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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 $\forall\delta>0$.

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    $\begingroup$ You negated it incorrectly. $\endgroup$
    – Git Gud
    Jan 27, 2015 at 17:24
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    $\begingroup$ More precisely, the negation of $p\to q$ is not $p\to\neg q$, but $p\land \neg q$. $\endgroup$
    – hmakholm left over Monica
    Jan 27, 2015 at 17:27
  • $\begingroup$ I understand why i am not correct but why is $p\to q \equiv p\land \neg q$? $\endgroup$
    – usainlightning
    Jan 27, 2015 at 18:59
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    $\begingroup$ @usainlightning $\neg (p \to q) \equiv p \wedge \neg q$. Make a truth table: $$\begin{array}{cc|cc} p&q&p\to q&p \wedge \neg q \\ \hline 0&0&1&0\\ 0&1&1&0\\ 1&0&0&1\\ 1&1&1&0 \end{array}$$ $\endgroup$
    – AlexR
    Jan 27, 2015 at 19:10
  • $\begingroup$ I haven't come across truth tables before. I don't understand why say: it is raining $\implies$ sky cloudy is equivalent to (it is not raining) or (the sky is cloudy) (or both) $\endgroup$
    – usainlightning
    Jan 27, 2015 at 19:18

1 Answer 1

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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.

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    $\begingroup$ $\sqrt{x}$ is uniformly continuous even though it is not Lipschitz. Indeed nothing will work on a compact domain. You could use $\tan(x)$ on $(0,\pi/2)$, or you could use $x^2$ on $\mathbb{R}$. Also, the tricky thing that is hard to explain in words is that there is a "bad" $\varepsilon$ such that every $\delta$ has a "bad" $x$. But the $x$ can depend on $\delta$, and generally will if $f$ is continuous but not uniformly continuous. $\endgroup$
    – Ian
    Jan 27, 2015 at 17:43
  • $\begingroup$ @Ian Thanks, fixed. (+1) $\endgroup$
    – AlexR
    Jan 27, 2015 at 17:59
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    $\begingroup$ Points near $0$ aren't bad either. The problem is just near $\pi/2$. (Perhaps you were thinking of the domain being $(-\pi/2,\pi/2)$. $\endgroup$
    – Ian
    Jan 27, 2015 at 18:28
  • $\begingroup$ @Ian Yes, I did. Fixing. I'd give you +10 for that :D Feeling stupid. $\endgroup$
    – AlexR
    Jan 27, 2015 at 18:29

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