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A is Uncountable and B is Countable. Does $\left | A \right | = \left | A \cup B \right |$ ?

I was thinking build a countable set of distinct ordered elements $A_{N}= \left \{ \right. a_{N}: N\epsilon \mathbb{N}\left. \right \} $ where $A_{N}\subset{A}$.

Then define $ f: A_{N}\rightarrow{B}, f(a_{N})=b_{N}$.

Next define $ f: A\setminus A_{N}\rightarrow{A}, f(a_{n})=a_{n-N},\forall n>N$.

So we have two bijections and thus by taking the union $\left |A_{N} \cup A\setminus A_{N}\right | = \left | {B}\cup A \right | $.

Does this work out?

Andrés E. Caicedo
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Mscntst
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2 Answers2

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You are on the right track. We are playing Hilbert's hotel here. Once we have a countably infinite subset $A_{\Bbb N}=\{\,a_n\mid n\in\Bbb N\,\}$ of $A$, we find a bijection of $A$ with $A\setminus A_{2\Bbb N}$, where $A_{2\Bbb N}=\{\,a_{2n}\mid n\in\Bbb N\,\}$ is another countably infinite subset of $A$: $$a\mapsto\begin{cases}a&\text{if }a\notin A_{\Bbb N}\\ a_{2n+1}&\text{if }a=a_n\end{cases} $$ This allows us to "stow away" the countably infinite set $B$.

If $B$ is countable and finite, use $$a\mapsto\begin{cases}a&\text{if }a\notin A_{\Bbb N}\\ a_{n+|B|}&\text{if }a=a_n\end{cases} $$ instead to "make room" for $B$.

Hagen von Eitzen
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  • What do you mean by "countable and finite"? Countable is defined as far as I know to be infinite. Thank you :) – Mscntst Sep 24 '16 at 20:21
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Your definition of $A_N$ is odd. $N$ can't be both an index inside the definition of the set, and a parameter outside. You should write $$A'=\{a_N;N\in\mathbb{N}\}$$ Then your second map does not make sense since you never define an integer $N$. On top of it, it seems that neither map takes values outside of $A'$.

Olivier Moschetta
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