Show that there exists a sequence of functions $\{f_n:[0,1]\to\mathbb C\}$ satisfying:
1) $f_n\to0$ pointwise; 2) $\gamma_nf_n\not\to0$, for all $\gamma_n\in\mathbb C$ such that $\gamma_n\to\infty$.
I'm asked to show by cardinality. I know that the cardinality of all complex sequences converging to $0$ is $c=\aleph$, and the cardinality of all complex functions on $[0,1]$ is $f=2^c>c$. But how can I prove the original claim?
Hint: if you can prove that the set $\Gamma = \{ \gamma_n \subset \mathbb{C} : \gamma_n \to \infty \}$ has cardinality $\mathfrak{c}$, then you have a bijection $b : [0,1] \to \Gamma$. Then define $f_n(x)=\frac{1}{(b(x))_n}$ (choosing arbitrarily when $(b(x))_n=0$).