TITLE: Understanding an example of "for all" and "for some" usage in statements.
QUESTION [6 upvotes]: I'm reading "Analysis I" by Tao and reviewing an appendix chapter on logic. In there he gives an example on how "for all x" is usually much stronger than just saying "for some x":

"$6<2x<4$ for all $3<x<2$" is vacuously true, but
"$6<2x<4$ for some $3<x<2$" is false.

I can see how the first statement is vacuously true: the hypothesis "for all $3<x<2$" is false as there is no $x$ that satisfies both $3<x$ and $x<2$, meaning the statement is true by default. 
But I don't see how the second statement works. I presume to say that statement 2 is false, one has to show that the implication ("$6<2x<4$") is false when the hypothesis ("for some $3<x<2$") is true. But is there a $x$ such that "$3<x<2$" is true? I'm obviously missing something here so clarification on this would be appreciated.

REPLY [12 votes]: A hypothesis of the form "For some ..." automatically assumes that there is such a .... In symbols, "$Q$ is true for some $x$ such that $P$" is written "$\exists x[P(x)\wedge Q(x)]$." This is in contrast with "$Q$ is true for all $x$ such that $P$," which is formalized as "$\forall x[P(x)\implies Q(x)]$." There isn't really a great short phrase for "$\exists x[P(x)\implies Q(x)]$." Roughly speaking, any time a sentence begins with an existential statement ("for some . . ."), it can't be vacuously true. "Some pink elephant conquered Canada" can only be true if there is some pink elephant in the first place, after all.
So the second statement means

There is an $x$ such that $3<x<2$ and $6<2x<4$,

which is false since in particular there is no $x$ such that $3<x<2$.
By the way, note that this is a case where the formalism may not match up with the natural language (for me it does, but this is ultimately a subjective question). But that's what it means, formally.