1. We'll need to establish the Schröder-Bernstein Theorem first. Assume there exists a 1 – 1 function f:X→Y and another function g : Y → X. Follow the steps to show that there exists a 1 – 1 and onto function h : X → Y and hence X ~ Y.

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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Write a rigorous proof for the following statement.
Statement: P(N)~R
follow the road map I have provided below, make sure the proof is
logically sound and that the logic is clear to the reader (not just you)
1 function f :X → Y and
1. We'll need to establish the Schröder-Bernstein Theorem first. Assume there exists a 1 –
another function g : Y → X. Follow the steps to show that there exists a 1 – 1 and onto function h : X → Y and hence
X - Y.
1.1. The strategy is to partition X and Y into components
X = AU A'
with ANA' = Ø and BNB' = Ø, in such a way that f maps A onto B, and g maps B' onto A'. Explain how this
would lead to a proof that X ~ Y.
and
Y = BUB'
1.2. Set Aj = X\g(Y) = {x EX|x¢ g(Y)} (What happens if A1 = Ø?) and inductively define a sequence of sets by
letting An+1 = g(S(A,)). Show that {A» |n EN} is a pairwise disjoint collection of subsets of X, while
{F(An) |n EN} is a similar collection in Y.
%3|
1.3. Let A = U14n and B = Uiƒ(An). Show that f maps A onto B.
%3D
1.4. Let A' =X\A and B' = Y\B. Show that
g maps B' onto A'.
2. Find a 1 -
1 function f : P(N) → R. Be careful not to break mathematics by finding a 1 – 1 function f : P(N) → N!
3. Find a 1 – 1 function f : R → P(N) and conclude that P(N) ~ R.
Transcribed Image Text:Write a rigorous proof for the following statement. Statement: P(N)~R follow the road map I have provided below, make sure the proof is logically sound and that the logic is clear to the reader (not just you) 1 function f :X → Y and 1. We'll need to establish the Schröder-Bernstein Theorem first. Assume there exists a 1 – another function g : Y → X. Follow the steps to show that there exists a 1 – 1 and onto function h : X → Y and hence X - Y. 1.1. The strategy is to partition X and Y into components X = AU A' with ANA' = Ø and BNB' = Ø, in such a way that f maps A onto B, and g maps B' onto A'. Explain how this would lead to a proof that X ~ Y. and Y = BUB' 1.2. Set Aj = X\g(Y) = {x EX|x¢ g(Y)} (What happens if A1 = Ø?) and inductively define a sequence of sets by letting An+1 = g(S(A,)). Show that {A» |n EN} is a pairwise disjoint collection of subsets of X, while {F(An) |n EN} is a similar collection in Y. %3| 1.3. Let A = U14n and B = Uiƒ(An). Show that f maps A onto B. %3D 1.4. Let A' =X\A and B' = Y\B. Show that g maps B' onto A'. 2. Find a 1 - 1 function f : P(N) → R. Be careful not to break mathematics by finding a 1 – 1 function f : P(N) → N! 3. Find a 1 – 1 function f : R → P(N) and conclude that P(N) ~ R.
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