Prove |A-B|=|A|: An Uncountable Set Solution

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To prove that |A-B|=|A| where B is a countable subset of an uncountable set A, it is essential to establish a bijection between A and A-B. The discussion highlights that A-B is uncountable, which is a crucial starting point. Participants suggest using a reordering of A, positioning B as an initial segment, to facilitate the creation of a one-to-one correspondence. The concept of reductio ad absurdum is also mentioned as a potential strategy for the proof. Ultimately, the goal is to explicitly demonstrate the bijection between the two sets.
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hello I am struggled with a qustion
let B be a countable subset of uncountable set A.
Prove |A-B|=|A|
i know how to prove that A-B is uncountable
but how do i show 1:1 with A?

thanks ahead guys
 
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welcome to pf!

hello ranykar! welcome to pf! :smile:

(no, you're struggling with the question! :wink:

or being strangled by it? )​

hint: assume the opposite :wink:

(ie, that A-B is countable)
 


tnx but that's not what i was asking.
i know that A-B is uncountable.
but how do i show 1:1 with A??
 
Try doing a reordering of A , so that B is an initial segment of A.
 
I think s/he wants to set up an actual bijection between A and A\B .
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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