Why does $x \in A_5$ imply that $\displaystyle x\in\bigcup_{i=1}^{\infty} A_i$? If it is the case that $x \in A_5$, why does it necessarily imply that $\displaystyle x\in\bigcup_{i=1}^{\infty} A_i$? It seems that $\displaystyle x\in\bigcup_{i=1}^{\infty} A_i$ is bigger than any element so it seems strange to imply this. 
 A: By definition, $x\in\bigcup_{i=1}^\infty A_i$ means that $x$ is an element of at least one of the $A_i$. This is true since $x$ is an element of $A_5$.
For example, if $A_i=\{i\}$, then $\bigcup_{i=1}^\infty A_i$ would equal $\{1,2,3,\dots\}$. If $x$ were an element of $A_5=\{5\}$, then $x$ would definitely be an element of $\bigcup_{i=1}^\infty A_i=\{1,2,3,\dots\}$.
A: Because it is necessarily so by definition of union. The notation $$\bigcup_{i=1}^{\infty} A_i$$ means infinite union of all $A_i$. Think of $\bigcup_{i=1}^{\infty} A_i$ as a bucket. To fill this bucket, you rip open each $A_i$ and dump its contents inside of $\bigcup_{i=1}^{\infty} A_i$.
So if $x \in A_5$ and all of $A_5$'s contents have been dumped into the bucket $\bigcup_{i=1}^{\infty} A_i$ then $x$ must be in the bucket as well.
A: $\displaystyle x\in\bigcup_{i=1}^{\infty} A_i$ implies that there is at least one $i \in I$ so that  $x \in A_i$. $\bigcup_{i=1}^{\infty} A_i$ has all the elements of $A_i$ for all $i \in I$ (in this case $I$ is the set of natural numbers). And of course $A_5 \subset \bigcup_{i=1}^{\infty} A_i$ therefore all $x \in A_5$ are in $ \bigcup_{i=1}^{\infty} A_i$ 
