# Meaning of : Set is closed under finite intersections and arbitrary unions

I have been working through "Set theory for working mathematician" and near the end of chapter about real numbers there is a small bit of topology.

Namely the natural topology $\tau$ on euclidean space $\mathbb{R}^n$ is defined as set of all open balls in $\mathbb{R}^n$ and then it says that such a set is closed under finite intersections and arbitrary unions.What does it mean?

Also what does it mean that some set is closed under some operation?

"closed under finite intersections" means that if $A_1,A_2,\ldots, A_k$ are each in the set, then their mutual intersection $A_1\cap A_2\cap \cdots \cap A_k$ is in the set. However this must be a finite collection of elements, i.e. $k<\infty$.

"closed under arbitrary unions" means something stronger. If $\mathcal{A}$ is a collection of sets, i.e. $\mathcal{A}=\{A_1, A_2,\ldots\}$, then the union of all of them is again in the set, i.e. $\cup \mathcal{A}=A_1\cup A_2\cup\cdots$ (the dots may conceal uncountably many elements of $\mathcal{A}$, and uncountably many unions in $\cup \mathcal{A}$).

In particular, if $\mathcal{A}$ is finite, this means closed under finite unions -- however $\mathcal{A}$ need not be finite.

Note that the natural topology is not closed under arbitrary intersections, for example let $A_i=(-\frac{1}{i},\frac{1}{i})$; the intersection of all of them is the single point $0$, which is not open.

• The use of $\ldots$ in the second paragraph may suggest that $\mathcal A$ is countable. However, $\mathcal A$ can be much larger, in fact arbitrary – Hagen von Eitzen Oct 10 '14 at 14:28
• Why doesn't $\bigcup_{i=1}^{\infty} (0, 1 - 2^{-i}) = (0, 1]$, which is not open? – wchargin Oct 10 '14 at 14:29
• Because $1$ is not in the union, since it is not in any of the sets. – vadim123 Oct 10 '14 at 14:31
• Indexing $\mathcal A = \{A_1, A_2, \ldots\}$ is at best misleading, because it strongly implies that $\mathcal A$ is countable. But the arbitrary unions of a topological space are not restricted to countable unions. – MJD Oct 10 '14 at 14:37
• 1. Nobody said the set of indices is countable. 2. I explicitly pointed out that the collection need not be countable. 3. My notation adds clarity for readers confused by "collection of sets". – vadim123 Oct 10 '14 at 14:40

A set being closed under some operation means that you can perform this operation, and the result will still be part of your set.

As an example: You have your set $X$, which is closed under finite intersections. This means that for every finite set $\left\{x_1,...,x_n\right\}\subseteq X$ we have $\cap_{i=1}^n{x_i}\in X$