solutionshw1-201141

solutionshw1-201141 - Solutions for HW 1 Problem 10;5. (1)...

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Unformatted text preview: Solutions for HW 1 Problem 10;5. (1) y ∈ f ( ∪ A ∈C A ) ⇐⇒ y = f ( x ) for some x ∈ ∪ A ∈C A ⇐⇒ y ∈ f ( A ) for some A ∈ C ⇐⇒ y ∈ ∪ A ∈C f ( A ). (2) Let y ∈ f ( ∩ A ∈C A ). This implies y = f ( x ) for some x ∈ ∩ A ∈C A . Hence y ∈ f ( A ) for all A ∈ C , i.e., y ∈ ∩ A ∈C f ( A ) Assume now f is injective. Let y ∈ ∩ A ∈C f ( A ). Then for all A ∈ C there exists x ∈ A such that f ( x ) = y . In general x depends on A , but since f is injective x does not depend on A , since if x 1 ∈ A 1 , x 2 ∈ A 2 such that f ( x 1 ) = y = f ( x 2 ) then x 1 = x 2 . Hence y = f ( x ) for some x ∈ ∩ A ∈C A . Thus y ∈ f ( ∩ A ∈C A ). Conversely, if f ( A ∩ B ) = f ( A ) ∩ f ( B ) for all A,B , then let A = { x } and B = { y } with x 6 = y . Then A ∩ B = ∅ implies that { f ( x ) }∩{ f ( y ) } = ∅ , i.e. f ( x ) 6 = f ( y ). Problem 20: 4. Assume that x k = { ξ ik } → x = { ξ i } in the Euclidean metric d . Let > 0. Then there exists a K such that d ( x k ,x ) < for all k ≥ K . Now | ξ ik- ξ i | ≤ n X j =1 ( ξ jk- ξ j ) 2 ! 1 2 implies that | ξ ik- ξ i | < for all k ≥ K . Hence ξ ik → ξ i for all i = 1 , ··· ,n ....
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This note was uploaded on 02/05/2012 for the course MATH 703 taught by Professor Schep during the Fall '11 term at South Carolina.

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solutionshw1-201141 - Solutions for HW 1 Problem 10;5. (1)...

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