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finding a set which is not equinumerous with series of sets

issacnewton

Member
Jan 30, 2012
61
Hi
Let \( A_1=\mathbb{Z^+} \) and \( \forall n\in \mathbb{Z^+}\) let \( A_{n+1}=\mathcal{P}(A_n) \)

I have to come up with an infinite set which is not equinumerous with \( A_n \) for any \( n\in \mathbb{Z^+} \).
Clearly \( \mathbb{R}\) will not fit the bill since \( \mathbb{R}\;\sim\; A_2 \). So I was thinking of
the set \( \mathbb{Z^+}\times \mathbb{R} \). I will need to use induction here. But does my test function seem
right ?

Thanks
(Emo)
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Hi
Let \( A_1=\mathbb{Z^+} \) and \( \forall n\in \mathbb{Z^+}\) let \( A_{n+1}=\mathcal{P}(A_n) \)
I have to come up with an infinite set which is not equinumerous with \( A_n \) for any \( n\in \mathbb{Z^+} \).
Have you considered $\displaystyle\bigcup\limits_n {A_n }~? $
 

issacnewton

Member
Jan 30, 2012
61
Ok, Plato I will try working on it. I think induction would be the way to go .........
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Ok, Plato I will try working on it. I think induction would be the way to go .........
Well $\forall n$ we know that $\left\| {A_n } \right\| \prec \left\| {A_{n + 1} } \right\|$.
 

issacnewton

Member
Jan 30, 2012
61
So what I have to prove is that

\[ \forall\; n\in \mathbb{Z^+}\left [ A_n \nsim \bigcup_{n\in \mathbb{Z^+}} A_n\right ] \]

I figured that this is can be easily done by assuming negation and getting a contradiction that some set is equinumerous with its power set.