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)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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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