I got a question about the cardinal of set of all function of $\mathbb N$ $\to$ $\mathbb R$. I think that the cardinal number of the set is C because the range of all function is R which the cardinal number of it is C, but I dont knowhow to prove it.
Thank you,
Let $B$ be the set of binary sequences. I take it as known that $B\sim \mathbb R,$ where $\sim$ means "same cardinality". Then your set $\mathbb R^{\mathbb N} \sim B^{\mathbb N}.$
An element of $B^{\mathbb N}$ can be viewed as an infinite matrix $(a_{mn}),$ with $m$ denoting the row, and $n$ denoting the column. We can associate with this matrix the binary sequence $a_{11},\,a_{21},\,a_{12},\, a_{31},\,a_{22},\,a_{13}, \dots,$ where we've used Cantor's trick of moving along the cross diagonals of the matrix. This establishes a 1-1 correspondence between $B^{\mathbb N}$ and $B.$ Thus $\mathbb R^{\mathbb N} \sim \mathbb R.$
You're correct that it's $\mathfrak{c}$, but the cardinality of the range isn't enough to conclude that. The most straightforward way to prove a question of cardinality is to construct a bijection; a good shortcut to this is to use Cantor-Bernstein and construct two injections.
The first is easy: for each $r \in \mathbb{R}$, the constant function $f(n) = r$ is a function from $\mathbb{N}$ to $\mathbb{R}$. This gives us an injection from $\mathbb{R}$ into the set of functions.
The second is trickier, but not too much so: Given a function $f$, take the binary expansions of the various $f(n)$ and "interleave" them to form a single number with all the digits of the $f(n)$ in it. I won't get into specifics, but think about the proof that $\mathbb{Q}$ and $\mathbb{N}$ have the same cardinality. That gives you a single real corresponding to each function, so we have an injection from the set of functions into $\mathbb{R}$.
In general, when dealing with questions of cardinality, you should always be thinking about bijections. If the word "bijection" does not appear in your thought process, then your solution is almost certainly wrong.
