This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 10. OCTOBER 5TH: FUNCTIONS. 25 10. October 5th: Functions. 10.1 . Let A , B be sets, and f : A → B a function. Let S and T be subsets of A . • Is f ( S ∩ T ) necessarily equal to f ( S ) ∩ f ( T )? Give a proof or find a coun terexample. • Is f ( S ∪ T ) necessarily equal to f ( S ) ∪ f ( T )? Proof or counterexample. Answer: • f ( S ∩ T ) and f ( S ) ∩ f ( T ) need not be equal. For example, consider the function f : {◦ , *} → { } defined by f ( ◦ ) = 0, f ( * ) = 0, and let S = {◦} , T = {*} . Then S ∩ T = ∅ , so f ( S ∩ T ) = ∅ ; while f ( S ) ∩ f ( T ) = { } = ∅ . Note that the inclusion f ( S ∩ T ) ⊆ f ( S ) ∩ f ( T ) always holds. (Why?) • It is true that f ( S ) ∪ f ( T ) = f ( S ∪ T ) for all A, B, f, S, T . Proof: —Since S and T are subsets of S ∪ T , then f ( S ) and f ( T ) are both subsets of f ( S ∪ T ); this shows f ( S ) ∪ f ( T ) ⊆ f ( S ∪ T ). —On the other hand, if b ∈ f ( S ∪ T ) then ∃ a ∈ S ∪ T such that f ( a ) = b . Since a ∈ S ∪ T , then a ∈ S or a ∈ T . If a ∈ S , then...
View
Full
Document
This note was uploaded on 12/14/2011 for the course MGF 3301 taught by Professor Aluffi during the Fall '11 term at FSU.
 Fall '11
 Aluffi

Click to edit the document details