Why is this intersection of sets empty? An example I saw online tells me this following set of arbitrary intersections are empty:
If we define a set $B_m =\{m,m+1,m+2,...\}$ where $m\in \mathbb{N}$, then $\bigcap_{m\in\mathbb{N}}B_m=\emptyset$$?$
Because I just cant get this idea correct as for instance:
$$B_1\cap B_2=B_2$$
$$B_2\cap B_3=B_3$$
$$B_3\cap B_4=B_4$$
$$.........$$
$$B_{m-1} \cap B_m=B_m $$
If my understanding is correct then by above:
$$\bigcap_{m\in\mathbb{N}}B_m=B_1\cap B_2 \cap ... \cap B_m\ne \emptyset$$
This tells me that the intersection are indeed non - empty, so how $\bigcap_{m\in\mathbb{N}}B_m=\emptyset$ hold true$?$
 A: An intersection of a family of sets consists of any element that is contained in all those sets.
In other words:
$$
x\in\bigcap_{m\in\Bbb N}B_m\iff \text{for all }m\text{, we have }x\in B_m
$$
There is no natural number that is an element of all the $B_m$. Which is to say, there is no natural number (or anything else) that can be contained in the intersection. This makes the intersession empty.
A: The correct definition is:
$$
\bigcap_{m \in \mathbb{N}} B_m = B_1 \cap B_2 \cap B_3 \cap \cdots
$$
in which the RHS does not terminate. If it's non-empty, say $n \in \bigcap_{m \in \mathbb{N}} B_m$, then $n \in B_m$ for all $m \in \mathbb{N}$. However, this is not possible, as $B_{n+1} = \{n+1,n+2,\dots\}$, so $n \notin B_{n+1}$.
A: Other question, can you give an element that is in the intersection?
For $n\in\bigcap_{m\in\mathbb{N}} B_m$ there has to be a $n\in B_m$ for every $m\in\mathbb{N}$.
But $n\notin B_{n+1}$ for example, so no such $n$ can exist.
Just make clear how these sets look.
$B_1=\{1,2,3,4,\dotso\}$
$B_2=\{2,3,4,\dotso\}$
$B_3=\{3,4,\dotso\}$ and so on.
Eventually every element will be sorted out, or more specifically for every element you can give easily a set that does not contain this element, as shown above.
A: Note that $B_m=B_1\cap...\cap B_m\ne\cap_{n\in\Bbb N}B_n=B_1\cap B_2\cap ...$.
The former is a finite intersection, the latter is not.
Next note that for every $m\in\Bbb N,m\notin B_{m+1}\implies m\notin\cap_{n\in\Bbb N}B_n$. Thus the intersection is empty.
A: Assume $\cap B_{n \in \mathbb{N}} \not=\emptyset$.
Then there is a $k \in \mathbb{N}$ s.t.
$k \in B_n$ for all $n \in \mathbb{N}$, e.g.
$ k \in B_1, B_2, .....$.
Consider
$B_{k+1}=$ {$k+1,k+2,......$};
$k \not \in B_{k+1}$, a contradiction.
