Uniform Convergence of a subsequence over an arbitrary interval I am getting ready for an entrance exam that is in August and I am trying to get a head start on the analysis section. I recently came across this problem and it is giving me some trouble.
Consider the sequence of functions 
$$f_k(x) := {\sin(kx)},\quad k = 1, 2,\dots ,$$ and observe
that $\sin(kx) = 0$ if $x = mπ/k$ for all integers $m$. Given an arbitrary interval $[a, b]$, show that {${f_k}$} has no subsequence that converges uniformly on $[a, b]$.
I am rather unsure how to begin this problem. I can see that this sequence doesn't converge pointwise because if $k=2n$ and $x= \pi/4$, then when $n$ is even, then $\sin(kx) = 0$ but if $n$ is odd, then $\sin(kx) = \pm 1$. Since this function doesn't converge pointwise, then it doesn't converge uniformly. I am not sure if implies that there is no subsequence that converges uniformly. Or is there a better way to approach this problem?
 A: To elaborate on the hints given in comments, suppose for contradiction that $\{f_k\}$ has a subsequence $\{ f_{k_j} \}$ that converges uniformly on $[a,b]$ to a limit function $f$.   Since the uniform limit of continuous functions is continuous, $f$ should be continuous.  However, I claim this is impossible.  
To show that the limit function cannot be continuous, one should show that there exists $\varepsilon_0 \geq 0$ so that for any $\delta >0$, there are $x,y \in [a,b]$ with $|y-x| < \delta$ and $|f(y) - f(x)| \geq \varepsilon_0$.  I claim that taking $\varepsilon_0 = 1$ works.
Let $\delta$ be fixed.  By our assumption that $\{ f_{k_j} \} \to f$ uniformly, we can choose $J$ so that if $j \geq J$, then $|f(x) - f_{k_j}(x)| \leq 1/2$ for all $x \in [a,b]$.  By taking $J$ bigger as needed, we assume that $\frac{2 \pi }{k_J} < \delta$.  This guarantees that any interval of width $\delta$ contains an entire period of $f_{k_J}$.  Can you see how to pick $x$ and $y$ to make $|f(y) - f(x)| \geq 1$?
