proof of exhaustion by compact sets for n

First consider An to be a bounded open set and designate the open ball centered at x with radius r by Br(x)

Construct Cn=xAB1n(x), where A is the boundary of A and define Kn=A\Cn.

  • Kn is compact.

    It is bounded since KnA and A is by assumptionPlanetmathPlanetmath bounded. Kn is also closed. To see this consider xKn but xKn. Then there exists yA and 0<r<1n such that xBr(y). But Br(y)Kn={} because B1n(y)Kn={} and 0<r<1nBr(y)B1n(y). This implies that xKn and we have a contradictionMathworldPlanetmathPlanetmath. Kn is therefore closed.

  • KnintKn+1

    Suppose xKn and xintKn+1. This means that for all yA, xB1n+1(y)¯xn\A. Since xKnxA we must have xB1n+1(y)¯. But xB1n+1(y)¯B1n(y)xKn and we have a contradiction.

  • n=1Kn=A

    Suppose xA, since A is open there must exist r>0 such that Br(x)A. Considering n such that 1n<r we have that xB1n(y) for all yA and thus xKn.

Finally if A is not bounded consider Ak=ABk(0) and define Kn=k=1nKk,n where Kk,n is the set resulting from the previous construction on the bounded set Ak.

  • Kn will be compact because it is the finite union of compact sets.

  • KnintKn+1 because Kk,nintKk,n+1 and int(AB)intAintB

  • n=1Kn=A

    First find k such that xAk. This will always be possible since all it requires is that k>|x|. Finally since n>kKk,nKn,n by construction the argument for the bounded case is directly applicable.

Title proof of exhaustion by compact sets for n
Canonical name ProofOfExhaustionByCompactSetsFormathbbRn
Date of creation 2013-03-22 15:51:38
Last modified on 2013-03-22 15:51:38
Owner cvalente (11260)
Last modified by cvalente (11260)
Numerical id 5
Author cvalente (11260)
Entry type Proof
Classification msc 53-00