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...
Equivalence of two sets
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elementary-set-theory
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1You're right. My answer only dealt with the case where $A$ and $B$ were disjoint. Use Joriki's answer. – 2012-04-10
2 Answers
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$A\cup B=A\cup(B\setminus A)$ 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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0thank you, forgive me I didn't see your post before. – 2012-04-10
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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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0@JanosAudron: Yes I noticed. We answered just about the same time, so his came in as I posted mine. – 2012-04-10