solutionshw10-554-2010

# solutionshw10-554-2010 - Solutions homework 10 Page 64...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions homework 10. Page 64 Problem 3. Let a = sup A . Then for all n ≥ 1 the number a- 1 n is not an upperbound of A , so there exists a n ∈ A such that a- 1 n < a n ≤ a . Then | a- a n | < 1 n , so a n → a . Thus a ∈ A . The proof for inf A ∈ A is similar. Page 64 Problem 6. (i) First let x ∈ A ∪ B . Then there exists x n ∈ A ∪ B such that x n → x . Either x n ultimately in A or x n frequently in B . In the first case x n ∈ A for all n ≥ N , so x ∈ A . In the second case there exists a subsequence x n k ∈ B . Then x ∈ B . Either way x ∈ A ∪ B . Another way to prove this inclusion is: note A ⊂ A and B ⊂ B . Hence A ∪ B ⊂ A ∪ B . As the union of two closed sets is closed, the righthand side is closed. This implies that A ∪ B ⊂ A ∪ B . For the opposite inclusion observe A ⊂ A ∪ B implies A ⊂ A ∪ B . Similarly B ⊂ A ∪ B . Hence A ∪ B ⊂ A ∪ B (ii) Let A = Q and B = I . Then A ∩ B = ∅ ., so A ∩ B = ∅ . On other hand....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online