TITLE: Cardinal equalities: $\aleph_0^\mathfrak c=2^\mathfrak c$
QUESTION [8 upvotes]: $\aleph_0^\mathfrak c=2^\mathfrak c$
This is a question on my homework, and I am not sure how to show they equal each other, I tried Cantor-Schroeder-Bernstein but got stuck, any help would be great.

REPLY [3 votes]: Here is an alternate "more elementary" proof that $\aleph_0^\mathfrak c \leq 2^\mathfrak c$:
We use that $c$ is the cardinality of both $\mathbb{R}$ and $(0,1)$. 
Let $f: (0,1) \rightarrow \mathbb{N}$ be any function.
We define $F_f := \{ x+ f(x) | x \in (0,1) \}$.
Then it is easy to check that any function $f$ is uniquely identified by $F_f$ and thus
$f \to F_f$ is a 1-1 function from $\mathbb{N}^{(0,1)}$ to ${\mathcal P}(\mathbb{R})$.