Completeness

Let $(X,d)$ be a metric space. A sequence $\left(x_n\right)$ in $X$ is a Cauchy sequence if and only if its terms become arbitrarily close to one another.

$$\forall\varepsilon>0\,\exists N\in\mathbb{N}:d\left(x_m,x_n\right)\leq\varepsilon\,\,\forall\,m,n\geq N$$

Every convergent sequence is a Cauchy sequence and every Cauchy sequence is bounded. A Cauchy sequence with a convergent subsequence converges to the same limit. The metric space $X$ is complete if and only if every Cauchy sequence in $X$ converges to a point of $X.$

The real numbers are complete and so, since convergence in $\mathbb{R}^n$ is coordinatewise, is $\mathbb{R}^n$ with any norm. The rational numbers are not complete, since a sequence of rational numbers converging to $\sqrt{2}$ is a Cauchy sequence with no limit in $\mathbb{Q},$ and neither is the open interval $\left(0,1\right),$ in which $\left(\frac{1}{n}\right)$ has no limit. Completeness is preserved by strong equivalence of metrics but not by equivalence. Let $A\subseteq X.$ If $X$ is complete then the metric subspace $A$ is complete if and only if $A$ is closed in $X,$ and a complete metric subspace of any metric space is closed.

A Banach space is a complete normed vector space and a Hilbert space is a complete inner product space. Every finite-dimensional normed vector space is a Banach space and $\mathbb{R}^n$ with the dot product is a Hilbert space. Every metric space $X$ has a completion, a complete metric space containing an isometric copy of $X$ as a dense subset, which is unique up to isometry; the real numbers are the completion of the rational numbers.

Let $T:X\rightarrow X$ be a function from a metric space to itself. A point $x^*\in X$ is a fixed point of $T$ if $T\left(x^*\right)=x^*.$ $T$ is a contraction if it is Lipschitz continuous with a constant $q<1,$ so that it brings every pair of points closer together by at least the factor $q.$

$$d\left(T(x),T(y)\right)\leq q\,d(x,y)\,\,\forall\,x,y\in X,\qquad 0\leq q<1$$

The contraction mapping theorem states that a contraction $T$ on a non-empty complete metric space has exactly one fixed point $x^*,$ and that for any $x_0\in X$ the sequence of iterates $x_{n+1}:=T\left(x_n\right)$ converges to $x^*$ with

$$d\left(x_n,x^*\right)\leq\frac{q^n}{1-q}\,d\left(x_1,x_0\right).$$

The iterates form a Cauchy sequence because $d\left(x_{n+1},x_n\right)\leq q^n d\left(x_1,x_0\right)$ is summable, completeness gives the limit, continuity of $T$ makes the limit a fixed point and the contraction property rules out a second one. The theorem is also called the Banach fixed-point theorem. The constant $q<1$ is essential: $T(x)=x+\frac{1}{x}$ on $\left[1,\infty\right)$ satisfies $d\left(T(x),T(y)\right)<d(x,y)$ for $x\neq y$ but has no fixed point. If $T$ is not a contraction but some iterate $T^k=T\circ\cdots\circ T$ is, then $T$ still has exactly one fixed point, since a fixed point of $T^k$ is unique and $T$ maps it to another fixed point of $T^k.$