Proving $(0,1)^{\mathbb{N}} \cong \mathbb{R}^{\mathbb{N}}$ It's straightforward to show that $(0,1) \cong \mathbb{R}$. The map $x \longmapsto \tan\left(\pi\left(x - \frac{1}{2}\right)\right)$ does the trick. This should imply that $(0,1)^{\mathbb{N}} \cong \mathbb{R}^{\mathbb{N}}$, i.e., that there are as many functions from $\mathbb{N} \to (0,1)$ as there are $\mathbb{N} \to \mathbb{R}$, but I can't figure out a suitable bijection. My second thought was that I could use Shroder Bernstein. One injection is obvious. If I have a function $f: \mathbb{N} \to (0,1)$, since $(0,1) \subset \mathbb{R}$, I can identify $f$ with a function $f': \mathbb{N} \to \mathbb{R}$ with $f'(n) = f(n)$, so its codomain is $\mathbb{R}$, but its image $(0,1)$. A bijection in the opposite direction is less obvious. I want to in some way use the fact that there exists a bijection $g: \mathbb{R} \to (0,1)$ and say "take a function $f: \mathbb{N} \to \mathbb{R}$ and define $f'$ that sends $n$ to $g(n)$, i.e., its image under the bijection $g$. But that isn't sufficiently rigorous enough, nor does it construct a well-defined function.
Any help on the approach would be appreciated.
 A: Hint: see if you can prove the more general result that, for all sets $A,B,X$, if $A \cong B$, then $A^X \cong B^X$.
Bigger hint:

 Consider what happens when you compose a general function $X \to A$ with the bijection $A \xrightarrow{\cong} B$.

A: Let $h: (0,1) \to \mathbb{R}$ be your original bijection. Given a function $f: \mathbb{N} \to (0,1)$, consider $h \circ f: \mathbb{N} \to \mathbb{R}$.  This defines a map $(0,1)^\mathbb{N} \to \mathbb{R}^\mathbb{N}$.  Do the same thing with $h^{-1}$ to get a function in the other direction.  Now show that they compose to the respective identities.
To be super explicit, the bijection $h: (0,1) \to \mathbb{R}$ induces a function
$$
h_*: (0,1)^\mathbb{N} \to \mathbb{R}^\mathbb{N}
$$
defined by $h_*(f) = h \circ f$. By the exact same logic, $h^{-1}: \mathbb{R} \to (0,1)$ induces a function
$$
(h^{-1})_*: \mathbb{R}^\mathbb{N} \to (0,1)^\mathbb{N}
$$
by $(h^{-1})_*(g) = h^{-1}\circ g$.  Now show that $h_*$ and $(h^{-1})_*$ compose to the respective identities in both directions.
