real analysis - Disprove uniform convergence of sequence of functions $f_n: mathbb{R} to mathbb{R}^2$, $f_n(x) = (sinfrac{x}{n}, cosfrac{x}{n})$

I have already found the pointwise limit of $f_n(x) = (0, 1)$.

I have a theorem that states "Let $D\subset\mathbb{R}^q$, and $D$ compact. Let $f, f_n:D\to\mathbb{R}^p$ and $f_n$ continuous for all $n\in\mathbb{N}$. Then $f_n$ converges uniformly to $f$ if and only if $\lim_{n\to\infty}\left\lVert f_n - f \right\rVert_D = 0$."

So, following an example from my professor, I let $f = (0, 1)$ and found that $\lim_{n\to\infty}\left\lVert f_n - f \right\rVert = \lim_{n\to\infty}\left\lVert (\sin\frac{x}{n}, \cos\frac{x}{n}) - (0, 1) \right\rVert = \lim_{n\to\infty}\left\lVert (\sin\frac{x}{n}, \cos\frac{x}{n} - 1) \right\rVert = \lim_{n\to\infty} \sqrt{\sin^2\frac{x}{n} + (\cos\frac{x}{n} - 1)^2} = \sqrt{0 + (1-1)^2} = 0$

Shouldn't this prove that in fact the function DOES converge uniformly? I'm supposed to prove that it does not. What am I missing here?

Answer

What you have proved is pointwise convergence, not uniform convergence. Suppose the convegence is uniform. Then there must be an integer $m$ such that $\|(\sin (\frac x n),\cos (\frac x n))-(0,1)\| <\frac 1 2$ for all $x \in \mathbb R$ for all $n \geq m$. Take $x=\frac {m\pi} 2$ and $n=m$ to get a contradiction.