Why is $\bigcap_{x\in\mathbb{R}}\mathbb{N}=\mathbb{N}$? $\bigcap_{x\in\mathbb{R}}\mathbb{N}=\mathbb{N}$
The generaldefinition of Intersection/Union is, that you have a indexset and an Array of sets $(A_i)$ that can be directly assigned to the index:
Let $I$ be the indexset then $\bigcap_{x\in I}\ A_i := \{x|\forall_{i \in I} x \in A_i \}  $ If $\mathbb{R}$ is the indexset $\mathbb{N}$ has no such Index. For example I can take $\sqrt{2}$ for my x, then there is no set with such an index that I can assign my x to it. It doesn't matter that I picked an irrational number, I also could have Chosen $2$ as my Index. The Problem is that we have $\mathbb{N}$ but not $\mathbb{N}_x$ written 
 A: The expression you're asking about is somewhat similar to the definition of a constant function, say
$$
f(x) = 12.
$$
Usually the definition of a function depending on $x$ should have an $x$ somewhere on the right hand side, but in this case that is not happening. In particular, if we then define something based on $f$, say
$$
n = \sum_{i = 1}^{10} f(i) = \sum_{i=1}^{10} 12,
$$
you see that you get a summation where the index of the summation does not appear in the summand.
The same thing is happening in your case. We could define $A_x = \mathbb N$. Note that this is just a function definition: we could also have written it $A(x) = \mathbb N$. Then we could consider
$$
A:=\bigcap_{x \in \mathbb R} A_x = \bigcap_{x \in \mathbb R} \mathbb N = \mathbb N.
$$
Such an expression is a bit silly -- without context, it is trivial but confusing, while in context it will generally be clear what's going on.
A: What the notation$$\bigcap_{x\in\mathbb R}\mathbb N\tag1$$means is the intersection of a family of sets indexed by $\mathbb R$ such that each of its elements is $\mathbb N$. In other words, it's the family $\{A_x\,|\,x\in\mathbb{R}\}$ in which, for each $x\in\mathbb R$, $A_x=\mathbb N$. So, yes, $(1)=\mathbb N$.
