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Homework 1 Solutions

LetXandYbe sets and letf:X→Ybe a mapping. LetA,B?XandC,D?Ybe subsets, and let A i?X,i?I, be a family of subsets, indexed by some setI. Remember: the goal in writing proofs is not only to be right, but to be understood. Never turn in an

assignment without proofreading it. (As Truman Capote said in reference to the Beat writers, "That isn"t

writing at all, it"s typing.") And when you proofread your proofs, check not only for correctness but ask

yourself: is it clear? Is it unambiguous? Could I say the same thing more simply? Could I shorten it and

still be clear? Would a few extra words be helpful? Also remember that proof writing is a form of writing

and thus requires sentences. Proofs with no words are hard to read. A specific general rule is never to begin

a sentence with a symbol. So every proof should have at least one sentence and thus at least one word!

Each problem is worth 20 points. Solutions will be graded for correctness, clarity and style.

(1) Prove that (?i?IAi)c=∩i?IAci. (IfSis a subset of a setT, recall thatScdenotes the complement

ofSinT; i.e.,Sc=T\S. Note that the notationScis ambiguous ifSis a subset of two different sets.) Solution: First verify (?i?IAi)c? ∩i?IAci. Leta?(?i?IAi)c. Thusa?? ?i?IAi, soacannot be in any of the setsAi; i.e., for alli?I, we havea??Ai, hencea?Acifor alli?I. Thusa? ∩i?IAci. Now verify (?i?IAi)c? ∩i?IAci. Leta? ∩i?IAci. Thusa?Acifor alli?I, hencea??Aifor all i?I, soa?? ?i?IAi, hencea?(?i?IAi)c. (2) Prove thatf-1(C?D) =f-1(C)?f-1(D). Solution: First verifyf-1(C?D)?f-1(C)?f-1(D). Letx?f-1(C?D). Thusf(x)?C?D, so eitherf(x)?Corf(x)?D, hence eitherx?f-1(C) orx?f-1(D), sox?f-1(C)?f-1(D). Now verifyf-1(C?D)?f-1(C)?f-1(D). Letx?f-1(C)?f-1(D). Thus eitherx?f-1(C) orx?f-1(D), so eitherf(x)?Corf(x)?D, sof(x)?C?D, hencex?f-1(C?D). (3) Isf(A?B) =f(A)?f(B) always true? If so, prove it. If not, give a specific example for which equality does not hold. Solution: Yes,f(A?B) =f(A)?f(B) is always true. We first verifyf(A?B)?f(A)?f(B). Lety?f(A?B). Thus there is anx?A?Bsuch thatf(x) =y. Eitherx?Aorx?B, so either f(x)?f(A) orf(x)?f(B), hencey=f(x)?f(A)?f(B). Now we verifyf(A?B)?f(A)?f(B). Lety?f(A)?f(B). Then eithery?f(A) (in which case y=f(a) for somea?A) ory?f(B) (in which casey=f(b) for someb?B). Butf(a)?f(A?B) andf(b)?f(A?B), so either wayy?f(A?B).

(4) Isf(A∩B) =f(A)∩f(B) always true? If so, prove it. If not, give a specific example for which

equality does not hold. Solution: It is not always true thatf(A∩B) =f(A)∩f(B). LetX=Y=Rbe the reals, and letfbef(x) =x2. Consider the intervalsA= [-1,0] and letB= [0,1]. ThenA∩B={0}, so f(A∩B) ={0}, butf(A) = [0,1] =f(B), sof(A)∩f(B) = [0,1]. (5) Which off(f-1(C))?C,f(f-1(C)) =C, andf(f-1(C))?Cis always true? Prove those which are true and give specific examples showing the others can fail. Solution: It is always true thatf(f-1(C))?C, butf(f-1(C))?Cand hencef(f-1(C)) =C can fail. For example, letfbe as in the solution to [4], sof(x) =x2and letC= [-1,1]. Then f -1(C) = [-1,1] =Candf(C) = [0,1]?C, sof(f-1(C))?C, hencef(f-1(C))?Cdoes not hold. Now we showf(f-1(C))?Calways holds. Letc?f(f-1(C)). Thenc=f(d) for some d?f-1(C), andf(d)?C; i.e.,c=f(d)?C, as we wanted to show. 1quotesdbs_dbs14.pdfusesText_20