This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full
Document
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.
 Fall '11
 Schep

Click to edit the document details