Why is $\exists x \forall y (x+y=0)$ false?

Question

In a textbook example about nested quantifiers, we had $P(x,y): x+y=0, x,y \in \mathbb{R}$. The examples stated the following

1.$\forall x \forall y P(x,y)$ is False, $(3+2 = 0)$

2.$\forall x \exists yP(x,y)$ is Correct ,$x + (-x) = 0$

3.$\exists x \forall yP(x,y)$ is False, $\forall y (c+y=0)$

I cannot fully understand the third example however. Where did $c$ came from? And why is $\forall y (c+y=0)$ given as example to prove that it is incorrect? Also, why does it differ so much from $\forall x \exists yP(x,y)$? I appreciate the help!

Answer 1

Just translate it into English. What is the third example saying? There exists an $x$ such that for all $y$, $x+y = 0$. It is saying that for all real numbers $y$, there is an element of the real numbers that "cancels" out $y$.

So, for example, there is an $x$ such that $x+1 = 0$ and $x+2 = 0$. This is, of course, nonsense. We know that it's not true (from your studies with the real numbers blah blah blah).

How does it differ from:

$$\forall x \exists y P(x,y)$$?

Again, look at what's being said in English. It's saying that for each real number $x$, there exists a real number $y$ such that $x+y = 0$. This is saying that every real number has a "negative", so to speak. In formal language, every real number has an additive inverse. The two statements are very different in the assertions they are making. The point is that interchanging quantifiers results in different statements entirely.

Answer 2

This is not a counterexample situation. To negate an existential quantifier, you need to prove a universal negation, which you can't do with a counterexample.

The point here is that for any specific number $x=c$, there will be values of $y$ for which $c+y\ne 0$

Answer 3

Order matters. In the third example, there is only one fixed x for all y. Since x is fixed we can say it is equal to some constant, c. In the second example, y is not fixed, but can be chosen for each x. (In this case we choose y = -x). We do not have that luxury in the third example.

Answer 4

The third proposition says that there exist a real $ x_0 $ such that for any real $ y $ , we have $$x_0+y=0$$ observe that $ x_0 $ is fixed and it doesn't depend on $ y $.

So, the condition $ x_0+y=0$ must be satisfied for all $ y\in \Bbb R $, and in particular for $ y=0$ which gives $ x_0=0$ and for $ y=-1 $ which gives $ x_0=1$.

This is a contradiction because $ x_0 $ cannot take two different values..

The difference between $$\forall x \exists y$$ and $$\exists x \forall y$$ is the fact that in the first, $ y $ depends on $ x $ and in the second $ x $ is fixed.