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 $\u03f5>0$, there exists a finite subset $\{{s}_{1},{s}_{2},\mathrm{\dots},{s}_{n}\}$ of $A$ such that $A\subseteq {\bigcup}_{k=1}^{n}B({s}_{k},\u03f5)$, where $B({s}_{k},\u03f5)$ denotes the open ball around ${s}_{k}$ with radius $\u03f5$.
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 |
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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 |