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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?

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"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.

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    $\begingroup$ 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 $\endgroup$ – Hagen von Eitzen Oct 10 '14 at 14:28
  • $\begingroup$ Why doesn't $\bigcup_{i=1}^{\infty} (0, 1 - 2^{-i}) = (0, 1]$, which is not open? $\endgroup$ – wchargin Oct 10 '14 at 14:29
  • $\begingroup$ Because $1$ is not in the union, since it is not in any of the sets. $\endgroup$ – vadim123 Oct 10 '14 at 14:31
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    $\begingroup$ 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. $\endgroup$ – MJD Oct 10 '14 at 14:37
  • $\begingroup$ 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". $\endgroup$ – vadim123 Oct 10 '14 at 14:40
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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 $\cap_{i=1}^n{x_i}\in X$

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