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solutionshw1-201141

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

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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 . Now we assume that ξ ik ξ i for all i = 1 , · · · , n . Let > 0. Then there exists K i such that | ξ ik - ξ i | < n for all k K i . Now let K = max { K 1 , · · · , K n } . Then n X j =1 ( ξ jk - ξ j ) 2 !

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