# 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 TotallyBounded 2013-03-22 13:09:54 2013-03-22 13:09:54 Mathprof (13753) Mathprof (13753) 12 Mathprof (13753) Definition msc 54E35 MetricSpace Bounded Subset