Equivalent statements not giving equivalent negations

Question

Consider the statement $\forall \epsilon >0 \exists y\ldots.$ Negating it gives $\exists \epsilon>0 \forall y\ldots.$ I understand this.

$\forall \epsilon:\epsilon>0 \exists y\ldots.$ is equivalent to the first statement, but negating it gives $\exists\epsilon:\epsilon\leq 0 \forall y\ldots.$

What is my error? It makes sense why first one is right, so the first statement has to be different from last statement, but why is it different isn't $\epsilon>0=\epsilon:\epsilon>0 ?$

Any answer or reference to a website is appreciated.

Answer 1

The negation of 'All of these are such and so' is 'Some of these are not such and so'

So note: the 'of these' part is still the same.

Of course! Quantifiers range over some domain. If not all objects of that domain have some property, then some objects of that very domain lack that property.

All that a restricted quantifier does is to further restrict that domain, and so what you get is that if not all objects of that restricted domain have some property, then some objects of that very same restricted domain lack that property.

Applied to your case: to say that it is not true that for all $\epsilon > 0$ we have [blah blah blah] is to say that for some $\epsilon > 0$ we have not [blah blah blah]. In both cases, the restricted domain is all $\epsilon > 0$

Answer 2

Actually the “equivalent”**statement you made does not make any sense(it is not well formed), really it should be: $\forall \epsilon( \epsilon>0 \implies \exists …)$ negating this we get $\exists \epsilon (\epsilon>0 \land \forall …)$ which I hope you will agree is the negation of the original statement.

** I am referring to this statement that you made: $\forall \epsilon:\epsilon>0 \exists y\ldots.$

It might also help to remember that $\forall \epsilon >0 (blah)$ is defined as $\forall \epsilon (\epsilon >0\to blah)$

Answer 3

Since your "$\forall x:x>0\;P(x)$" (equivalent to the more usual $\forall x>0\;P(x)$) means $$\forall x\left(x>0\implies P(x)\right),$$ its negation is $$\exists x\left(x>0\land\neg P(x)\right),$$ also writen $\exists x>0\;\neg P(x).$