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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$?$

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  • $\begingroup$ For every $m\in\Bbb N,m\notin B_{m+1}\implies m\notin\cap_{n\in\Bbb N}B_n$ $\endgroup$ – Shubham Johri Oct 27 at 6:48
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    $\begingroup$ The statement $\bigcap_{m\in \Bbb N}B_m=B_1\cap\cdots\cap B_m$ makes no sense. What is $m$ on the right hand side? $\endgroup$ – Elliot G Oct 27 at 7:01
  • $\begingroup$ The "obvious" answer which is both trivial and profound is "because it is a set with no elements". If you try to exhibit an element of the set, you may see what goes wrong. It is often useful to cut out some of the apparent complexity of a situation by going back to basic definitions. $\endgroup$ – Mark Bennet Oct 27 at 7:31
  • $\begingroup$ Aurora, if you found one of the answers below satisfactory, consider accepting it. $\endgroup$ – Shubham Johri Oct 28 at 11:34
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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.

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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 $n \notin B_{n+1}$.

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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.

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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.

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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.

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