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hw5sol

# hw5sol - Honors Introduction to Analysis I Homework V...

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Unformatted text preview: Honors Introduction to Analysis I Homework V Solution March 10, 2009 Problem 1 Let A be an open set. Show that if a finite number of points are removed from A , the remaining set is still open. Is the same true if a countable number of points are removed? Solution. If we order the points removed from A , x 1 < x 2 < ... < x k we can write A- { x 1 ,x 2 ,...,x k } = A ∩ ((-∞ ,x 1 ) ∪ ( x 1 ,x 2 ) ∪ ( x 2 ,x 3 ) ∪ ... ( x k- 1 ,x k ) ∪ ( x k , ∞ )) B := (-∞ ,x 1 ) ∪ ( x 1 ,x 2 ) ∪ ( x 2 ,x 3 ) ∪ ... ( x k- 1 ,x k ) ∪ ( x k , ∞ ) is a union of open sets, hence open. A ∩ B is a finite intersection of open sets, hence it is also open. The statement is not true for countable number of points. Consider R- Q . This is not open, since given any irrational point x , by the density of rational, we can find for every n a rational y ∈ ( x- 1 /n,x + 1 /n ), hence we couldn’t find a neighborhood around x completely in R- Q . Problem 2 Give an example of a set A that is not closed but such that every point of A is a limit-point. Solution. Consider the open interval A = (0 , 1). For any x ∈ (0 , 1) there is an open interval ( x- 1 /n,x +1 /n ) for some n ≥ N such that ( x- 1 /n,x + 1 /n ) ⊂ (0 , 1). But in ( x- 1 /n,x + 1 /n ) there are infintely points from (0 , 1), hence x is a limit point. On the other hand 0 and 1 are limit points, but they are not in A , hence A is not closed. Problem 3 Show that the set of limit-points of a sequence is a closed set. Solution. Let L be the set of limit points of the sequence { x n } . Let a be a limit point of the set L . Then for all errors 1 /n , there exists an x ∈ L such that | x- a | < 1 2 n ,x 6 = a . But x is a limit point of the sequence, hence for all n there exist infinitely many x k ’s such that | x- x k | < 1 2 n . From the above we have that for all n ∈ N , | a- x k | = | a- x + x- x k | < | a- x | + | x- x k | < 1 / 2 n + 1 / 2 n < 1 /n for infinitely many x k ’s. Hence a is also a limit point of the sequence, so a ∈ L . Therefore L is closed....
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hw5sol - Honors Introduction to Analysis I Homework V...

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