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I've just begun an introductory text on probability. In the first chapter there is a preview/review of set theory, which I am not familiar with. One of the examples has me a little confused.

I do not understand "how to read" the mathematics due to my unfamiliarity with set theoretic notation. I presume the "big cup" and "big cap" are similar to $\Sigma$ and $\prod$. Thus, I am supposing that rather than addition and multiplication, a union or intersection is done each iteration.

Anyways, from the book:

For example, if for every positive integer $n$, we are given a set $S_n$, then

$\bigcup^\infty_{n=1} S_n = S_1 \cup S_2 \cup ... = \{x \mid x \in S_n $ for some $ n \} $

   and

$\bigcap^\infty_{n=1} S_n = S_1 \cap S_2 \cap ... = \{x \mid x \in S_n $ for all $ n \} $

I can't even tell if $S_3$, for example, $=\{3\}$ or something else. Very confused.

If I can't tell the the composition of any $S_n$ where $n$ is known and finite, I do not know how to test union/intersection of infinite such sets.

Help is appreciated, and especially thoroughness, for this easy question.

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A set $S$ is a collection of elements. When you are referring to a set $S_n$, you are simply referring to to some set which can be identified by $n$. Thus, $S_0$, $S_1$, $S_2$, ... $S_n$ are just different sets which can be identified by the number index $n$.

Saying that some element $x$ is in $\bigcup\limits_{n=1}^{\infty}S_{n}$ means that $x$ is in at least one of the sets $S_0$, $S_1$, $S_2$, ... . Therefore, for some $n \in \mathbb{N}$, $x$ is in the set $S_n$. Rewritting this we have $$x\in \bigcup\limits_{n=1}^{\infty}S_{n} \Longleftrightarrow x \in \{y \mid y \in S_n \text{ for some }n \in \mathbb{N}\}$$

Which is equivalent as $$\bigcup\limits_{n=1}^{\infty}S_{n} = \{y \mid y \in S_n \text{ for some }n \in \mathbb{N}\}$$

Saying that some element $x$ is in $\bigcap\limits_{n=1}^{\infty}S_{n}$ means that $x$ is in all of the sets $S_0$, $S_1$, $S_2$, ... . Therefore, for all $n \in \mathbb{N}$, $x$ is in the set $S_n$. Rewritting this we have $$x\in \bigcap\limits_{n=1}^{\infty}S_{n} \Longleftrightarrow x \in \{y \mid y \in S_n \text{ for all }n \in \mathbb{N}\}$$

Which is equivalent as $$\bigcap\limits_{n=1}^{\infty}S_{n} = \{y \mid y \in S_n \text{ for all }n \in \mathbb{N}\}$$

You should take a look at this book if you want to have a good introduction to set theory (and other stuff). It is very well written and straightforward.

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  • $\begingroup$ I find your answer well written and helpful. Special thanks for the link to that book--I will certainly be using it. $\endgroup$ – d0rmLife Nov 30 '13 at 16:56
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With $S_n$ they mean a family of sets. Suppose we have that $S_n$ equals the interval open interval $(0,n)$. By taking the unions $\bigcup^\infty_{n=1} S_n = (0,1) \cup (0,2) \cup (0,3) \cup ... = (0,\infty)$. When taking the intersections we will get the following: $\bigcap^\infty_{n=1} S_n = (0,1) \cap (0,2) \cap (0,3) \cap ... = (0,1)$, because the open interval $(0,1)$ is in every set. I hope this example will make things more clear.

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  • $\begingroup$ Ummm... your second example is wrong. $0$ is in none of those intervals, but the interval $(0, 1)$ is in all of them. $\endgroup$ – apnorton Nov 30 '13 at 1:32
  • $\begingroup$ oh my bad, meant to use closed intervals. $\endgroup$ – user112167 Nov 30 '13 at 1:33

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