Proving sequence of functions has no convergent subsequence

Question

Define the following sequence of functions in $[0,1]$: $$f_n(x)=\begin{cases} (-1) &,0\leq x\leq1/n \\(-1)^{k+1} &,k/n<x\leq \frac{k+1}{n},k=1,...,n-1 \end{cases} $$ Prove there's no convergent subsequence. My attempt was: it's easy to check that in order for the subsequence to be convergent, it needs to be constant at every $x$ after some $N_x$. In order for $f_{n_k}(1/n)$ to be constant, every $n_k$ needs to be congruent modulo $2n$, for every $n$ (also easy to check). But that just implies that the $n_k's$ need to be increasingly really big... How do I go from here? I suspect I need to be using irrationals somehow...

Answer 1

Suppose $f_{n_k}$ is a subsequence. For each natural number $m$ let $A_m$ be the set of $x \in [0,1]$ such that there are $k > m$ and $j$ even with $j-1 < n_k x < j$ (implying $f_{n_k}(x) = 1$), and similarly $B_m$ with "even" replaced by "odd". $A_m$ and $B_m$ are unions of open intervals, therefore open sets, and are dense (given an open interval $J$ of length $\epsilon$, take $n_k$ with $k > m$ large enough that $1/n_k < \epsilon$, and we find that $J$ contains points of both $A_m$ and $B_m$). By the Baire category theorem, $C = \bigcap_m A_m \cap \bigcap_m B_m$ is dense; the subsequence fails to converge at members of $C$.