Let $X$ be a set and $d:X\times X\rar\R_0^+$ a function having the following properties: 1. For all $x,y\in X$: $d(x,x)=0$ and $d(x,y)=d(y,x)$. 2. For all $x,y,z\in X$: $d(x,z)\leq d(x,y)+d(y,z)$. Then $d$ is called a pseudo-metric on $X$. If moreover $d(x,y)=0$ implies $x=y$, then $(X,d)$ is said to be a metric space.
Suppose $(X,d)$ is a metric space, then the subsets
$$
B_r(x)\colon=\{y\in X:d(x,y)< r\}\quad\mbox{and}\quad
B_r^\prime(x)\colon=\{y\in X:d(x,y)\leq r\}
$$
are called the open ball of radius $r$ and center $x$ and the closed ball respectively.
Suppose $(X_n,d_n)$ is a sequence of metric spaces. On the product space $X\colon=\prod X_n$ we can define a metric $d$ with the following property: A function $F:Y\rar X$, $y\mapsto(F_n(y))$ on any metric space $Y$ is continuous if and only if all functions $F_n:X\rar X_n$ are continuous: for $x=(x_n), y=(y_n)\in X$ put:
$$
d(x,y)\colon=\sum_n2^{-n}\frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}
$$
A function $p:E\rar\R_0^+$ on a vector space $E$ is said to be a semi-norm on $E$, if 1. for all $\l\in\R$ ($\l\in\C$) and all $x\in E$: $p(\l x)=|\l|p(x)$. 2. for all $x,y\in E$: $p(x+y)\leq p(x)+p(y)$. If moreover $p(x)=0$ implies $x=0$, then $p$ is said to be a norm on $E$ and $(E,p)$ is called a normed space.
Norms on a vector-space $E$ are commonly denoted by $\norm{.}$ and $(E,\norm{.})$
is said to be a normed space; obviously, $d(x,y)\colon=\norm{x-y}$ is a metric
on $E$ and thus every normed space can be considered as a metric space! The open (or closed) ball of radius $1$ with center $0$ is called the unit ball of $E$.
Completion of metric spaces
A metric space is said to be complete if every Cauchy-sequence converges.
Suppose $(X,d)$ is a metric space and $x_0$ any fixed point in $X$. Define a mapping $J:X\rar B(X,\R)$ by $J(x)\colon=d(x,.)-d(x_0,.)$. Then
$$
\norm{J(x)-J(y)}
=\sup\{|d(x,z)-d(y,z)|:z\in X\}
=d(x,y)
$$
i.e. $J$ is an isometry from $X$ into the Banach space $B(X,\R)$, i.e. $X$ is isometric to the dense supspace $J(X)$ of the complete space $\cl{J(X)}$.
For every metric space $(X,d)$ there exists a complete metric space $(\wh X,\wh d)$ such that $X$ is a dense subset of $X$ and $\wh d|X\times X=d$. Moreover every complete metric space $Y\sbe X$, which contains a dense subset isometric to $X$ is itself isometric to $\wh X$. We say: $\wh X$ is the completion of $X$.