Completeness'mathematics/functional-analysis/metric-spaces/completeness.md
+        - 'Completeness': mathematics/functional-analysis/metric-spaces/completeness.md
+        - 'Completion': mathematics/functional-analysis/metric-spaces/completion.md
       - 'Normed spaces':
         - 'Vector spaces': mathematics/functional-analysis/normed-spaces/vector-spaces.md
         - 'Definition':  mathematics/functional-analysis/normed-spaces/definition.md

docs/en/mathematics/functional-analysis/metric-spaces/completeness.md

@@ -0,0 +1,173 @@
+# Completeness
+
+> *Definition 1*: a sequence $(x_n)_{n \in \mathbb{N}}$ in a metric space $(X,d)$ is a **Cauchy sequence** if 
+>
+> $$
+>   \forall \varepsilon > 0 \exists N \in \mathbb{N} \forall n,m > N: \quad d(x_n, x_m) < \varepsilon.
+> $$
+
+A convergent sequence $(x_n)_{n \in \mathbb{N}}$ in a metric space $(X,d)$ is always a Cauchy sequence since 
+
+$$
+    \forall \varepsilon > 0 \exists N \in \mathbb{N}: \quad d(x_n, x) < \frac{\varepsilon}{2}, 
+$$
+
+for all $n > N$. By axiom 4 of the definition of a metric space we have for $m, n > N$
+
+$$
+    d(x_m, x_n) \leq d(x_m, x) + d(x, x_n) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon,
+$$
+
+showing that $(x_n)$ is Cauchy.
+
+> *Definition 2*: a metric space $(X,d)$ is **complete** if every Cauchy sequence in $X$ is convergent.
+
+Therefore, in a complete metric space every Cauchy sequence is a convergent sequence.
+
+> *Proposition 1*: let $M \subset X$ be a nonempty subset of a metric space $(X,d)$ and let $\overline M$ be the closure of $M$, then
+>
+> 1. $x \in \overline M \iff \exists (x_n)_{n \in \mathbb{N}} \text{ in } M: x_n \to x$,
+> 2. $M \text{ is closed } \iff M = \overline M$. 
+
+??? note "*Proof*:"
+
+    To prove statement 1, let $x \in \overline M$. If $x \notin M$ then $x$ is an accumulation point of $M$. Hence, for each $n \in \mathbb{N}$ the ball $B(x,\frac{1}{n})$ contains an $x_n \in M$ and $x_n \to x$ since $\frac{1}{n} \to 0$ as $n \to \infty$. Conversely, if $(x_n)_{n \in \mathbb{N}}$ is in $M$ and $x_n \to x$, then $x \in M$ or every neighbourhood of $x$ contains points $x_n \neq x$, so that $x$ is an accumulation point of $M$. Hence $x \in \overline M$. 
+
+    Statement 2 follows from statement 1.
+
+We have that the following statement is equivalent to statement 2: $x_n \in M: x_n \to x \implies x \in M$. 
+
+> *Proposition 2*: let $M \subset X$ be a subset of a complete metric space $(X,d)$, then
+>
+> $$
+>   M \text{ is complete} \iff M \text{ is a closed subset of } X
+> $$
+
+??? note "*Proof*:"
+
+    Let $M$ be complete, by proposition 1 statement 1 we have that 
+
+    $$
+        \forall x \in \overline M \exists (x_n)_{n \in \mathbb{N}} \text{ in } M: x_n \to x.
+    $$
+
+    Since $(x_n)$ is Cauchy and $M$ is complete, $x_n$ converges in $M$ with the limit being unique by statement 1 in [lemma 1](). Hence, $x \in M$ which proves that $M$ is closed because $x \in \overline M$ has been chosen arbitrary.
+
+    Conversely, let $M$ be closed and $(x_n)$ Cauchy in $M$. Then $x_n \to x \in X$ which implies that $x \in \overline M$ by statement 1 in proposition 1, and $x \in M$ since $M = \overline M$ by assumption. Hence, the arbitrary Cauchy sequence $(x_n)$ converges in $M$.
+
+> *Proposition 3*: let $T: X \to Y$ be a map from a metric space $(X,d)$ to a metric space $(Y,\tilde d)$, then
+> 
+> $$
+>   T \text{ is continuous in } x_0 \in X \iff x_n \to x_0 \implies T(x_n) \to T(x_0),
+> $$
+>
+> for any sequence $(x_n)_{n \in \mathbb{N}}$ in $X$ as $n \to \infty$. 
+
+??? note "*Proof*:"
+
+    Suppose $T$ is continuous at $x_0$, then for a given $\varepsilon > 0$ there is a $\delta > 0$ such that
+
+    $$
+        \forall \varepsilon > 0 \exists \delta > 0: \quad d(x, x_0) < \delta \implies \tilde d(Tx, Tx_0) < \varepsilon.
+    $$
+
+    Let $x_n \to x_0$ then 
+
+    $$
+        \exists N \in \mathbb{N} \forall n > N: \quad d(x_n, x_0) < \delta.
+    $$
+
+    Hence,
+
+    $$
+        \forall n > N: \tilde d(Tx_n, Tx_0) < \varepsilon.
+    $$
+
+    Which means that $T(x_n) \to T(x_0)$. 
+
+    Conversely, suppose that $x_n \to x_0 \implies T(x_n) \to T(x_0)$ and $T$ is not continuous. Then
+
+    $$
+        \exists \varepsilon > 0: \forall \delta > 0 \exists x \neq x_0: \quad d(x, x_0) < \delta \quad \text{ however } \quad \tilde d(Tx, Tx_0) \geq \varepsilon,
+    $$
+
+    in particular, for $\delta = \frac{1}{n}$ there is a $x_n$ satisfying 
+
+    $$
+        d(x_n, x_0) < \frac{1}{n} \quad \text{ however } \quad \tilde d(Tx_n, Tx_0) \geq \varepsilon,
+    $$
+
+    Clearly $x_n \to x_0$ but $(Tx_n)$ does not converge to $Tx_0$ which contradicts $Tx_n \to Tx_0$. 
+
+## Completeness proofs
+
+To show that a metric space $(X,d)$ is complete, one has to show that every Cauchy sequence in $(X,d)$ has a limit in $X$. This depends explicitly on the metric on $X$.
+
+The steps in a completeness proof are as follows
+
+1. take an arbitrary Cauchy sequence $(x_n)_{n \in \mathbb{N}}$ in $(X,d)$,
+2. construct for this sequence a candidate limit $x$,
+3. prove that $x \in X$,
+4. prove that $x_n \to x$ with respect to metric $d$.
+
+> *Proposition 4*: the Euclidean space $\mathbb{R}^n$ with $n \in \mathbb{N}$ and the metric $d$ defined by
+>
+> $$
+>   d(x,y) = \sqrt{\sum_{j=1}^n \big(x(j) - y(j) \big)^2},
+> $$
+>
+> for all $x,y \in \mathbb{R}^n$ is complete.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+A similar proof exists for the completeness of the Unitary space $\mathbb{C}^n$. 
+
+> *Proposition 5*: the space $C([a,b])$ of all **real-valued continuous functions** on a closed interval $[a,b]$ with $a<b \in \mathbb{R}$ with the metric $d$ defined by
+>
+> $$
+>   d(x,y) = \max_{t \in [a,b]} |x(t) - y(t)|,
+> $$
+>
+> for all $x, y \in C$ is complete.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+While $C$ with a metric $d$ defined by 
+
+$$
+    d(x,y) = \int_a^b |x(t) - y(t)| dt,
+$$
+
+for all $x,y \in C$ is incomplete.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+> *Proposition 6*: the space $l^p$ with $p \geq 1$ and the metric $d$ defined by
+>
+> $$
+>   d(x,y) = (\sum_{j \in \mathbb{N}} | x(j) - y(j) |^p)^\frac{1}{p},
+> $$
+>
+> for all $x,y \in l^p$ is complete.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+> *Proposition 7*: the space $l^\infty$ with the metric $d$ defined by
+>
+> $$
+>   d(x,y) = \sup_{j \in \mathbb{N}} | x(j) - y(j) |,
+> $$
+>
+> for all $x,y \in l^\infty$ is complete.
+
+??? note "*Proof*:"
+
+    Will be added later.

docs/en/mathematics/functional-analysis/metric-spaces/completion.md

@@ -0,0 +1,20 @@
+# Completion
+
+> *Definition 1*: let $(X,d)$ and $(\tilde X, \tilde d)$ be metric spaces, then
+>
+> 1. a mapping $T: X \to \tilde X$ is an **isometry** if $\forall x, y \in X: \tilde d(Tx, Ty) = d(x,y)$.
+> 2. $(X,d)$ and $(\tilde X, \tilde d)$ are **isometric** if there exists a bijective isometry $T: X \to \tilde X$. 
+
+Hence, isometric spaces may differ at most by the nature of their points but are indistinguishable from the viewpoint of the metric. 
+
+Or in other words, the metric space $(\tilde X, \tilde d)$ is unique up to isometry.
+
+> *Theorem 1*: for every metric space $(X,d)$ there exists a complete metric space $(\tilde X, \tilde d)$ that contains a subset $W$ that satisfies the following conditions
+>
+> 1. $W$ is a metric space isometric with $(X,d)$.
+> 2. $W$ is dense in $X$.
+
+??? note "*Proof*:"
+
+    Will be added later.
+

docs/en/mathematics/functional-analysis/metric-spaces/convergence.md

@@ -0,0 +1,59 @@
+# Convergence
+
+> *Definition 1*: a sequence $(x_n)_{n \in \mathbb{N}}$ in a metric space $(X,d)$ is **convergent** if there exists an $x \in X$ such that
+>
+> $$
+>   \lim_{n \to \infty} d(x_n, x) = 0.
+> $$
+>
+> $x$ is the **limit** of $(x_n)$ and is denoted by 
+>
+> $$
+>   \lim_{n \to \infty} x_n = x,
+> $$
+>
+> or simply by $x_n \to x$, $(n \to \infty)$.
+
+We say that $(x_n)$ *converges to* $x$ or *has the limit* $x$. If $(x_n)$ is not convergent then it is **divergent**. 
+
+We have that the limit of a convergent sequence must be a point of $X$.
+
+> *Definition 2*: a non-empty subset $M \subset X$ of a metric space $(X,d)$ is **bounded** if there exists an $x_0 \in X$ and an $r > 0$ such that $M \subset B(x_0,r)$. 
+
+Furthermore, we call a sequence $(x_n)$ in $X$ a **bounded sequence** if the corresponding point set is a bounded subset of $X$.
+
+> *Lemma 1*: let $(X,d)$ be a metric space then
+>
+> 1. a convergent sequence in $X$ is bounded and its limit is unique,
+> 2. if $x_n \to x$ and $y_n \to y$ then $d(x_n, y_n) \to d(x,y)$, $(n \to \infty)$. 
+
+??? note "*Proof*:"
+
+    For statement 1, suppose that $x_n \to x$. Then, taking $\varepsilon = 1$, we can find an $N$ such that $d(x_n, x) < 1$ for all $n > N$. Which shows that $(x_n)$ is bounded. Suppose that $x_n \to x$ and $x_n \to z$ then by axiom 4 of the definition of a metric space we have 
+
+    $$
+        d(x_n, x) \leq d(x_n, z) + d(x, z) \to 0,
+    $$
+
+    as $n \to \infty$ and by axiom 2 of the definition of a metric space it follows that $x = z$. 
+
+    For statement 2, we have that 
+
+    $$
+        d(x_n,y_n) \leq d(x_n, x) + d(x, y) + d(y, y_n),
+    $$
+
+    by axiom 4 of the definition of a metric space. Hence we obtain
+
+    $$
+        d(x_n, y_n) - d(x, y) \leq d(x_n, x) + d(y_n, y),
+    $$
+
+    such that
+
+    $$
+        |d(x_n, y_n) - d(x, y)| \leq d(x_n, x) + d(y_n, y) \to 0
+    $$
+
+    as $n \to \infty$. 
+