Proving that $\mathbb{R} \cong \mathbb{R}^{\mathbb{N}}$ I am trying to prove that the real numbers have the same cardinality as the set of real sequences.
I'm treating this as proving that $\mathbb{R} \cong \mathbb{R}^{\mathbb{N}}$ since the set of real sequences can be identified with the set of functions $\mathbb{N} \to \mathbb{R}$. Indeed, given such a sequence $(s_n)$ in $\mathbb{R}$, we identify it with the mapping $s: \mathbb{N} \to \mathbb{R}$ such that $s(n) = s_n$.
I don't know of a way to prove this directly, so my only thought is to use the Schröder–Bernstein theorem. One direction is rather obvious. Given $c \in \mathbb{R}$, we can define the constant function (or the constant sequence) $f: \mathbb{N} \to \mathbb{R}$ sending $f(n) = c$ for all $n$. So $\mathbb{R} \hookrightarrow \mathbb{R}^{\mathbb{N}}$.
It suffices to find an injection in the opposite direction $\mathbb{R}^{\mathbb{N}} \hookrightarrow \mathbb{R}$, but I can't figure out how to do. I could perhaps identify $\mathbb{R}$ with $\mathcal{P}(\mathbb{N})$, but that would require proving that $\mathbb{R} \cong \mathcal{P}(\mathbb{N})$, wich I know to be true, but also don't know how to prove.
Any help or guidance would be appreciated.
 A: For simplicity, we'll inject $(0,1)^\mathbb{N} \hookrightarrow (0,1)$. This suffices since we can use $\arctan$ or something to squish the whole real line into this interval.
What's the idea, then? Well, we know there are infinitely many primes. And each prime has infinitely many powers. So we can code up the digits of our primes by putting them in the prime power digits. This is a famous trick, and is worth keeping in the back of your mind.
Explicitly, say we have a sequence of reals:

*

*$0.d_0^1 d_1^1 d_2^1 d_3^1 d_4^1 \ldots$

*$0.d_0^2 d_1^2 d_2^2 d_3^2 d_4^2 \ldots$

*$0.d_0^3 d_1^3 d_2^3 d_d^3 d_4^3 \ldots$

*$\vdots$
Here $d^i_j$ is the $j$th digit in the binary expansion of the $i$th number. Let's agree to always end with an infinite string of $0$s rather than $9$s when we have that ambiguity.
How, then do we inject this into $(0,1)$? Well remember the prime trick! We have all the space in the world as long as we put things really far apart.
We'll make a new number by specifying its digits. Throughout, $p_n$ will be the $n$th prime.

We put $d_j^i$ in the $p_j^i$th position. We put $0$ (say) everywhere else.

So the digits of our first number $0.d_0^1 d_1^1 d_2^1 d_3^1 d_4^1 \ldots$ get spread out to the power-of-two positions in our new number.
$d_0^1$ is the $2$nd digit. $d_1^1$ is the $4$th digit. $d_2^1$ is the $8$th digit, and so on.
Similarly, we put $d_0^2$ as the $3$rd digit, $d_1^2$ as the $9$th, $d_2^2$ as the $27$th, and so on.
Notice we have infinitely much room to store these infinitely many sequences! Moreover (perhaps counterintuitively) most of our digits are $0$!
I'll leave it to you to work out that this is really an injection, but that should be somewhat clear by construction. If we start with $2$ different sequences $s_n$ and $t_n$, than some fixed $s_i \neq t_i$. So they disagree on a digit...

I hope this helps ^_^
A: It's easy to see that $|\mathbb{R}^{\mathbb{N}}|=|(0,1)^{\mathbb{N}}|.$ Then define an injection $\iota:(0,1)^{\mathbb{N}}\to (0,1)$ that takes the first digit of the first number in the sequence, then the first two unpicked digits of the first two numbers in the sequence, then the first three unpicked digits of the first three numbers, so that
$$
\iota(a_1a_2a_3...,b_1b_2b_3...,c_1c_2c_3...,...)=a_1a_2b_1a_3b_2c_1...
$$
Then obviously $|(0,1)|\leq |\mathbb{R}|.$
