Difference between arbitary collection and finite collection

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An arbitrary collection of open sets is just that: any collection of open sets at all, finite or infinite, with no restriction save that the sets be open. If you change finite to arbitrary in the second statement, it becomes false. For example, in $\Bbb R$ the sets $(-x,x)$ for $x>0$ are all open, but their intersection is $\{0\}$, which is not open: $\bigcap_{x>0}(-x,x)=\{0\}$. Here are two more examples.

• For each $q\in\Bbb Q$ let $U_q=\Bbb R\setminus\{q\}=(\leftarrow,q)\cup(q,\to)$; this is an open set. But $\bigcap_{q\in\Bbb Q}U_q=\Bbb R\setminus\Bbb Q$, the set of irrational numbers, which is definitely not an open set.

• For each $n\ge 2$ let $U_n=\left(\frac1n,1+\frac1n\right)$; then \begin{align*}\bigcap_{n\ge 2}U_n&=\left(\frac12,\frac32\right)\cap\left(\frac13,\frac43\right)\cap\left(\frac14,\frac54\right)\cap\ldots\\ &=\left(\frac12,1\right]\;.\end{align*} You may have to think a bit about that intersection to see that it really is $\left(\frac12,1\right]$.

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Updated on August 08, 2022