MHB Finding a set which is not equinumerous with series of sets

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The discussion centers on finding an infinite set that is not equinumerous with any set \( A_n \), defined recursively with \( A_1 = \mathbb{Z^+} \) and \( A_{n+1} = \mathcal{P}(A_n) \). The user considers \( \mathbb{Z^+} \times \mathbb{R} \) as a potential candidate but acknowledges the need for induction to prove its non-equinumerosity. Another participant suggests considering the union of all \( A_n \) sets, prompting further exploration. The main goal is to show that no \( A_n \) is equinumerous with the union of all \( A_n \), using a proof by contradiction regarding the relationship between a set and its power set. The conversation emphasizes the complexity of cardinality and the properties of infinite sets.
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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)
 
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IssacNewton said:
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 }~? $
 
Ok, Plato I will try working on it. I think induction would be the way to go ...
 
IssacNewton said:
Ok, Plato I will try working on it. I think induction would be the way to go ...
Well $\forall n$ we know that $\left\| {A_n } \right\| \prec \left\| {A_{n + 1} } \right\|$.
 
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.
 

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