Complete space

In calculus, a complete space is a metric space in which every Cauchy sequence of elements of space converges . For example, the space of the rational numbers with the absolute value metric is not complete, because the number is not rational, but there are Cauchy sequences of rational numbers which, when the rational numbers are embedded in the real numbers, do not converge against and thus against any rational number. But it is always possible to fill in the holes, i.e. to complete an incomplete metric space . In the case of rational numbers, this gives the space of real numbers . ${\displaystyle {\sqrt {2}}}$ ${\displaystyle {\sqrt {2}}}$

definition

A sequence of elements of a metric space is called a Cauchy sequence , if ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ ${\displaystyle (M,d)}$

${\displaystyle \forall \varepsilon >0\quad \exists N\in \mathbb {N} \quad \forall n,m\geq N\quad d(x_{n},x_{m})<\varepsilon }$

applies. Further , a sequence converges to an element if ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ ${\displaystyle x\in M}$

${\displaystyle \forall \varepsilon >0\quad \exists N\in \mathbb {N} \quad \forall n\geq N\quad d(x_{n},x)<\varepsilon }$

applies.

A metric space is now called complete if every Cauchy sequence converges in it.

Remarks

• A convergent sequence is always a Cauchy sequence, but the reverse direction does not necessarily have to be true. In a complete space a sequence has a limit if and only if it is a Cauchy sequence; the two terms coincide.

• In the definition of completeness one often demands that every Cauchy sequence converges to an element “in ”. The addition “in ” is not absolutely necessary, since , according to the definition of convergence, only elements from can be considered as limit values for sequences in . Only when several metric spaces are considered, between which there are intersections, limit values ​​from another space are usually taken into account. A typical example of this is that a subspace of a metric space is treated. ${\displaystyle M}$ ${\displaystyle M}$ ${\displaystyle M}$ ${\displaystyle M}$

Examples

• The set of rational numbers is with the amount metric ${\displaystyle \mathbb {Q} }$

${\displaystyle d(x,y)=|x-y|}$

not complete, because the sequence of rational numbers is a Cauchy sequence whose limit (see Heron method ) is the irrational number that is not in . ${\displaystyle x_{1}=1,x_{n+1}={\tfrac {x_{n}}{2}}+{\tfrac {1}{x_{n}}}}$ ${\displaystyle {\sqrt {2}}}$ ${\displaystyle \mathbb {Q} }$

• The closed real interval , the set of real numbers and the set of complex numbers are each complete with the real or complex amount metric. ${\displaystyle [0,1]}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle \mathbb {C} }$

• The open real interval is not complete with the absolute value metric, because the limit value of the harmonic sequence does not lie in the interval. There are, however, complete metrics that produce the same topology as the amount metric, for example ${\displaystyle (0,1)}$ ${\displaystyle 0}$ ${\displaystyle \textstyle \left({\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},{\frac {1}{5}},\dots \right)}$ ${\displaystyle (0,1)}$

${\displaystyle d(x,y):=|x-y|+{\frac {1}{x}}+{\frac {1}{y}}+{\frac {1}{1-x}}+{\frac {1}{1-y}}}$   for   . ${\displaystyle x\not =y}$

• The space of the p -adic numbers is complete for every prime number . This space is the completion of with respect to the metric of the p -adic amount ${\displaystyle \mathbb {Q} _{p}}$ ${\displaystyle p}$ ${\displaystyle \mathbb {Q} }$

${\displaystyle d(x,y)=|x-y|_{p}}$,

as is the completion of for the absolute value metric. ${\displaystyle \mathbb {R} }$ ${\displaystyle \mathbb {Q} }$

• Every finite dimensional scalar product space , for example the Euclidean vector space or the unitary vector space with the standard scalar product , has the metric derived from the scalar product ${\displaystyle \mathbb {R} ^{n}}$ ${\displaystyle \mathbb {C} ^{n}}$

${\displaystyle d(x,y)={\sqrt {\langle x-y,x-y\rangle }}}$

Completely. A complete scalar product space is called a Hilbert space .

• Every finite-dimensional normed space , for example the space of real or complex matrices or with a matrix norm , has the metric derived from the norm ${\displaystyle \mathbb {R} ^{m\times n}}$ ${\displaystyle \mathbb {C} ^{m\times n}}$

${\displaystyle d(x,y)=\|x-y\|}$

Completely. A completely normalized space is called a Banach space .

• If there is any non-empty set, then the set of all sequences in can be made into a complete metric space by taking the distance between two sequences on ${\displaystyle S}$ ${\displaystyle S^{\mathbb {N} }}$ ${\displaystyle S}$ ${\displaystyle (x_{n}),(y_{n})}$

${\displaystyle d((x_{n}),(y_{n}))={\frac {1}{N}}}$

sets, where is the smallest index for which is different from , and where is the distance of a sequence to itself . ${\displaystyle N}$ ${\displaystyle x_{N}}$ ${\displaystyle y_{N}}$ ${\displaystyle 0}$

• For further examples of complete spaces of infinite dimension see the articles Banach space and Hilbert space .

A few sentences

Every compact metric space is complete. A metric space is compact if and only if it is complete and totally bounded .

A subset of a complete space is complete if and only if it is closed .

If a non-empty set and a complete metric space, then the space of bounded functions is from to with the metric ${\displaystyle X}$ ${\displaystyle (M,d)}$ ${\displaystyle B(X,M)}$ ${\displaystyle X}$ ${\displaystyle M}$

${\displaystyle d(f,g):=\sup _{x}d(f(x),g(x))}$

a full metric space.

If a topological space and a complete metric space, then the set of bounded continuous functions from to is a closed subset of and as such with the above metric complete. ${\displaystyle X}$ ${\displaystyle (M,d)}$ ${\displaystyle C_{b}(X,M)}$ ${\displaystyle X}$ ${\displaystyle M}$ ${\displaystyle B(X,M)}$

In Riemann's geometry , the statement of metric completeness is equivalent to geodetic completeness ( Hopf-Rinow's theorem ).

completion

Any metric space with a metric can be completed, that is, there is a complete metric space with a metric and an isometric so that is close in . The space is called completion of . Since all completions of isometrically isomorphic , one also speaks of the completion of . ${\displaystyle M}$ ${\displaystyle d}$ ${\displaystyle {\hat {M}}}$ ${\displaystyle {\hat {d}}}$ ${\displaystyle \varphi \colon M\rightarrow {\hat {M}}}$ ${\displaystyle \varphi (M)}$ ${\displaystyle {\hat {M}}}$ ${\displaystyle ({\hat {M}},{\hat {d}})}$ ${\displaystyle (M,d)}$ ${\displaystyle (M,d)}$ ${\displaystyle (M,d)}$

construction

The completion of can be constructed as a set of equivalence classes of Cauchy sequences in . ${\displaystyle M}$ ${\displaystyle M}$

First, let the set of Cauchy sequences in , and let the distance between two Cauchy sequences be through ${\displaystyle {\tilde {M}}}$ ${\displaystyle {\tilde {x}}:=\left(x_{m}\right)_{m\in \mathbb {N} }}$ ${\displaystyle M}$ ${\displaystyle {\tilde {d}}({\tilde {x}},{\tilde {y}})}$ ${\displaystyle {\tilde {x}},{\tilde {y}}\in {\tilde {M}}}$

${\displaystyle {\tilde {d}}({\tilde {x}},{\tilde {y}}):=\lim _{m,n\in \mathbb {N} }d(x_{m},y_{n})}$

Are defined. This distance is well defined and a pseudo metric on . The property ${\displaystyle {\tilde {M}}}$

${\displaystyle {\tilde {d}}({\tilde {x}},{\tilde {y}})=0\quad \Longleftrightarrow \quad {\bigl (}\forall \varepsilon >0\;\exists N\in \mathbb {N} \;\forall m,n\geq N\colon \;d(x_{m},y_{n})<\varepsilon {\bigr )}\quad \Longleftrightarrow :\quad {\tilde {x}}\sim {\tilde {y}}}$

defines an equivalence relation on . The distance can be transferred to the quotient set as follows: ${\displaystyle {\tilde {M}}}$ ${\displaystyle {\tilde {d}}}$ ${\displaystyle {\hat {M}}:={\tilde {M}}/\!\sim }$

If two equivalence classes and and are two (arbitrary) representatives, then one defines ${\displaystyle {\hat {x}},{\hat {y}}\in {\hat {M}}}$ ${\displaystyle {\tilde {x}}\in {\hat {x}}}$ ${\displaystyle {\tilde {y}}\in {\hat {y}}}$

${\displaystyle {\hat {d}}({\hat {x}},{\hat {y}}):={\tilde {d}}({\tilde {x}},{\tilde {y}})}$

as distance in . It is well-defined and is if and only if are equivalent. ${\displaystyle {\hat {M}}}$ ${\displaystyle {\hat {d}}({\hat {x}},{\hat {y}})={\tilde {d}}({\tilde {x}},{\tilde {y}})=0}$ ${\displaystyle {\tilde {x}}\sim {\tilde {y}}}$

This is a metric space. ${\displaystyle ({\hat {M}},{\hat {d}})}$

One can assign the stationary sequence to each element , because it is a Cauchy sequence. The equivalence class is in . In this way, the original metric space can in embedding . ${\displaystyle x\in M}$ ${\displaystyle \varphi (x):=(x)_{m\in \mathbb {N} }\in {\tilde {M}}}$ ${\displaystyle {\hat {x}}:=\{\xi \in {\tilde {M}}\mid \xi \sim \varphi (x)\}}$ ${\displaystyle {\hat {M}}}$ ${\displaystyle (M,d)}$ ${\displaystyle ({\hat {M}},{\hat {d}})}$

Since the elements are all Cauchy sequences , there is an approximating with for each ${\displaystyle {\tilde {x}}=:\left(x_{m}\right)_{m\in \mathbb {N} }\in {\tilde {M}}}$ ${\displaystyle M}$ ${\displaystyle \varepsilon >0}$ ${\displaystyle x_{n}\in M}$

${\displaystyle {\tilde {d}}({\tilde {x}},\varphi (x_{n}))=\lim _{m\in \mathbb {N} }d(x_{m},x_{n})<\varepsilon }$ .

So the picture lies close in , and that can be transferred to. ${\displaystyle \varphi (M)}$ ${\displaystyle {\tilde {M}}}$ ${\displaystyle {\hat {M}}={\tilde {M}}/\!\sim }$

In the following, the function name is omitted for the sake of brevity . ${\displaystyle \varphi }$

${\displaystyle ({\hat {M}},{\hat {d}})}$ is also complete.

proof

Let be a Cauchy sequence of elements . To show is: ${\displaystyle {\hat {x}}:=\left({\hat {x}}_{\mu }\right)_{\mu \in \mathbb {N} }}$ ${\displaystyle {\hat {M}}}$ ${\displaystyle {\hat {x}}}$owns in a Limes. ${\displaystyle {\hat {M}}}$ In the following, one of its representatives is taken instead of the equivalence class . This is possible because and behave equivalently under the metric. ${\displaystyle {\hat {x}}_{\mu }}$ ${\displaystyle \left(x_{\mu m}\right)_{m\in \mathbb {N} }:={\tilde {x}}_{\mu }\in {\hat {x}}_{\mu }}$ ${\displaystyle {\hat {x}}_{\mu }}$ ${\displaystyle {\tilde {x}}_{\mu }}$ For the sake of simplicity it is assumed that two successive representatives are not equivalent to one another. (If that is not the case, then the duplicate-free partial sequence is formed, the convergence of which results in that of the initial sequence; or the sequence becomes stationary , then is .) ${\displaystyle {\tilde {x}}_{\mu }\nsim {\tilde {x}}_{\mu +1}}$ ${\displaystyle \exists \mu \in \mathbb {N} \colon \forall \nu >\mu \implies {\tilde {x}}_{\nu }\sim {\tilde {x}}_{\mu }}$ ${\displaystyle \lim _{\nu \in \mathbb {N} }{\tilde {x}}_{\nu }\sim {\tilde {x}}_{\mu }\in \lim _{\mu \in \mathbb {N} }{\hat {x}}_{\mu }}$ Translation: . ${\displaystyle \varepsilon _{\mu }:={\tilde {d}}\left({\tilde {x}}_{\mu },{\tilde {x}}_{\mu +1}\right)}$ Because is a Cauchy sequence, and is a null sequence . ${\displaystyle \left({\tilde {x}}_{\mu }\right)_{\mu \in \mathbb {N} }}$ ${\displaystyle \left(\varepsilon _{\mu }\right)_{\mu \in \mathbb {N} }}$ ${\displaystyle \forall \mu \implies \varepsilon _{\mu }>0}$ Since each is itself a Cauchy sequence with terms , an approximating one with the property ${\displaystyle {\tilde {x}}_{\mu }=:\left(x_{\mu m}\right)_{m\in \mathbb {N} }}$ ${\displaystyle M}$ ${\displaystyle {\tilde {x}}_{\mu }}$ ${\displaystyle x_{\mu m_{\mu }}\in M}$ ${\displaystyle {\tilde {d}}(x_{\mu m_{\mu }},{\tilde {x}}_{\mu })<\varepsilon _{\mu }}$ can be chosen, analogous to the sequential term an approximating with . And since Cauchy is, for every one , so ${\displaystyle {\hat {x}}_{\nu }\ni {\tilde {x}}_{\nu }=:\left(x_{\nu n}\right)_{n\in \mathbb {N} }}$ ${\displaystyle x_{\nu n_{\nu }}\in M}$ ${\displaystyle {\tilde {d}}(x_{\nu n_{\nu }},{\tilde {x}}_{\nu })<\varepsilon _{\nu }}$ ${\displaystyle {\hat {x}}:=\left({\hat {x}}_{\mu }\right)_{\mu \in \mathbb {N} }}$ ${\displaystyle \varepsilon >0}$ ${\displaystyle J\in \mathbb {N} }$ ${\displaystyle \forall \mu ,\nu >J\implies {\hat {d}}({\hat {x}}_{\mu },{\hat {x}}_{\nu })={\tilde {d}}\left({\tilde {x}}_{\mu },{\tilde {x}}_{\nu }\right)<{\tfrac {\varepsilon }{3}}}$. There is also one and one , so that ${\displaystyle I\in \mathbb {N} }$ ${\displaystyle K\in \mathbb {N} }$ ${\displaystyle \forall \mu >I\implies \varepsilon _{\mu }={\tilde {d}}(x_{\mu m_{\mu }},{\tilde {x}}_{\mu })<{\tfrac {\varepsilon }{3}}}$   and   ${\displaystyle \forall \nu >K\implies \varepsilon _{\nu }={\tilde {d}}(x_{\nu n_{\nu }},{\tilde {x}}_{\nu })<{\tfrac {\varepsilon }{3}}}$ is. With are for the three distances , and all , well ${\displaystyle L:=\max\{I,J,K\}}$ ${\displaystyle \mu ,\nu >L}$ ${\displaystyle {\tilde {d}}\left(x_{\mu m_{\mu }},{\tilde {x}}_{\mu }\right)}$ ${\displaystyle {\tilde {d}}\left({\tilde {x}}_{\mu },{\tilde {x}}_{\nu }\right)}$ ${\displaystyle {\tilde {d}}\left({\tilde {x}}_{\nu },x_{\nu n_{\nu }}\right)}$ ${\displaystyle <{\tfrac {\varepsilon }{3}}}$ ${\displaystyle {\begin{array}{ll}d\left(x_{\mu m_{\mu }},x_{\nu n_{\nu }}\right)={\tilde {d}}\left(x_{\mu m_{\mu }},x_{\nu n_{\nu }}\right)&\leq {\tilde {d}}\left(x_{\mu m_{\mu }},{\tilde {x}}_{\mu }\right)+{\tilde {d}}\left({\tilde {x}}_{\mu },{\tilde {x}}_{\nu }\right)+{\tilde {d}}\left({\tilde {x}}_{\nu },x_{\nu n_{\nu }}\right)\\&<{\tfrac {\varepsilon }{3}}+{\tfrac {\varepsilon }{3}}+{\tfrac {\varepsilon }{3}}=\varepsilon .\end{array}}}$ Thus Cauchy is and . His equivalence class is . There as well ${\displaystyle {\tilde {y}}:=\left(x_{\mu m_{\mu }}\right)_{\mu \in \mathbb {N} }}$ ${\displaystyle \in {\tilde {M}}}$ ${\displaystyle {\hat {y}}:=\{\eta \in {\tilde {M}}\mid \eta \sim {\tilde {y}}\}\in {\hat {M}}}$ ${\displaystyle \forall \mu >I\implies {\hat {d}}({\hat {y}},{\hat {x}}_{\mu })<{\tfrac {\varepsilon }{3}}}$, surrendered ${\displaystyle {\hat {y}}=\lim _{\mu \in \mathbb {N} }{\hat {x}}_{\mu }}$ .

In this way, the expectation resulting from the word “completed” is actually fulfilled “completely”, and the completion of an already complete space does not bring anything new.

If a normalized space is , its completion can also be formed more easily by using ${\displaystyle M}$

${\displaystyle {\hat {M}}:={\overline {\varphi (M)}}\subseteq M^{\prime \prime }}$

as the conclusion of the image from in the bidual space under the canonical embedding . ${\displaystyle M}$ ${\displaystyle M^{\prime \prime }}$ ${\displaystyle \varphi \colon M\rightarrow M^{\prime \prime }}$

properties

Cantor's construction of the real numbers from the rational ones is a special case of this. However, since the metric already presupposes the existence of the real numbers, the equivalence relation must be defined in that the difference sequence of two Cauchy sequences is a zero sequence. ${\displaystyle {\tilde {d}}}$

If a standardized vector space is completed, a Banach space is obtained which contains the original space as a dense subspace. Therefore a Hilbert space is obtained when completing a Euclidean vector space, because the parallelogram equation remains fulfilled in the completion as a normalized space and the complete scalar product then results from the polarization formula .

Uniformly continuous mappings of a metric space in a complete metric space can always be unambiguously continued (automatically also uniformly) continuous mappings on the completion with values ​​in . ${\displaystyle M}$ ${\displaystyle X}$ ${\displaystyle {\hat {M}}}$ ${\displaystyle X}$

Fully metrizable rooms

Completeness is a property of metric , not topology , that is, a complete metric space can be homeomorphic to an incomplete metric space. For example, the real numbers are complete but homeomorphic to the open interval that is not complete (for example, a homeomorphism is from to ). Another example are the irrational numbers, which are not complete, but homeomorphic to the space of natural number sequences (a special case of an example from above). ${\displaystyle (0,1)}$ ${\displaystyle \tan((x-1/2)\pi )}$ ${\displaystyle (0,1)}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle \mathbb {N} ^{\mathbb {N} }}$

In topology one considers completely metrizable spaces , that is, spaces for which at least one complete metric exists that generates the existing topology.

Uniform rooms

Like many other terms from the theory of metric spaces, the concept of completeness can also be generalized to the class of uniform spaces : A uniform space is called complete if every Cauchy network converges. Most of the above statements remain valid in the context of uniform rooms, for example every uniform room also has a unique completion. ${\displaystyle (X,\Phi )}$

Topological vector spaces have a naturally uniform structure and they are called complete if they are complete with regard to this uniform structure. They are called quasi-complete if every bounded Cauchy net converges, that is, if every bounded, closed set is complete.

A topological group is called complete if it is complete with regard to its left uniform structure (or equivalent: to its right uniform structure ).

literature

• Dirk Werner : functional analysis , Springer-Verlag, Berlin, 2007, ISBN 978-3-540-72533-6

Individual evidence

1. ↑ ^{a b} Dirk Werner: functional analysis . 2005, p. 2 .
2. ↑ BL van der Waerden Algebra I . 8th edition. Springer, 1971 p. 243f