open set in n contains an open rectangle

Theorem Suppose n is equipped with the usual topology induced by the open ballsPlanetmathPlanetmath of the Euclidean metric. Then, if U is a non-empty open set in n, there exist real numbers ai,bi for i=1,,n such that ai<bi and [a1,b1]××[an,bn] is a subset of U.

Proof. Since U is non-empty, there exists some point x in U. Further, since U is a topological spaceMathworldPlanetmath, x is contained in some open set. Since the topology has a basis consisting of open balls, there exists a yU and ε>0 such that x is contained in the open ball B(y,ε). Let us now set ai=yi-ε2n and bi=yi+ε2n for all i=1,,n. Then D=[a1,b1]××[an,bn] can be parametrized as

D={y+(λ1,,λn)ε2nλi[-1,1]for alli=1,,n}.

For an arbitrary point in D, we have

|y+(λ1,,λn)ε2n-y| = |(λ1,,λn)ε2n|
= ε2nλ12++λn2

so DB(y,ϵ)U, and the claim follows.

Title open set in n contains an open rectangleMathworldPlanetmath
Canonical name OpenSetInmathbbRnContainsAnOpenRectangle
Date of creation 2013-03-22 14:07:46
Last modified on 2013-03-22 14:07:46
Owner matte (1858)
Last modified by matte (1858)
Numerical id 5
Author matte (1858)
Entry type Theorem
Classification msc 54E35
Related topic IntervalMathworldPlanetmathPlanetmath