2

I am looking for a mathematical example that illustrates why

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

Can anybody give me any advice on how to approach such a problem?

Oliphaunt
  • 933
jmoore00
  • 619
  • 4
    It is not that (as in the title) existential quantifiers must be in a certain order. That doesn't matter at all. A block of universal ones can be permuted at will as well. It is that existential and universal quantifiers do not commute. – Ross Millikan Mar 19 '17 at 03:28

3 Answers3

8

HINT: Take $P(x, y)$ to be "$x<y$," and think about whether there is a biggest number . . .

Noah Schweber
  • 245,398
6

In words, said out loud, they sound very similar. But note that in the first case, we are saying that each $x$ has its own personal $y$ such that $P$ occurs. I really wish people would say $y_x$ in this case. It would clarify things so well.

In the second case, we are saying that there is a $y$ which covers any $x$.

A good example is continuity as opposed to uniform continuity. I suggest looking up those and studying the differences carefully. For example. The function $f(x)=1/x$ is continuous on $(0,\infty)$, but it is not uniformly continuous there.

The difference between continuity and uniform continuity is precisely the concern you have.

The Count
  • 3,620
5

Here is a simple example:

$$\forall x \exists y(x =y)$$ That is, every object $x$ has some object $y$ which is equal to it. This is true, since $x = x$.

$$\exists y \forall x(x=y)$$ That is, there is some object out there which everything equals. I.e., there is only one object in this entire mathematical theory. Definitely false.

As The Count has already said, in the first case, $y$ can depend on $x$. In the second case, it is picked first, so it has to be independent of $x$.

Paul Sinclair
  • 43,643