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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.

Xander Henderson
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mathsdummie1234
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    This is where abbreviations are hurting you. The negation of $\forall\epsilon\in A\exists y\in B(\varphi(\epsilon,,y))$ is $\exists\epsilon\in A\forall y\in B(\neg\varphi(\epsilon,,y))$; the negation of $\forall\epsilon\exists y\in B(\epsilon\in A\to\varphi(\epsilon,,y))$ is $\exists\epsilon\forall y\in B(\epsilon\in A\land\neg\varphi(\epsilon,,y))$. There's no way for $\epsilon\notin A$ to come out of this. – J.G. Feb 27 '23 at 10:14
  • Firstly: Do not change significantly your question. Given answers are for old version. Instead of this set new one. Secondly: in answers is already said, to use clean logical expressions. $\forall \varepsilon (\varepsilon>0)$ is not same with $\forall \varepsilon >0 (\varepsilon>0)$. – zkutch Feb 27 '23 at 10:45
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    I strongly recommend not using this ambiguous notation "$\forall \varepsilon ,:, \varepsilon > 0$". It looks like you're saying: "For all epsilon, epsilon is negative." which is obviously false. I would say "on this simple example we understand what you mean, but on a more complicated example it will be extremely confusing" but actually, this whole question shows that you are already confused with this example. So, don't use this notation. Either write "$\forall \varepsilon > 0,\ldots$", or "$\forall \varepsilon,, (\varepsilon > 0 \implies ...)$", but not something inbetween. – Stef Feb 27 '23 at 21:19

3 Answers3

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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 $\epsilon > 0$

Bram28
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  • I kind of understand it but my question was more behind the technical reasoning behind it. Of course if I say it works for all human this statement is apparently false if there one human for whom it does not work. I was looking for a techincal reasoning(/proof)... – mathsdummie1234 Mar 01 '23 at 09:35
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    @mathsdummie1234 Many of the comments and other answers address this. You are confusing the notation of a restricted quantifier with the notation of an unrestricted quantifier followed by a predicate. That is: if we are to read $\forall x : x > 0$ as a restricted quantifier, we don’t even have a claim or statement … we just have the beginning of one: “ for all $x$ that are greater than 0 ….. “. So we cannot negate it like you keep trying to do. If it is an unrestricted quantifier, however, then it is a claim: “All $x$ are greater than $0$ … and its negation is indeed $\exists x : x \leq 0$ – Bram28 Mar 01 '23 at 12:31
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    @mathsdummie1234 In sum: you have seen how you negate statements with unrestricted quantifiers, but you try to apply that to a statement that contains restricted quantifiers, and where unfortunately your particular notation mixes things up so that instead of negating the actual claim you end up ‘negating’ the restriction, and that’s where things go completely wrong. – Bram28 Mar 01 '23 at 12:34
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    The good news is that you can convert a restricted quantifier into an unrestricted one, and then you can apply the quantifier negation laws as usual. Again, several other people already explained this to you, but just to be sure: A claim that involves a restricted quantifier like $\forall x: x > 0 \ P(x)$ (which involves an actual claim: $P(x)$ in addition to just a restriction $x > 0$) can be written as a claim using an unrestricted quantifier as follows: $\forall x (x > 0 \to P(x))$. Likewise, $\exists x > 0 \ P(x)$ can be written as: $\exists x (x > 0 \land P(x))$. (...continued) – Bram28 Mar 01 '23 at 13:23
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    So if you want to negate $\forall x : x > 0 \ P(x)$ and want to use the equivalence for unrestricted quantifiers that $\neg \forall x \ P(x) \Leftrightarrow \exists x \neg P(x)$, you'll first have to convert $\forall x : x > 0 \ P(x)$ to $\forall x (x > 0 \to P(x))$. Negating that, we get $\neg \forall x (x > 0 \to P(x))$, and now we can apply the equivalence to get $\exists x \neg (x > 0 \to P(x))$. And since $\neg (P \to Q)$ is equivalent to $P \land \neg Q$, this gives us $\exists x (x > 0 \land \neg P(x))$, and that we can write back as a restricted $\exists x : x > 0 \ \neg P(x)$ – Bram28 Mar 01 '23 at 13:30
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    And so you see that as general equivalence principle for restricted quantifiers we have that $\neg \forall x : x \in S \ P(x) \Leftrightarrow \exists x : x \in S \ \neg P(x)$, where the $x \in S$ is the restriction. So, as I said in my Answer: Keep your hands off the restriction! – Bram28 Mar 01 '23 at 13:32
  • really appreciate your time – mathsdummie1234 Mar 02 '23 at 10:37
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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 $\exists x>0\;\neg P(x).$

Anne Bauval
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  • Thanks but what I mean basically is this −(∀:>0)=∃:≤0 That is my whole problem, why this doesnt hold. With "-" i mean negating – mathsdummie1234 Feb 27 '23 at 10:43
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    You are really unclear: you wrote (in a comment to another answer) "What i mean [with (∀:>0)] is for all epsilon with the property of epsilon being greater than 0" so: what you mean with (∀:>0)P() is: "for all epsilon with the property of epsilon being greater than 0, P(epsilon) is satified". And this is exactly what I rewrote formally, and then negated. – Anne Bauval Feb 27 '23 at 10:50
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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 $\forall \epsilon (\epsilon >0\to blah)$

Vivaan Daga
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