Difference between arbitary collection and finite collection

real-analysis

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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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Idonknow

Updated on August 08, 2022

Comments

Idonknow over 1 year

In basic topology of real number ( real analysis) , there is these two theorems which states : The union of arbitrary collection of open sets is open. Another one is : The intersection of finite collection of open sets is open. Here, I would like to ask what is the difference between arbitrary and finite ? If I change finite in the second theorem, will it still hold ? If it does not hold, can anyone give some examples to illustrate ?
Idonknow over 10 years

thx for your explanation and examples. I understand now
Brian M. Scott over 10 years

@Idonknow: You’re welcome; glad it helped.
Idonknow over 10 years

Since set of irrational numbers is an open set, is the set of rational numbers also an open set ?
Brian M. Scott over 10 years

@Idonknow: The set of irrational numbers is not an open set, and neither is the set of rational numbers. Neither of them even contains an open interval.