totally bounded

Let $A$ be a subset of a topological vector space $X$ .

$A$ is called totally bounded if , for each neighborhood $G$ of 0, there exists a finite subset $S$ of $A$ with $A$ contained in the sumset $S+G$ .

The definition can be restated in the following form when $X$ is a metric space:

A set $A\subseteq X$ is said to be totally bounded if for every $\epsilon>0$ , there exists a finite subset $\{s_{1},s_{2},\ldots,s_{n}\}$ of $A$ such that $A\subseteq\bigcup_{k=1}^{n}B(s_{k},\epsilon)$ , where $B(s_{k},\epsilon)$ denotes the open ball around $s_{k}$ with radius $\epsilon$ .

References

1 G. Bachman, L. Narici, Functional analysis, Academic Press, 1966.
2 A. Wilansky, Functional Analysis, Blaisdell Publishing Co., 1964
3 W. Rudin, Functional Analysis, 2nd ed. McGraw-Hill , 1973

Title	totally bounded
Canonical name	TotallyBounded
Date of creation	2013-03-22 13:09:54
Last modified on	2013-03-22 13:09:54
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	12
Author	Mathprof (13753)
Entry type	Definition
Classification	msc 54E35
Related topic	MetricSpace
Related topic	Bounded
Related topic	Subset