How to solve the paradox in the negation to definition of limit? While learning a proof of the statement "A continuous function on a closed bounded interval is uniformly continuous", I faced with a hypothesis that the book had considered. 
(Proof by contradiction of convergence, i.e. negation to "for any $\epsilon>0$ there exists some $N$ s.t. for all $n\ge N$, $|f(u_n)-f(v_n)|<\epsilon$") ... so there is some $\epsilon>0$ such that $|f(u_n)-f(v_n)|\ge \epsilon$ for all $n\in \mathbb N$. 
Why contradiction to the first statement is the second one? I mean, contradiction to the first statement should be "for some $\epsilon>0$ there never exists some $N$ s.t. for all $n\ge N$; how this is equivalent to the book's statement (the second one)? (especially why $n\in \mathbb N$?)
EDIT - According to @par's answer, The negation of the statement
$$
\forall\epsilon>0:\exists N\in\mathbb{N}\colon\forall n\geq N\colon\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|<\epsilon
$$
is
$$
\exists\epsilon>0\colon\forall N\in\mathbb{N}\colon\exists n\geq N\colon\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|\geq\epsilon.
$$
which I am convinced by the negation; but it is intrinsically inconsistent:
The negation means $\exists\epsilon>0\colon\exists n\geq 1 \ \text {and}\ \exists n\geq 2\ \text {and}\ \dots \colon\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|\geq\epsilon$. Considering them together (some kind of intersection) $\infty\ne n\in \emptyset$, so again $\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|<\epsilon$, even in negation! Am I right? 
I would appreciate a clear explanation.    
 A: Sentence 
$$\forall \epsilon >0\exists \delta>0 :\forall  |x-y|< \delta :|f(x)-f(y)|< \epsilon $$
Negation
$$\exists \epsilon>0 \forall \delta>0 : \exists |x-y|< \delta , |f(x)-f(y)|\geq \epsilon $$
consider $\delta = 1/n$
so you will find $x_n, y_n$ such that $|x_n-y_n|<1/n$ and $|f(x_n)- f(y_n)|>\epsilon$.
Since you are in a compact set, you can consider without loss of generality (suffices to take a subsequence) that $x_n \to x^*$ now $y_n \to x^*$ but the continuity in $x^*$ implies that
$$|f(x_n) - f(y_n)|\leq |f(x_n)- f(x^*)| + |f(x^*) - f(y_n)| \to 0 $$
this is a contradiction with the assumption $|f(x_n) - f(y_n)| > \epsilon$
A: 
The interpretation is not correct. Instead of using the quantor $\forall$ and state
\begin{align*}
\exists\epsilon>0\colon\forall N\in\mathbb{N}\colon\exists n\geq N\colon\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|\geq\epsilon\tag{1}
\end{align*}
you could consider the countably infinite set of statements
\begin{align*}
\exists\epsilon>0\colon\exists n\geq 1\colon&\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|\geq\epsilon\\
\exists\epsilon>0\colon\exists n\geq 2\colon&\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|\geq\epsilon\tag{2}\\
\exists\epsilon>0\colon\exists n\geq 3\colon&\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|\geq\epsilon\\
&\ldots
\end{align*}
  which is logically equivalent with (1).

Informally the meaning of the quantor $\forall$ in (1) is whenever you take a value $N$ in $\mathbb{N}$ there exists an $n$ fulfilling the corresponding statement in (2).
A: The negation of the statement
$$
\forall\epsilon>0:\exists N\in\mathbb{N}\colon\forall n\geq N\colon\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|<\epsilon
$$
is
$$
\exists\epsilon>0\colon\forall N\in\mathbb{N}\colon\exists n\geq N\colon\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|\geq\epsilon.
$$
Certainly you are right in saying that it is not necessarily true
for all $n$.
