Want to understand a set notation I cannot picture $$\bigcap \{X\subseteq A| F(X)\subseteq X\}$$ in my head. 
Perhaps I could better understand it in symbolic logic. How do you write it in symbolic logic ? 
 A: Edited. It's unlikely that $F$ is the induced function of a map $F\colon A\to A$, because in that case the given intersection would necessarily be empty. Edited to reflect that.
Okay, presumably you have a set $A$.
Now, remember that whenever you have a set $A$, you also have the power set of $A$, which is a set whose elements are the subsets of $A$. The power set of $A$ is often represented something like $\mathcal{P}(A)$, or $\mathfrak{P}(A)$.
For example, if $A=\{1,2,3\}$, then 
$$\mathcal{P}(A) = \Bigl\{ \emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\Bigr\},$$
because those are exactly the subsets of $A$.
So, what we have is that a set $X$ satisfies $X\in\mathcal{P}(A)$ if and only if it satisfies $X\subseteq A$.
When you write:
$$\bigl\{ X\subseteq A\mid \text{...stuff...}\}$$
you are saying that you want to consider those elements of $\mathcal{P}(A)$ which satisfy ...stuff...
You probably also have a function $F\colon\mathcal{P}(A)\to\mathcal{P}(A)$. If $X\in\mathcal{P}(A)$, then $F(X)$ is also a subset of $A$, so it makes sense to ask whether $F(X)\subseteq X$. For example, let's take the $A$ I take above, and define a function $F\colon\mathcal{P}(A)\to\mathcal{P}(A)$ as follows:
$$\begin{array}{rcl}
\emptyset & \longmapsto & \{1,3\}\\
\{1\} &\longmapsto & \{2\}\\
\{2\} &\longmapsto & \emptyset\\
\{3\} &\longmapsto & \{1,2,3\}\\
\{1,2\} &\longmapsto & \{1\}\\
\{1,3\} &\longmapsto & \{1,2,3\}\\
\{2,3\} &\longmapsto & \{1,3\}\\
\{1,2,3\} & \longmapsto & \{3\}.
\end{array}$$
So, suppose we had that $A$ and that $F$. What would be the set
$$\Bigl\{ X\subseteq A \mid F(X)\subseteq X\Bigr\}\quad?$$
Looking at the description above, we see that the sets that are mapped to subsets of themselves are: $\{2\}$, $\{1,2\}$,  and $\{1,2,3\}$.
Finally, what does it mean to take the intersection of a single set? If $R$ is a set (whose elements are sets), then $\cap R$ is the result of intersecting all the elements of $R$; that is:
$$\bigcap R = \Bigl\{ x \mid x\text{ is in }S\text{ for every element }S\in R\Bigr\}.$$
So, for my example above, we would have:
$$\begin{align*}
\bigcap\Bigl\{X\subseteq A\mid F(X)\subseteq X\Bigr\} &= \bigcap\Bigl\{ \{2\},\ \{1,2\}\, \{1,2,3\}\Bigr\}\\
&= \{2\}\cap\{1,2\}\cap\{1,2,3\}\\
&= \{2\}.
\end{align*}$$

It's also possible that your $F$ is actually a function $A\to A$. Remember that whenever you have a function $F\colon R\to S$, you automatically get another function $\overline{F}\colon \mathcal{P}(R)\to\mathcal{P}(S)$ (called the "direct image") which is defined by
$$\overline{F}(X) = \{ F(x)\mid x\in R\}.$$
(You also get another function, $\underline{F}\colon \mathcal{P}(S)\to\mathcal{P}(R)$, called the "inverse image", which is defined by 
$$\underline{F}(Y) = \{ x\in R \mid F(x)\in Y\},$$
but this does not matter now). The direct image function is often, by abuse of notation, written just $F$ instead of $\overline{F}$. For example, going back to $A=\{1,2,3\}$, suppose you define $F\colon A\to A$ by $F(1)=2$, $F(2)=2$, and $F(3)=1$. Then the direct image function would have the following values:
$$\begin{align*}
F(\emptyset) &= \emptyset,\\
F(\{1\}) &= \{F(1)\} = \{2\}\\
F(\{2\}) &= \{F(2)\} = \{2\}\\
F(\{3\}) &= \{F(3)\} = \{1\}\\
F(\{1,2\}) &= \{F(1),F(2)\} = \{2,2\} = \{2\}\\
F(\{1,3\}) &= \{F(1), F(3)\} = \{2,1\} = \{1,2\}\\
F(\{2,3\}) &= \{F(2),F(3)\} = \{2,1\} = \{1,2\}\\
F(\{1,2,3\}) &= \{F(1),F(2),F(3)\} = \{2,2,1\} = \{1,2\}.
\end{align*}$$
In this case, you would have that the set
$$\Bigl\{ X\subseteq A\mid F(X)\subseteq X\Bigr\}$$
consists exactly of $\emptyset$, $\{2\}$, $\{1,2\}$, and $\{1,2,3\}$, so that
$$\begin{align*}
\bigcap\Bigl\{ X\subseteq A\mid F(X)\subseteq X\Bigr\} &= \bigcap\Bigl\{ \emptyset,\ \{2\},\ \{1,2\},\ \{1,2,3\}\Bigr\}\\
&= \emptyset\cap\{2\}\cap\{1,2\}\cap\{1,2,3\}\\
&=\emptyset.
\end{align*}$$

The reason I don't think (upon reflection) that this is the situation you have is that if your $F$ is the direct image function of some map that goes from $A$ to $A$, then the expression you have will always be the empty set (try to prove that, it's pretty easy).
A: I assume $F$ is a function from $A$ to $A$. Rasmus's explanation is correct, the notation
$$\bigcap \{X\subseteq A| F(X)\subseteq X\}$$ 
refers to the intersection (i.e. $\bigcap$) of all subsets $X$ of $A$ (i.e. $X\subseteq A$) such that (this is the symbol $\mid$; see set-builder notation) the image of $X$ under $F$ is contained in $X$ (i.e. $F(X)\subseteq X$).
If $F$ is a function from $A$ to $A$, this set is empty, as $\varnothing\subseteq A$ is a subset of $A$ for which $F(\varnothing)\subseteq\varnothing$.

yunone makes the point in the comments on this post that $F$ might be a function from ${\mathcal P}(A)$ to ${\mathcal P}(A)$. In that case, the notation
$$\bigcap \{X\subseteq A| F(X)\subseteq X\}$$ 
refers to the intersection (i.e. $\bigcap$) of all subsets $X$ of $A$ (i.e. $X\subseteq A$) such that the set $F(X)\subseteq A$ that $F$ sends $X$ to is contained in $X$ (i.e. $F(X)\subseteq X$).
A: The notation $\bigcap T$, for a non-empty $T$, is the set $\{x\mid \forall a\in T(x\in a)\}$.
That is, all the elements which are in all the elements of $A$ (which are sets themselves).
If $T$ is finite, e.g. $T=\{A,B,C\}$ then $\bigcap T = A\cap B\cap C$. If it is infinite, then it is a bit shorter and clearer to use $\bigcap T$ instead.
In your specific case, it means the intersection of all $X\in\mathcal P(A)$ with a certain property.
A: $$\bigcap \{X\subseteq A| F(X)\subseteq X\}$$
Is the intersection of all sets $X$ such that $X$ is a subset of $A$ and $F(X) \subseteq X$.
So in terms of logic:
$$x \in \left( \bigcap \{X\subseteq A| F(X)\subseteq X\} \right) \Leftrightarrow (\forall X | X \subseteq A \land F(X) \subseteq X )( x \in X )$$
