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Please I need help proving the following question: Given an infinite set A and a countable set B, prove that ${A}\cup{B}\sim{A}$. I already proved the case when ${A}\cap{B}=\emptyset$ and when ${B}\subset{A}$. I need help proving what happens when ${B}\nsubseteq{A}$ and ${A}\cap{B}\neq\emptyset$ 10x in advance...

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You're right. My answer only dealt with the case where $A$ and $B$ were disjoint. Use Joriki's answer. –  David Mitra Apr 10 '12 at 18:55
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2 Answers 2

up vote 3 down vote accepted

$A\cup B=A\cup(B\setminus A)$ and $A\cap(B\setminus A)=\emptyset$. Since $B$ is countable, $B\setminus A$ is countable (with any enumeration of $B$ inducing an obvious enumeration of $B\setminus A$), so you can apply the case you've already treated.

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thank you, forgive me I didn't see your post before. –  JanosAudron Apr 10 '12 at 19:04
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Because $B$ is countable, you can find a bijection with $\mathbb N$, so call the elements $b_i$. Similarly there is a countable subset $C$ of $A$, for which there is a bijection with $\mathbb N$, call them $c_i$. Take $b_i \to c_{2i}, c_i \to c_{2i+1},$ rest of $A$ to itself.

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Thanks @Ross, however what if ${A}\cap{B}\neq\emptyset$? does your mapping handle that case? If not how wold you define it? –  JanosAudron Apr 10 '12 at 18:53
    
@JanosAudron: I think you are right that if $A \setminus B$ is finite this doesn't work. –  Ross Millikan Apr 10 '12 at 19:49
    
joriki had provided a good solution, take a look –  JanosAudron Apr 10 '12 at 22:28
    
@JanosAudron: Yes I noticed. We answered just about the same time, so his came in as I posted mine. –  Ross Millikan Apr 10 '12 at 22:58
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