s6 - Math 5615H: Introduction to Analysis I. Fall 2011...

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Unformatted text preview: Math 5615H: Introduction to Analysis I. Fall 2011 Homework #6. Problems and Solutions. #1. Show that for an arbitrary set E in a metric space ( X,d ), the set E of its limit point is closed. Proof. Let p be a limit point of E . Then ∀ r > 0, the set G r := N r ( p ) \ { p } is open and contains a point q ∈ E . Therefore, N ε ( q ) ⊆ G r for some ε > 0. Since q ∈ E , there is a point s ∈ ( E ∩ N ε ( q ) ) \ { q } ⊆ E ∩ G r . This means that p ∈ E , and E is closed. #2. Show that any set E ⊂ R 1 has at most countably many isolated points. Proof. Let S ⊆ E denote the set of all isolated points of E . For each x ∈ S , there is a neighborhood N r ( x ) := ( x- r,x + r ) of x , where r = r ( x ) > 0, such that E ∩ N r ( x ) = { x } . We must have max ' r ( x ) ,r ( y ) “ ≤ | x- y | ∀ x,y ∈ S, x 6 = y. Indeed, if r ( x ) > | x- y | , then N r ( x ) contains at least two different points x,y ∈ S , which is not the case. Therefore, r ( x ) ≤ | x- y | , and similarly,...
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s6 - Math 5615H: Introduction to Analysis I. Fall 2011...

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