- #1
cpsinkule
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From Baby Rudin
"Thm: Let P be a non-empty, perfect subset of R^k. Then P is uncountable.
Pf: Since P has limit points, P must be infinite. Suppose P is countable, list the point of P {x1 ...xn }. Construct a sequence of nbhds. as follows. Let V1 be any nbhd of x1 . Suppose Vn has been constructed so that Vn ∩ P is non-empty. Since every point of P is a limit point of P, there is a nbhd Vn+1 such that it's closure is in Vn , xn is NOT in the closure of Vn+1, and Vn+1 ∩ P is non-empty. Let Kn=closure(Vn)∩P, then Kn is closed, bndd., therefore compact. The Kn are clearly nested compact sets, thus their arbitrary intersection is non-empty. But xn∉Kn+1 implies that ∩n∞Kn is empty, thus a contradiction. "
My confusion is in the claim of the contradiction. How does xn∉Kn+1 imply the intersection is empty? My thoughts are that, since we assumed P to be countable, xn must be an element of any inductive subset of P indexed by n+1. Is my logic behind this statement true? It just seemed a little artificial to me as I have not seen a proof which uses induction to arrive at a contradiction in this manner before.
"Thm: Let P be a non-empty, perfect subset of R^k. Then P is uncountable.
Pf: Since P has limit points, P must be infinite. Suppose P is countable, list the point of P {x1 ...xn }. Construct a sequence of nbhds. as follows. Let V1 be any nbhd of x1 . Suppose Vn has been constructed so that Vn ∩ P is non-empty. Since every point of P is a limit point of P, there is a nbhd Vn+1 such that it's closure is in Vn , xn is NOT in the closure of Vn+1, and Vn+1 ∩ P is non-empty. Let Kn=closure(Vn)∩P, then Kn is closed, bndd., therefore compact. The Kn are clearly nested compact sets, thus their arbitrary intersection is non-empty. But xn∉Kn+1 implies that ∩n∞Kn is empty, thus a contradiction. "
My confusion is in the claim of the contradiction. How does xn∉Kn+1 imply the intersection is empty? My thoughts are that, since we assumed P to be countable, xn must be an element of any inductive subset of P indexed by n+1. Is my logic behind this statement true? It just seemed a little artificial to me as I have not seen a proof which uses induction to arrive at a contradiction in this manner before.