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What is this? Can you show with circles? For real function f A = f(f(A)) for en every why is A subset of the inverse? Can you show with for every real functim f: IR→IR, A = ² (F(A)) for every A CIR =) To show A is contained in f(f(A)), let a EA; we need to show that a Ef(f(A)). But this holds ift f(a) Ef (A), which holds since a EA and +(A) = {f(x)=xEA} The other inclusion i.e. f-1 (+(A)) S.A does not hold in general. So I have the following: Let fix→y be a function. Then of is one-to-one (injective) iff for every A CX, we have A = f(f(A)). pref= Assume first that of is injective, Let A CX. So we We already know that A ≤ f($(A)). -need to show that f-1 (F(A)) SA. only Let x Ef(f(A)). i.e. f(x) Ef(A) So 3 EA st. f(x) = f(a). Since of is one-to-one (injective), so f(x) = f(x) =)x=a SO REA = f(f(A)) ≤A Hence A = f(f(A)) Now, Conversely; Assume that A = f(f(A)) for all A EX. can you explain the proof (injective)

What is this? Can you show with circles? For real function f A = f(f(A)) for en every why is A subset of the inverse? Can you show with for every real functim f: IR→IR, A = ² (F(A)) for every A CIR =) To show A is contained in f(f(A)), let a EA; we need to show that a Ef(f(A)). But this holds ift f(a) Ef (A), which holds since a EA and +(A) = {f(x)=xEA} The other inclusion i.e. f-1 (+(A)) S.A does not hold in general. So I have the following: Let fix→y be a function. Then of is one-to-one (injective) iff for every A CX, we have A = f(f(A)). pref= Assume first that of is injective, Let A CX. So we We already know that A ≤ f($(A)). -need to show that f-1 (F(A)) SA. only Let x Ef(f(A)). i.e. f(x) Ef(A) So 3 EA st. f(x) = f(a). Since of is one-to-one (injective), so f(x) = f(x) =)x=a SO REA = f(f(A)) ≤A Hence A = f(f(A)) Now, Conversely; Assume that A = f(f(A)) for all A EX. can you explain the proof (injective)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.1: Inverse Functions
Problem 18E
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What is this? Can you show with circles? For real function f A = f(f(A)) for en every why is A subset of the inverse? Can you show with for every real functim f: IR→IR, A = ² (F(A)) for every A CIR =) To show A is contained in f(f(A)), let a EA; we need to show that a Ef(f(A)). But this holds ift f(a) Ef (A), which holds since a EA and +(A) = {f(x)=xEA} The other inclusion i.e. f-1 (+(A)) S.A does not hold in general. So I have the following: Let fix→y be a function. Then of is one-to-one (injective) iff for every A CX, we have A = f(f(A)). pref= Assume first that of is injective, Let A CX. So we We already know that A ≤ f($(A)). -need to show that f-1 (F(A)) SA. only Let x Ef(f(A)). i.e. f(x) Ef(A) So 3 EA st. f(x) = f(a). Since of is one-to-one (injective), so f(x) = f(x) =)x=a SO REA = f(f(A)) ≤A Hence A = f(f(A)) Now, Conversely; Assume that A = f(f(A)) for all A EX. can you explain the proof (injective)
Transcribed Image Text:What is this? Can you show with circles? For real function f A = f(f(A)) for en every why is A subset of the inverse? Can you show with for every real functim f: IR→IR, A = ² (F(A)) for every A CIR =) To show A is contained in f(f(A)), let a EA; we need to show that a Ef(f(A)). But this holds ift f(a) Ef (A), which holds since a EA and +(A) = {f(x)=xEA} The other inclusion i.e. f-1 (+(A)) S.A does not hold in general. So I have the following: Let fix→y be a function. Then of is one-to-one (injective) iff for every A CX, we have A = f(f(A)). pref= Assume first that of is injective, Let A CX. So we We already know that A ≤ f($(A)). -need to show that f-1 (F(A)) SA. only Let x Ef(f(A)). i.e. f(x) Ef(A) So 3 EA st. f(x) = f(a). Since of is one-to-one (injective), so f(x) = f(x) =)x=a SO REA = f(f(A)) ≤A Hence A = f(f(A)) Now, Conversely; Assume that A = f(f(A)) for all A EX. can you explain the proof (injective)
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