# sol2 - Problem Set 2 18.100B/C Fall 2011 Michael Andrews...

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Unformatted text preview: Problem Set 2, 18.100B/C, Fall 2011 Michael Andrews Department of Mathematics MIT September 21, 2011 1 Problem 15 from page 23. Let a = ( a 1 ,...,a n ) , b = ( b 1 ,...,b n ) ∈ C n . Put A = n X i =1 | a j | 2 , B = n X j =1 | b j | 2 and C = n X j =1 a j b j . Then it was proved in Theorem 1 . 35 that B ( AB- | C | 2 ) = n X j =1 | Ba j- Cb j | 2 ≥ . The Cauchy-Schwarz inequality is | C | 2 ≤ AB. When B = 0 we have b = 0; thus C = 0, Ba j = Cb j for all j , and we have equality. When B 6 = 0 we have equality if and only if n X j =1 | Ba j- Cb j | 2 = 0 ⇐⇒ Ba j = Cb j for each j ∈ { 1 ,...n } . Thus, in all cases we have equality in the Cauchy Schwarz inequality if and only if Ba j = Cb j for all j . Now, Ba j = Cb j for all j is equivalent to λa = μb for some ( λ,μ ) ∈ C 2 \ { } . You should check this. 2 Let X be a metric space and E ⊂ X . 1 a Let p ∈ E ◦ . By definition, p is an interior point of E , so there exists an r > 0 such that N r ( p ) ⊂ E . If we can show N r ( p ) ⊂ E ◦ it will follow that p is an interior point of E ◦ , and thus E ◦ is open. But for any q ∈ N r ( p ) we have N r- d ( p,q ) ( q ) ⊂ N r ( p ) ⊂ E, which implies q ∈ E ◦ , as required. (For the above inclusion we use the triangle inequality: x ∈ N r- d ( p,q ) ( q ) = ⇒ d ( x,q ) < r- d ( p,q ) = ⇒ d ( x,p ) ≤ d ( x,q ) + d ( p,q ) < r = ⇒ x ∈ N r ( p ) . ) b E is open ⇐⇒ every point of E is an interior point of E ⇐⇒ E ⊂ E ◦ . It is clear that we always have E ◦ ⊂ E (since a neighborhood of a point contains the point). Hence, E is open if and only if E ◦ = E . c Let G ⊂ E and suppose G is open. Given p ∈ G , there exists an r > 0 such that N r ( p ) ⊂ G . Since G ⊂ E we have N r ( p ) ⊂ E and so p ∈ E ◦ . d By definition, x ∈ E ◦ if and only if there exists an r > 0 such that N r ( x ) ⊂ E . Thus, x / ∈ E ◦ if and only if for all r > 0, N r ( x ) ∩ ( X \ E ) 6 = ∅ . Suppose that for all r > 0, N r ( x ) ∩ ( X \ E ) 6 = ∅ . Then either x ∈ X \ E or x is a limit point of X \ E , i.e. x ∈ X \ E . Conversely, if x ∈ X \ E , then either x ∈ X \ E or x is a limit point of X \ E and in either case N r ( x ) ∩ ( X \ E ) 6 = ∅ , for all r > 0. e, f No, in both cases. Let X = R and E = Q ....
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sol2 - Problem Set 2 18.100B/C Fall 2011 Michael Andrews...

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