Uniform convergence of supremum If a sequence $\{f_n\}$ converges uniformly to a limit $f$ on the domain $D$, then the sequence $\{M_n\}$, with $M_n = \sup_{x} |f_n(x)-f(x)| $, converges to zero.

So what I thought was since $\{f_n\}$ converges uniformly, given $\epsilon <0$ there exists an $N$ so that for $N<n$, $|f_n(x)-f(x)|<\epsilon $ for all $x$. Since $N$ isn't dependent of our choice of $x$, it follows that the supremum converges to zero (since the supremum is just a specific choice for x).
But the answer states that we can find a $N$ so that $|f_n(x)-f(x)|<\frac{\epsilon}{2} $ and therefore $\sup_{x} |f_n(x)-f(x)| \leq \frac{\epsilon}{2} < \epsilon$.
I don't understand how the supremum can be equal to $\frac{\epsilon}{2}$ since the supremum is just a special case of the normal sequence. But my reasoning is probably all wrong. Could somebody help me out?

If e.g. $f$ is prescribed by $x\mapsto0$ and $f_n$ by $x\mapsto\frac1{n}\arctan x$ then: $$\sup_{x\in\mathbb R}|f_n(x)-f(x)|=\frac{\pi}{2n}$$
while for every $x\in\mathbb R$: $$|f_n(x)-f(x)|<\frac{\pi}{2n}$$
So there is no specific $x_0\in\mathbb R$ with $|f_n(x_0)-f(x_0)|=\sup_{x\in\mathbb R}|f_n(x)-f(x)|$ here.

In order to prove for some $\epsilon>0$ that $$\sup_{x\in\mathbb R}|f_n(x)-f(x)|<\epsilon$$  it is used that some $N$ exists with $n>N$ implying that for every $x$: $$|f_n(x)-f(x)|<\frac12\epsilon$$
Based on that it can be concluded that $$\sup_{x\in\mathbb R}|f_n(x)-f(x)|\leq\frac12\epsilon<\epsilon$$
and that was intended to prove.
