Let $X= [0,1]^{[0,1]} = \{ f : f: [0,1] \rightarrow [0,1]\}$ be endowed with the topology $T$ of pointwise convergence.
For every $n \in \mathbb{N}$, consider the function $f_n \in X$ defined by the following formula: $f_n(t)=10^nt- \lfloor10^nt \rfloor$, for all $t\in [0,1]$.
Here $\lfloor x \rfloor$ is the floor function.
Show that int is not possible to find some indices $n(1) < n(2)< \cdots < n(k) < \cdots$ in $\mathbb{N}$ such that the sequence $(f_{n(k)})_{k=1}^{\infty}$ is pointwise convergent.
Note that we have $f_n(0)=f_n(1)=0$ for any $n \in \mathbb{N}$.
I tried to prove the by contradiction. Suppose $f_{n(k)} \rightarrow f$. Then there exists some $N>0$ so that for all $n(k)>N$, $|f_{n(k)} -f|< \epsilon$. But from here, I couldn't get any where to find a contradiction.
Any help will be appreciated.