From 8063dd20d5526f2ad97c3a02d848ade574ecfa09 Mon Sep 17 00:00:00 2001
From: Levi Pearson <levi.pearson@harman.com>
Date: Sat, 25 Jan 2014 16:00:39 -0700
Subject: [PATCH] More notes

---
 math/proof/notes.page | 21 ++++++++++++++++++++-
 1 file changed, 20 insertions(+), 1 deletion(-)

diff --git a/math/proof/notes.page b/math/proof/notes.page
index f24740f..686a984 100644
--- a/math/proof/notes.page
+++ b/math/proof/notes.page
@@ -236,8 +236,27 @@ $$\begin{align}
 \bigcap_{i=1}^{n} A_i &= A_1 \cap A_2 \cap \dots \cap A_n\\
 \end{align}$$
 
+For example, take the following infinite list of sets:
+
+$$
+A_1 = \{-1,0,1\}, A_2 = \{-2,0,2\}, \dots, A_i = \{-i,0,i\}, \dots
+$$
+
+We can see that:
+
 $$
-A_1 = \{-1,0,1\}, A_2 = \{-2,0,2\}, \dots, A_i = \{-i,0,i\}, \dots\\
 \bigcup_{i=1}^{\inf} A_i = \Bbb{Z} \text{ and }
 \bigcap_{i=1}^{\inf} A_i = \{0\}\\
 $$
+
+**Definition 1.8**
+
+If we have a set $A_{\alpha}$ for every $\alpha$ in some index set $I$,
+then
+
+$$\begin{align}
+\bigcup_{\alpha \in I} A_{\alpha} &= \{ x : x \in A_{\alpha} \text {
+for at least one set $A_{\alpha}$ with } \alpha \in I \}\\
+\bigcap_{\alpha \in I} A_{\alpha} &= \{ x : x \in A_{\alpha} \text {
+for every set $A_{\alpha}$ with } \alpha \in I \}\\
+\end{align}$$