- #1

jecharla

- 24

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Let P be a nonempty perfect set in R^k. Since P has limit points, P is infinite. Suppose P is countable and denote the points of P by x_1, x_2, x_3, ... We shall construct a sequence of neighborhoods {V_n} as follows.

Let V_1 be any neighborhood of x_1. If V_1 consists of all y in R^k such that |y-x_1| < r, the closure closure(V_1) is the set of all y in R^k such that |y-x_1| </= r.

Suppose V_n has been constructed, so that V_n intersect P is not empty. Since every point of P is a limit point of P, there is a neighborhood V_(n+1) such that (i) closure[v(n+1)] is a subset of V_n, (ii) x_n is not an element of closure[V(n+1)], (iii) V_(n+1) intersect P is not empty. By (iii) v_(n+1) satisfies our induction hypothesis, and the construction can proceed.

Put K_n = closure(V_n) intersect P. Since closure(V_n) is closed and bounded, closure(V_n) is compact. Since x_n is not an element of K_(n+1), no point of P lies in the intersection from 1 to infinity of K_n. Since K_n is a subset of P this implies that the intersection from 1 to infinity of K_n is empty. But each K_n is not empty and by (iii), and K_n contains K_(n+1), by (i); this contradicts the Corollary to theorem 2.36.

For reference the corollary to theorem 2.36 is:

If {K_n} is a sequence of nonempty compact sets such that K_n contains K_(n+1) (n = 1, 2, 3, ...), then the intersection from 1 to infinity of K_n is not empty.

My question is this: Rudin first assumes P is countable and that the elements of P are x_1, x_2, x_3, ... But it seems to me that when he specifies conditions for each x_n greater than x_1, these elements are no longer arbitrary, and no longer illustrate the general case of assuming P is countable. Where am I going wrong?