I was wondering what is the cardinality of $\mathbb{N} \times \mathbb{R}$. My guess is that it is the same as $\# \mathbb{R}=c$ but I haven't been able to prove it.
I was thinking that one can easily find a surjective function $f: \mathbb{N} \times \mathbb{R} \rightarrow \mathbb{R}$. For example taking $f(x, y) = y$. This would prove that $\# \mathbb{N} \times \mathbb{R} \geq \# \mathbb{R}$. But is there a bijective function between the two sets?
If $f: \mathbb{N} \times \mathbb{Z} \rightarrow \mathbb{Z}$ is a bijection, you could use $(n, x) \mapsto f(n, \lfloor x \rfloor) + \{x\}$ to get the bijection you want.
For $f$, you could have $f(n,0) =-n$, $f(n, z)= (z+n)^2+(z-n)$ for $z>0$ and $f(n, z)= (n-z)^2-(z+n)-1$ for $z<0$.