I need to prove that:
$$\mathbb{R}^{\mathbb{N}} \sim \mathcal{P}( \mathbb{N})$$
We know that
1. $\mathbb{N} \times \mathbb{N} \sim \mathbb{N}$
2. And that $A^{B^C} \sim A^{B \times C}$
And so I start the proof:
I start by proving that $\mathbb{R}^{\mathbb{N}} \sim \mathbb{R}$ by:
We know that $\mathbb{R} \sim \mathcal{P}(\mathbb{N}) \sim 2^{\mathbb{N}}$ (Because $| \mathbb{R} | = \aleph$ and $|\mathcal{P}(\mathbb{N})| = |2^{\mathbb{N}}| = 2^{\aleph_0})$
And so: $\mathbb{R}^{\mathbb{N}} \sim (2^{\mathbb{N}})^{\mathbb{N}}$
And by #2 above we know that $(2^{\mathbb{N}})^{\mathbb{N}} \sim 2^{\mathbb{N} \times \mathbb{N}}$
by #1 we know that:
$2^{\mathbb{N} \times \mathbb{N}} \sim 2^{\mathbb{N}} \sim \mathbb{R} \sim \mathcal{P}(\mathbb{N})$
So we are done.
I am completely unsure if it's true or not, I would like to get a proof review from you, I will be so thankful..
Thank you have a great day!
Your proof is fine. But we can prove something stronger if we use a third fact for inequalities: any $\kappa$ with $2\le\kappa\le2^{\aleph_0}$ satisfies$$2^{\aleph_0}\le\kappa^{\aleph_0}\le(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0^2}=2^{\aleph_0}\implies\kappa^{\aleph_0}=2^{\aleph_0}.$$Your proof strategy already handles the case $\kappa=2^{\aleph_0}$ and then implies the more general case, as we can show $\kappa_1\le\kappa_2\implies\kappa_1^\lambda\le\kappa_2^\lambda$. However, you'll see the above argument in inequalities in many sources, as it doesn't require checking a special case first.