Show that if #A < #B then #P(A) < #P(B). f : A → B. g : P(A) → P(B). Proof. Assume #A < #B. Then there exists a(n) To show #P(A)< #P(B), we need to provide a(n) For each Se P(A), we define g(S) = {f(x) | x E S}. For each x E S, we have f(x) E __. So g(S). B. Thus g is a mapping from to To show is we assume g(S) g(T) for some S, T e P(A) and show S T below. We have x E S= f(x) E » f(x) E = 3y E T, = f(y) = E T. Similarly, we have Hence S T.
Show that if #A < #B then #P(A) < #P(B). f : A → B. g : P(A) → P(B). Proof. Assume #A < #B. Then there exists a(n) To show #P(A)< #P(B), we need to provide a(n) For each Se P(A), we define g(S) = {f(x) | x E S}. For each x E S, we have f(x) E __. So g(S). B. Thus g is a mapping from to To show is we assume g(S) g(T) for some S, T e P(A) and show S T below. We have x E S= f(x) E » f(x) E = 3y E T, = f(y) = E T. Similarly, we have Hence S T.
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:Show that if #A < #B then #P(A) < #P(B).
f : A → B.
g : P(A) → P(B).
Proof. Assume #A < #B. Then there exists a(n)
To show #P(A)< #P(B), we need to provide a(n)
For each Se P(A), we define g(S) = {f(x) | x E S}.
For each x E S, we have f(x) E __. So g(S).
B. Thus g is a mapping from
to
To show
is
we assume g(S)
g(T) for some S, T e P(A) and show S
T below.
We have x E S= f(x) E
» f(x) E
= 3y E T,
= f(y) =
E T.
Similarly, we have
Hence S
T.
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