proof of exhaustion by compact sets for
Rn∗
cvalente†
2013-03-21 20:25:38
First consider A ⊂ Rn to be a bounded open set and designate the open ball
centered at x with radius
r by Br (x)
S
Construct Cn = x∈∂A B n1 (x), where ∂A is the boundary of A and define
Kn = A\Cn .
• Kn is compact.
It is bounded since Kn ⊂ A and A is by assumption bounded. Kn is also
closed. To see this consider x ∈ ∂Kn but x ∈
/ Kn . Then there exists
y ∈ ∂A and 0 < r < n1 such that x ∈ Br (y). But Br (y) ∩ Kn = {} because
B n1 (y) ∩ Kn = {} and 0 < r < n1 =⇒ Br (y) ⊂ B n1 (y). This implies that
x∈
/ ∂Kn and we have a contradiction. Kn is therefore closed.
• Kn ⊂ int Kn+1
Suppose x ∈ Kn and x ∈
/ int Kn+1 . This means that for all y ∈ ∂A,
n
1 (y) ∨ x ∈ R \A.
Since x ∈ Kn =⇒ x ∈ A we must have
x ∈ B n+1
1 (y). But x ∈ B 1 (y) ⊂ B 1 (y)
x ∈ B n+1
=⇒ x ∈
/ Kn and we have a
n+1
n
contradiction.
S∞
• n=1 Kn = A
Suppose x ∈ A, since A is open there must exist r > 0 such that Br (x) ⊂
/ B n1 (y) for all y ∈ ∂A
A. Considering n such that n1 < r we have that x ∈
and thus x ∈ Kn .
Sn Finally if A is not bounded consider Ak = A ∩ Bk (0) and define Kn =
k=1 Kk,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.
∗ hProofOfExhaustionByCompactSetsFormathbbRni created: h2013-03-21i by: hcvalentei
version: h37850i Privacy setting: h1i hProofi h53-00i
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• Kn ⊂ int Kn+1 because Kk,n ⊂ int Kk,n+1 and int(A ∪ B) ⊂ int A ∪ int B
S∞
• n=1 Kn = A
First find k such that x ∈ Ak . This will always be possible since all it
requires is that k > |x|. Finally since n > k =⇒ Kk,n ⊂ Kn,n by
construction the argument for the bounded case is directly applicable.
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