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# 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*:"
Let $(x_m)_{m \in \mathbb{N}}$ be a Cauchy sequence in $(\mathbb{R}^n, d)$, then we have
$$
\forall \varepsilon > 0 \exists N \in \mathbb{N}: \forall m, k > N: d(x_m, x_k) = \sqrt{\sum_{j=1}^n \big(x_m(j) - x_k(j) \big)^2} < \varepsilon,
$$
obtains for all $j \in \mathbb{N}$: $|x_m(j) - x_k(j)| < \varepsilon$.
Which shows that $(x_m(j))_{m \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$. Suppose that it converged by $x_m(j) \to x(j)$ as $(m \to \infty)$ then $x \in \mathbb{R}^n$ since $x = \big(x(1), \dots, x(n)\big)$.
Thus for $(k \to \infty)$ we have
$$
d(x_m, x) < \varepsilon \implies x_m \to x,
$$
which implies that $\mathbb{R}^n$ is complete.
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*:"
Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $(C,d)$, then we have
$$
\forall \varepsilon > 0 \exists N \in \mathbb{N}: \forall n, m > N: d(x_n, x_m) = \max_{t \in [a,b]} |x_n(t) - x_m(t)| < \varepsilon,
$$
obtains for all $t \in [a,b]$: $|x_n(t) - x_m(t)| < \varepsilon$.
Which shows that $(x_m(t))_{m \in \mathbb{N}}$ for fixed $t \in [a,b]$ is a Cauchy sequence in $\mathbb{R}$. Since $\mathbb{R}$ is complete the sequence converges; $x_m(t) \to x(t)$ as $m \to \infty$.
Thus, for $m \to \infty$ we have
$$
d(x_n, x) = \max_{t \in [a,b]} | x_n(t) - x(t) | < \varepsilon,
$$
hence $\forall t \in [a,b]: | x_n(t) - x(t) < \varepsilon$, obtaining convergence to $x_n \to x$ as $n \to \infty$ and $x \in C$ which implies that $C$ is complete.
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
>
> $$
> d(x,y) = \Big(\sum_{j \in \mathbb{N}} | x(j) - y(j) |^p\Big)^\frac{1}{p},
> $$
>
> for all $x,y \in l^p$ is complete.
??? note "*Proof*:"
Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $(l^p,d)$, then we have
$$
\forall \varepsilon > 0 \exists N \in \mathbb{N}: n, m > N: d(x_n, x_m) = \Big(\sum_{j \in \mathbb{N}} |x_n(j) - x_m(j)|^p\Big)^\frac{1}{p} < \varepsilon,
$$
obtains for all $j \in \mathbb{N}$: $|x_n(j) - x_m(j)| <\varepsilon$.
Which shows that $(x_m(j))_{m \in \mathbb{N}}$ for fixed $j \in \mathbb{N}$ is a Cauchy sequence in $\mathbb{C}$. Since $\mathbb{C}$ is complete the sequence converges; $x_m(j) \to x(j)$ as $m \to \infty$.
Thus, for $m \to \infty$ we have
$$
d(x_n, x) = \Big(\sum_{j \in \mathbb{N}} |x_n(j) - x(j)|^p\Big)^\frac{1}{p} < \varepsilon,
$$
implies that $x_n - x \in l^p$ and $x = x_n - (x_n - x) \in l^p \implies x \in l^p$ and $x_n \to x$ as $n \to \infty$ which implies that $l^p$ is complete.
> *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*:"
Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $(l^\infty,d)$, then we have
$$
\forall \varepsilon > 0 \exists N \in \mathbb{N}: n, m > N: d(x_n, x_m) = \sup_{j \in \mathbb{N}} | x_n(j) - x_m(j) | < \varepsilon,
$$
obtains for all $j \in \mathbb{N}$: $|x_n(j) - x_m(j)| <\varepsilon$.
Which shows that $(x_m(j))_{m \in \mathbb{N}}$ for fixed $j \in \mathbb{N}$ is a Cauchy sequence in $\mathbb{C}$. Since $\mathbb{C}$ is complete the sequence converges; $x_m(j) \to x(j)$ as $m \to \infty$.
Thus, for $m \to \infty$ we have
$$
d(x_n, x) = \sup_{j \in \mathbb{N}} | x_n(j) - x(j) | < \varepsilon \implies |x_n(j) = x(j)| < \varepsilon.
$$
Since $x_n \in l^\infty$ there exists a $k_n \in \mathbb{R}: |x_n(j)| \leq k_n$ for all $j \in \mathbb{N}$. Hence
$$
|x(j)| \leq |x(j) - x_n(j)| + |x_n(j)| < \varepsilon + k_n,
$$
for all $j \in \mathbb{N}$ which implies that $x \in l^\infty$ and $x_n \to x$ as $n \to \infty$ obtaining that $ l^\infty$ is complete.

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# 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.

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# 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 $n \to \infty$.

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# Metric spaces
> *Definition 1*: a **metric space** is a pair $(X,d)$, where $X$ is a set and $d$ is a metric on $X$, which is a function on $X \times X$ such that
>
> 1. $d$ is real, finite and nonnegative,
> 2. $\forall x,y \in X: \quad d(x,y) = 0 \iff x = y$,
> 3. $\forall x,y \in X: \quad d(x,y) = d(y,x)$,
> 4. $\forall x,y,z \in X: \quad d(x,y) \leq d(x,z) + d(y,z)$.
The metric $d$ is also referred to as a distance function. With $x,y \in X: d(x,y)$ the distance from $x$ to $y$.
## Examples of metric spaces
For the **Real line** $\mathbb{R}$ the usual metric is defined by
$$
d(x,y) = |x - y|,
$$
for all $x,y \in \mathbb{R}$. Obtaining a metric space $(\mathbb{R}, d)$.
??? note "*Proof*:"
Will be added later.
For the **Euclidean space** $\mathbb{R}^n$ with $n \in \mathbb{N}$, the usual metric is defined by
$$
d(x,y) = \sqrt{\sum_{j=1}^n (x(j) - y(j))^2},
$$
for all $x,y \in \mathbb{R}^n$ with $x = (x(j))$ and $y = (y(j))$. Obtaining a metric space $(\mathbb{R}^n, d)$.
??? note "*Proof*:"
Will be added later.
Similar examples exist for the complex plane $\mathbb{C}$ and the unitary space $\mathbb{C}^n$.
For the space $C([a,b])$ of all **real-valued continuous functions** on a closed interval $[a,b]$ with $a<b \in \mathbb{R}$ the metric may be defined by
$$
d(x,y) = \max_{t \in [a,b]} |x(t) - y(t)|,
$$
for all $x,y \in C([a,b])$. Obtaining a metric space $(C([a,b]), d)$.
??? note "*Proof*:"
Will be added later.
> *Definition 2*: let $l^p$ with $p \geq 1$ be the set of sequences $x \in l^p$ of complex numbers with the property that
>
> $$
> \sum_{j \in \mathbb{N}} | x(j) |^p \text{ is convergent},
> $$
>
> for all $x \in l^p$.
We have that a metric $d$ for $l^p$ may be defined by
$$
d(x,y) = (\sum_{j \in \mathbb{N}} | x(j) - y(j) |^p)^\frac{1}{p},
$$
for all $x,y \in l^p$.
??? note "*Proof*:"
Will be added later.
From definition 2 the sequence space $l^\infty$ follows, which is defined as the set of all bounded sequences $x \in l^\infty$ of complex numbers. A metric $d$ of $l^\infty$ may be defined by
$$
d(x,y) = \sup_{j \in \mathbb{N}} | x(j) - y(j) |,
$$
for all $x, y \in l^\infty$.
??? note "*Proof*:"
Will be added later.

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# Topological notions
> *Definition 1*: let $(X,d)$ be a metric space and let $x_0 \in X$ and $r > 0$, the following may be defined
>
> 1. **open ball**: $B(x_0, r) = \{x \in X \;|\; d(x,x_0) < r\}$,
> 2. **closed ball**: $\tilde B(x_0,r) = \{x \in X \;|\; d(x,x_0) \leq r\}$,
> 3. **sphere**: $S(x_0,r) = \{x \in X \;|\; d(x,x_0) = r\}$.
In all three cases $x_0$ can be thought of as the center and $r$ as the radius.
> *Definition 2*: a subset $M \subset X$ of a metric space $(X,d)$ is **open** if $\forall x_0 \in M \exists r > 0: B(x_0,r) \subset M$.
>
> $M$ is **closed** if $X \backslash M$ is open.
Therefore, one may observe that an open ball is an open set and a closed ball is a closed set.
## Neigbourhoods
> *Definition 3*: let $(X,d)$ be a metric space and let $x_0 \in X$, then $B(x_0, \varepsilon)$ is an **$\varepsilon$-neighbourhood** of $x_0$ for some $\varepsilon > 0$.
Using definition 3 we may define the following.
> *Definition 4*: a **neighbourhood** of $x_0$ is a set that contains an $\varepsilon$-neighbourhood of $x_0$ for some $\varepsilon > 0$.
Therefore $x_0$ is an element of each of its neighbourhoods and if $N$ is a neighbourhood of $x_0$ and $N \subset M$, then $M$ is also a neighbourhood of $x_0$.
> *Definition 5*: let $(X,d)$ be a metric space and let $M \subset X$, a point $x_0 \in M$ is an **interior point** of $M$ if $M$ is a neighbourhood of $x_0$.
One may think of an interior point of a subset as a point that lies within the interior of $M$.
> *Definition 6*: let $(X,d)$ be a metric space and let $M \subset X$, the **interior** of $M$, denoted by $M^\circ$ is the set of all interior points of $M$.
One may observe that $M^\circ$ is open and is the largest open set contained in $M$.
> *Lemma 1*: let $(X,d)$ be a metric space and let $\mathscr{T}$ be the set of all open subsets of $X$, then
>
> 1. $\empty \in \mathscr{T} \land X \in \mathscr{T}$,
> 2. the union of a collection of sets in $\mathscr{T}$ is itself a set in $\mathscr{T}$,
> 3. the intersection of a finite collection of sets in $\mathscr{T}$ is a set in $\mathscr{T}$.
??? note "*Proof*:"
Statement 1 follows by noting that $\empty$ is open since $\empty$ has no elements and $X$ is open.
For statement 2 we have that for any point $x$ of the union $U$ of open sets belongs to at least one of these sets $M$ and $M$ contains a ball $B$ about $x$. Then $B \subset U$, by the definition of a union.
For statement 3 we have that if $y$ is any point of the intersection of open sets $M_1, \dots, M_n$ with $n \in \mathbb{N}$ then each $M_j$ contains a ball about $y$ and the smallest of these balls is contained in that intersection.
From statements 1 and 3 from *lemma 1* we may define a topological space $(X,\mathscr{T})$ to be a set $X$ and a collection $\mathscr{T}$ of subsets of $X$ such that $\mathscr{T}$ satisfies the axioms 1 and 3. The set $\mathscr{T}$ is a topology for $X$, and it follows that a metric space is a topological space.
## Continuity
> *Definition 7*: let $(X,d)$ and $(Y,\tilde d)$ be metric spaces and let $T: X \to Y$ be a map. $T$ is **continuous in** $x_0 \in X$ if
>
> $$
> \forall \varepsilon > 0 \exists \delta > 0 \forall x \in X: \quad d(x,x_0) < \delta \implies \tilde d \big(T(x), T(x_0) \big) < \varepsilon.
> $$
>
> A mapping $T$ is **continuous** if it is continuous in all $x_0 \in X$.
Continuous mappings can be characterized in terms of open sets as follows.
> *Theorem 1*: let $(X,d)$ and $(Y,\tilde d)$ be metric spaces, a mapping $T: X \to Y$ is continuous if and only if the inverse image of any open subset of $Y$ is an open subset of $X$.
??? note "*Proof*:"
Suppose that $T$ is continuous. Let $S \subset Y$ be open and $S_0$ the inverse image of $S$. If $S_0 = \empty$, it is open. Let $S_0 = \empty$. For any $x \in S_0$ let $y_0 = T(x_0)$. Since $S$ is open, it contains an $\varepsilon$-neighbourhood $N$ of $y_0$. Since $T$ is continuous, $x_0$ has a $\delta$-neighbourhood $N_0$ which is mapped into $N$. Since, $N \subset S$ we have $N_0 \subset S_0$ so that $S_0$ is open because $x_0 \in S_0$ is arbitrary.
Suppose that the inverse image of every open set in $Y$ is an open set in $X$. Then for every $x_0 \in X$ and any $\varepsilon$-neighbourhood $N$ of $T(x_0)$, the inverse image $N_0$ of $N$ is open, since $N$ is open, and $N_0$ contains $x_0$. Hence, $N_0$ also contains a $\delta$-neighbourhood of $x_0$, which is mapped into $N$ because $N_0$ is mapped into $N$. Consequently, $T$ is continuous at $x_0$. Since $x_0 \in X$ was chosen arbitrary, $T$ is continuous.
## Accumulation points
> *Definition 8*: let $M \subset X$ be a subset of a metric space $(X,d)$. A point $x_0 \in X$ is an **accumulation point** of $M$ if
>
> $$
> \forall \varepsilon > 0 \exists y \in M \backslash \{x_0\}: d(x_0,y) < \varepsilon.
> $$
An accumulation point of a subset $M$ is also sometimes called a limit point of $M$. Implying the nature of these points.
> *Definition 9*: the set consisting of all points of $M$ and all accumulation points of $M$ is the **closure** of $M$, denoted by $\overline M$.
Therefore, $\overline M$ is the smallest closed set containing $M$.
> *Definition 10*: let $(X,d)$ be a metric space and let $M$ be a subset of $X$. The set $M$ is dense in $X$ if $\overline M = X$.
Hence if $M$ is dense in $X$, then every ball in $X$, no matter how small, will contain points of $M$.
> *Definition 11*: a metric space $(X,d)$ is separable if $X$ contains a countable subset $M$ that is dense in $X$.
For example the real line $\mathbb{R}$ is separable, since the set $\mathbb{Q}$ of all rational numbers is countable and is dense in $\mathbb{R}$.
Furthermore, $l^\infty$ is not separable while $l^p$ is indeed separable.
??? note "*Proof*:"
Will be added later.