Negation of uniform continuity The definition of uniform continuity is:
Given any $\varepsilon>0\ \exists\delta>0\ \forall x\in I \ \forall y\in I\ \left(\text{if }|x-y|<\delta\text{ then }\ |f(x)-f(y)|<\varepsilon\right)$ where $I=$ the interval on which $f$ is defined
so this means that $\delta$ must remain constant for a given $\epsilon$.
The negation statement is:
$\exists \varepsilon>0\ \forall\delta>0\ \exists x,y \in I\left(\text{if }|x-y|<\delta\text{ and}\ |f(x)-f(y)|\geq\varepsilon\right)$ where $I=$ the interval on which $f$ is defined.
I am trying to understand this logically without reverting to rules. The part which I don't understand is $\forall\delta>0$. 
 A: Let's first see how we arrive at the correct negation using predicate logic and then justify:
$$\neg \forall \epsilon > 0 \exists \delta > 0 \forall x, y \in I [ |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon] \\
\exists \epsilon > 0 \neg \exists \delta > 0 \forall x,y \in I [ |x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon] \\
\exists \epsilon > 0 \forall \delta > 0 \neg \forall x,y \in I [ |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon] \\
\exists \epsilon > 0 \forall \delta > 0 \exists x,y \in I \neg [|x-y| < \delta \Rightarrow |f(x)-f(y)|<\epsilon] \\
\exists \epsilon > 0 \forall \delta > 0 \exists x,y \in I [|x-y| < \delta \wedge \neg[|f(x)-f(y)|<\epsilon]] \\
\exists \epsilon > 0 \forall \delta > 0 \exists x,y \in I [|x-y| < \delta \wedge |f(x)-f(y)|\ge\epsilon]$$
What that means: There is some $\epsilon$ such that we can't bound the "change of $f$" around a point $x$ by $\epsilon$, no matter how close we stay to this $x$.
Visualize this with $\tan x$. If $I = (-\frac\pi2,\frac\pi 2)$ points within $\delta$ of $-\frac\pi2$ or $\frac\pi2$ will be "bad" points where the implication $|x-y|<\epsilon \Rightarrow |\tan x - \tan y| < \delta$ fails.
