Homework 5, Advanced Calculus 1

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Homework 5, Advanced Calculus 1

1. Rudin Chapter 2 Exercise 18.

2. Rudin Chapter 2 Exercise 19cd.

Solution: Let E be a closed subset of a separable metric space X. Let P be the set of condensation points of E, which is perfect by Exercise 27. It suffices to show that P ⊂ E, since then P = E ∪ (P ∩ E^c) and the second set in the union is at most countable by Exercise 27.

By definition, condensation points are clearly limit point (uncountable set is clearly infinite), and therefore P ⊂ E⁰. But since E is closed, we have E = E⁰ and the proof is done.

Solution: Note the the collection of open intervals

I := {(q − r, q + r) | q, r ∈ Q}

is a countable base for R. (why?) Every open subset is therefore a union of some collection of intervals {I_i⁰} ⊂ I. Since I is countable, there are at most countably many Ii:

E =

∞

[

i=1

I_i⁰.

Partition N into a disjoint union of at most countably many subsets N = ∪^αNα, where i, j ∈ Nαfor some α if I_i⁰∩ I_j⁰ 6= φ. Define

I_α:= [

i∈N_α

I_i⁰.

We may check that

• Each Iα = (aα, bα), where aα (resp. bα) is the infimum (resp. supremum) of the left (right) endpoints of open intervals in {I_i⁰| i ∈ Nα}.

• Iα∩ Iβ= φ if α 6= β.

Clearly, E = ∪αIα is the desired union.

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Solution: Note that we may assume that all Fnare nonempty, since empty set does not contribute to the union. Suppose that F_n^o= ∅ for all n. Take x1 ∈ F1 and some ball Br₁(x1) centered at x1. By our assumption x1 is not interior point of F1.

8. Rudin Chapter 2 Exercise 19ab.

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