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Difference between revisions of "Complete set"

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''in a topological vector space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c0238601.png" /> over a field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c0238602.png" />''
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''in a topological vector space $X$ over a field $K$''
  
A set <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c0238603.png" /> such that the set of linear combinations of the elements from <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c0238604.png" /> is (everywhere) dense in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c0238605.png" />, i.e. the closed subspace generated by the set <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c0238606.png" />, i.e. the closed linear hull of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c0238607.png" />, coincides with <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c0238608.png" />. For example, in the normed space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c0238609.png" /> of continuous functions on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386010.png" /> with values in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386011.png" /> the set <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386012.png" /> is a complete set. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386013.png" /> is a non-discretely normed field, each absorbing set (and in particular each neighbourhood of zero in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386014.png" />) is a complete set.
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A set $A$ such that the set of linear combinations of the elements from $A$ is (everywhere) dense in $X$, i.e. the closed subspace generated by the set $A$, i.e. the closed linear hull of $A$, coincides with $X$. For example, in the normed space $C$ of continuous functions on $[0,1]$ with values in $\mathbf C$ the set $\{x^n\}$ is a complete set. If $K$ is a non-discretely normed field, each absorbing set (and in particular each neighbourhood of zero in $X$) is a complete set.
  
In order for <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386015.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386016.png" />, to be a complete set in the weak topology <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386017.png" /> of a space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386018.png" /> it is necessary and sufficient that there exist an index <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386019.png" /> such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386020.png" /> for each <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386021.png" />; this means that no closed hyperplane contains all the elements <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386022.png" />, i.e. that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386023.png" /> is a total set. Moreover, if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386024.png" /> is a [[Locally convex space|locally convex space]], a complete set in the weak topology will be a complete set in the initial topology also.
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In order for $A=\{a_t\}$, $t\in T$, to be a complete set in the weak topology $\sigma(X,X^*)$ of a space $X$ it is necessary and sufficient that there exist an index $t$ such that $\langle a_t,\xi\rangle\neq0$ for each $\xi\in X^*$; this means that no closed hyperplane contains all the elements $a_t$, i.e. that $A$ is a total set. Moreover, if $X$ is a [[Locally convex space|locally convex space]], a complete set in the weak topology will be a complete set in the initial topology also.
  
  
  
 
====Comments====
 
====Comments====
Of course, a complete set in a topological vector space can also mean a set <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386025.png" /> such that every [[Cauchy sequence|Cauchy sequence]] in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386026.png" /> converges in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023860/c02386027.png" />; and this is by far the most frequently occurring meaning of the phrase. For the notion of an absorbing set cf. [[Topological vector space|Topological vector space]].
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Of course, a complete set in a topological vector space can also mean a set $A$ such that every [[Cauchy sequence|Cauchy sequence]] in $A$ converges in $A$; and this is by far the most frequently occurring meaning of the phrase. For the notion of an absorbing set cf. [[Topological vector space|Topological vector space]].

Latest revision as of 09:44, 5 August 2014

in a topological vector space $X$ over a field $K$

A set $A$ such that the set of linear combinations of the elements from $A$ is (everywhere) dense in $X$, i.e. the closed subspace generated by the set $A$, i.e. the closed linear hull of $A$, coincides with $X$. For example, in the normed space $C$ of continuous functions on $[0,1]$ with values in $\mathbf C$ the set $\{x^n\}$ is a complete set. If $K$ is a non-discretely normed field, each absorbing set (and in particular each neighbourhood of zero in $X$) is a complete set.

In order for $A=\{a_t\}$, $t\in T$, to be a complete set in the weak topology $\sigma(X,X^*)$ of a space $X$ it is necessary and sufficient that there exist an index $t$ such that $\langle a_t,\xi\rangle\neq0$ for each $\xi\in X^*$; this means that no closed hyperplane contains all the elements $a_t$, i.e. that $A$ is a total set. Moreover, if $X$ is a locally convex space, a complete set in the weak topology will be a complete set in the initial topology also.


Comments

Of course, a complete set in a topological vector space can also mean a set $A$ such that every Cauchy sequence in $A$ converges in $A$; and this is by far the most frequently occurring meaning of the phrase. For the notion of an absorbing set cf. Topological vector space.

How to Cite This Entry:
Complete set. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Complete_set&oldid=32720
This article was adapted from an original article by M.I. Voitsekhovskii (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article