Is $\exists x\in U,~\forall y\in U~~ f(x,y)$ and $\forall y\in U,~\exists x\in U~~ f(x,y)$ logically equivalent? Does order matter?

Question

anonymous · Answer

Suppose $f(x,y)$ is the statement that $x<y$.

To say "$\exists~x\in U$ such that $\forall~y\in U$, you have $x<y$" means there is a number $x$ that is smaller than any number $y$ in the set $U$.

To say "$\forall~y\in U,~\exists~x\in U$, you have $x<y$" means that for any number $y$ there is something smaller than $y$.

As a counter-example, consider the set of real numbers $\mathbb{R}$. Let $x\in\mathbb{R}$. No matter what $x$ is, $x$ is not smaller than any other number $y$. For instance, $x=0$ is smaller than every $y>0$, but this is not true if $y<0$ and thus not true *for any* $y\in\mathbb{R}$.

The second statement is true in this case, if $U=\mathbb{R}$ then there is always a real number $x$ smaller than any real number $y$.

To summarize, yes. Order matters for different logical quantifiers.

geerky42 · Answer

Very good explanation. Thanks!