Negation of "$\forall x >0, \exists n \in \mathbb{N}$ such that $1/n  x$" Given the statement, "$\forall x >0, \exists n \in \mathbb{N}$ such that $1/n <x$ and $n > x$".
I concluded that the negation of this statement would be, "$\exists x \leq 0, \forall n \in \mathbb{N}$ such that $1/n \geq x$ or $n \leq x$".
The correct answer is, "$\exists x > 0, \forall n \in \mathbb{N}$ such that $1/n \geq x$ or $n \leq x$"
The difference between my answer and the correct one simply being, "$\exists x > 0$" instead of "$\exists x \leq 0$". Why does the "$>$" not become "$\leq$"?
 A: I think the safest way to use $\forall$ and $\exists$ is with an explicit set element relation: the negation of
$$\forall x\in E, P(x)$$
is
$$\exists x\in E, \lnot P(x)$$
In your case the statement is (with an explicit set): $\forall x\in (0, \infty) ...$. The negation is $\exists x\in (0, \infty),...$
Edit: To answer more completely, I think it is the safest way but I don't think it is absolutely necessary to do so, because authors like Bourbaki for example use quantifiers without this. Nevertheless, the negation of "for all x satisfying property Q, property P holds" is "there exists an x satisfying property Q but not property P". The set could be replaced by an implication:
$$\forall x, x\in E \Rightarrow P(x)$$
which negation is
$$\exists x, \lnot(x\in E \Rightarrow P(x))$$
and this is the same as
$$\exists x, x\in E \text{ and } \lnot P(x)$$
In your case, the first statement could be written
$$\forall x, x > 0 \Rightarrow P(x)$$
It works because the negation of $(A\Rightarrow B)$ is $(A\text{ and }\lnot B)$, but it is a convoluted way to write things.
A: Let's look at a simpler example.  Suppose $A$ is some set of integers, and consider the statement $$\forall x\in A,\text{ $x$ is prime.}$$
In other words, all elements of $A$ are prime.
What's the negation of this?  Well, if not all elements of $A$ are prime, that means some number in $A$ is non-prime.  Using quantifiers, that's
$$\exists x\in A\text{ such that $x$ is not prime.}$$
Notice in particular that the condition $x\in A$ on the quantifier did not get negated: it stayed the same!  This makes sense, because we are talking about elements of $A$ the whole time.  Numbers that aren't in $A$ are irrelevant to the truth of the original statement, so we don't want to talk about them when forming the negation; instead, we still want to talk only about elements of $A$.
The same thing is going on in your example, just with $x>0$ being the condition on the quantifier instead of $x\in A$.
A: When negating a restricted quantifier, you do not negate the restriction.   That is implicitly handled.  $$\begin{align}\neg \forall x{\in}Q~P(x) &{~\iff~\neg\forall x~[x\in Q\to P(x)]\\~\iff~\exists x~\neg[x\in Q\to P(x)]\\~\iff~\exists x~[x\in Q\wedge\neg P(x)]}\\\neg \forall x{\in}Q~P(x) &~\iff~ \exists x{\in}Q~\neg P(x)\\[2ex]\neg\exists x{\in}Q~P(x)&~\iff~\forall x{\in}Q~\neg P(x)\end{align}$$ 
