# finding a set which is not equinumerous with series of sets

#### issacnewton

##### Member
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

#### Plato

##### Well-known member
MHB Math Helper
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
Ok, Plato I will try working on it. I think induction would be the way to go .........

#### Plato

##### Well-known member
MHB Math Helper
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
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.