rewrite in a mathematical format I have many sets containing three values like {1,−2,-5}. I want to write in mathematical form to filter set where there are only elements with same sign and also none of them is zero like {-1,−9,-5} or {4, 6, 2}. I tried this, is it correct?
$$
F = \{ s \in S \mid \forall a \in s, a \neq 0 \land \forall a \in s, a > 0 \vee \forall a \in s, a < 0 \}
$$
EDIT: 
$S$ is collection of all sets (s) and "a" denotes set member. 
 A: What you described is, put into words:

$F$ is the set of all sets which do not contain zero and which have all elements of the same sign.

It is unclear what $S$ is, althoutgh I imagine it should be either $\mathbb R$ or $\mathbb Z$. Without determining what $S$ is, your definition is not clear.
It is also worth noting that sets of different sizes will be in $F$. For example, if $S=\mathbb R$, then $\mathbb N\in S$ and $\emptyset\in S$

Afteer your edit, it seems your notation is correct.
Right now, I understand you already have some selection of sets in $S$, and you just want those that contain elements of the same sign (and do not contain zero). If that is what you tried to achieve, then you did it.
A: You need some brackets in your condition -- it is not clear whether you mean
$$ \bigl[(\forall a \in s, a \neq 0) \land (\forall a \in s, a > 0)\bigr] \vee (\forall a \in s, a < 0) $$
or
$$ (\forall a \in s, a \neq 0) \land \bigl[(\forall a \in s, a > 0) \vee (\forall a \in s, a < 0)\bigr] $$
In this particular case the two readings happen to work out to the same, but that is more by accident than design (namely because the $a\neq 0$ condition is superfluous; it is implied by either of $a<0$ or $a>0$).
But in general $(P\land Q)\lor R$ is not the same as $P\land(Q\lor R)$, and many people would read the unbracketed "$P\land Q\lor R$" as $(P\land Q)\lor R$, which I strongly suspect was not what you wanted to write here.
A: You stated $s\in S$, then $\forall s\in s$, but since $s$ is a number, then $a$ cannot be an element in $s$. This is one error, and I can see you've made a few.  You could simply write this is as $$F=\{(a,b,c) \ | \ (a,b,c\in\mathbb{R}^+) \ \vee \ (a,b,c\in\mathbb{R}^-)\}.$$
A: If I am interpreting what you are trying to do correctly, then you what the set of subsets of $\mathbb Z$ satisfying these properties. 
is $S$ the category Sets? how about
$F=\{s\in \mathcal{P}(\mathbb{Z}) : (\forall a \in s, a>0)\vee(\forall a \in s, a<0)\}$. 
The $a\neq 0$ is redundant by the inequalities you write for the two cases ($<,\: >$).
