Question

\[\neg\left[(\forall \epsilon > 0)(\exists \delta > 0)(\forall x)[|x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon]\right]\]

Answer 1

how to begin

Answer 2

Ugh! Epsilon Delta....

Answer 3

Well, slowly move the negation bar into the expression, except every time it passes a universal quantifier, change it to an existential,and every time it passes an existential quantifier, change it to universal :D

Answer 4

\[\neg\left[(\forall \epsilon > 0)(\exists \delta > 0)(\forall x)[|x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon]\right]\]\[\qquad\qquad\downarrow\]\[(\exists \epsilon > 0)(\forall \delta > 0)(\exists x)\left[\neg[|x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon]\right]\]

Answer 5

now how do i distribute the negation across the rest/

Answer 6

Negation of something of the form
p ⇒ q
is
p • ¬q

Answer 7

in this case
p is
|x - a| < delta
and
q is
|f(x) - f(a)| < epsilon

Answer 8

\[(\exists \epsilon > 0)(\forall \delta > 0)(\exists x)\left[\neg[|x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon]\right]\]\[\downarrow\]\[(\exists \epsilon > 0)(\forall \delta > 0)(\exists x)[|x - a| < \delta \wedge\neg[ |f(x) - f(a)| < \epsilon]]\]

Answer 9

Then, negation of
<
is
≥

Answer 10

\[\downarrow\]\[(\exists \epsilon > 0)(\forall \delta > 0)(\exists x)[|x - a| < \delta \wedge |f(x) - f(a)| ≥ \epsilon]\]

Answer 11

That's it, I can't go further than that...

Answer 12

Thank you terenzreignz

Answer 13

:D