# Help plz ! we say that 1) x\inuuu_(n\inN)A_n if x\inA_n for some n\inN and 2)x\innnn_(n\inN)A_n if x\inA_n for all n\inN prove or dispprove nnn_(n\inN)A_n\sube uuu_(n\inN)A_n for any index set N ?

Jun 21, 2018

True.

#### Explanation:

Let's call the intersection of all sets $A$ and their union $B$.

We want to prove that $A$ is a subset of $B$. By definition, this happens when every element of $A$ is also an element of $B$, whereas the contrary is not necessarily true.

So, we need to show that

$x \setminus \in A \setminus \implies x \setminus \in B$

We only need to recall the definition, and everything gets really simple: if we assume that $x \setminus \in A$, it means that $x \setminus \in {A}_{n}$ for all $n \setminus \in N$. On the other hand, in order to $x \setminus \in B$, it is sufficient that $x \setminus \in {A}_{n}$ for at least one $n \setminus \in N$.

So, if we assume that $x$ belongs to all the sets ${A}_{n}$, is it true that it belongs to at least one of the ${A}_{n}$? Well, of course it is.

On the other hand, if we know that $x$ belongs to at least one of the ${A}_{n}$, can we conclude that it belongs to all the ${A}_{n}$? Not necessarily.

This is enough to prove that the set given by the intersection is a subset of the set given by the union.