Prove $ \sin ^{2} \frac{\pi}{x} $not uniform convergence Negation of uniform convergence
$$
(\exists \varepsilon>0)(\forall N)(\exists x \in S)(\exists n>N)\left(\left|f_{n}(x)-f(x)\right| \geq \varepsilon\right)
$$
Let $f_{n}(x)=\left\{\begin{array}{l}0 \quad(x<1 /(n+1)) \\ \sin ^{2} \frac{\pi}{x} \quad(1 /(n+1) \leq x \leq 1 / n) \\ 0 \quad(1 / n<x)\end{array}\right.$
I need to prove $\left\{f_{n}\right\}$ converges to 0, but not uniformly. I've already proved it converges to 0.
My attempt:
This is not uniform convergence as $\exists \varepsilon=1, \forall n \in \mathbb{N}$ $\exists x=\frac{2}{2 n+1},\left|f_{n}(x)-0\right|=$ $\sin ^{2}\left(\frac{\pi}{\frac{2}{2 n+1}}\right)=\sin ^{2}\left(\frac{\pi(2 n+1)}{2}\right)=1$
However, Negation of uniform convergence should be $(\forall N\in\mathbb{N})$$(\exists n>N)$. I wonder if directly saying $\forall n \in \mathbb{N}$ is equivalent to $(\forall N\in\mathbb{N})(\exists n>N)$. Thanks
 A: To highlight the difference between $\forall n \in \mathbb{N}$ and $(\forall N\in\mathbb{N},\exists n>N)$, let $P(n)$ be the proposition "this problem solved at time $n$".

*

*$\forall n \in \mathbb{N}, P(n)$ means that it's already solved at the very beginning $n = 1$.

*$(\forall N\in\mathbb{N},\exists n>N: P(n))$ means that for every moment $N$, you think that "it can be  solved later OR it's already solved".  It might not be solved at the beginning $n = 1$.

Evening though (1) looks stronger than (2), in the context of proving uniform convergence, up to re-indexing of the sequence of functions $\{f_n\}_n$, it's possible to rephrase (2) as (1).  Note that in the definition of uniform convergence, $n$ is a function of $N$, so you may write $(n_N)_N$.  To skip "$\exists n > N$", you may

*

*replace $\left|f_{n}(x)-f(x)\right|$ with $\left|f_{n_N}(x)-f(x)\right|$

*replace $\{f_n\}_n$ with its subsequence $\{f_{n_N}\}_N$.  You may rename it to $\{g_N\}_N$ to make things clear.

Then the whole thing becomes
$$(\exists \varepsilon>0)(\forall N)(\exists x \in S)(\exists n>N)\left(\left|g_{N}(x)-f(x)\right| \geq \varepsilon\right).$$
As $n$ has no role in the last part of the above statement, you can safely cross out "$\exists n > N$", giving you a statement in (1) form.
