The Union of Indexed Sets Notation Considering the following question.
Let $\beta=[0,5)$ and for each $\alpha\in\beta$, let $F_\alpha= ${${1,\alpha}$}.
Find $\bigcup$ {${F_\alpha:\alpha\in\beta}$} and                 $\bigcap$ {$ {F_\alpha:\alpha\in\beta}$}.
I also feel I intuitively know what the answer is for the intersection but I am unsure how to write it. Any assistance would be greatly appreciated.
 A: Write a couple of the $F_\alpha$'s to get a feeling for it. Like $F_0 = \{1,0\}$, $F_{1/2} = \{1,1/2\}$, and so on. We have $$\bigcup \{F_\alpha \mid \alpha \in \beta\} = [0,5) \quad\mbox{and}\quad \bigcap \{F_\alpha \mid \alpha \in \beta\} = \{1\}.$$Rigorous proofs use the definition of union and intersection, as follows:


*

*Since $F_\alpha \subseteq [0,5)$ for all $\alpha \in \beta$, we have $\bigcup\{F_\alpha \mid \alpha \in \beta\} \subseteq [0,5)$. Conversely, given $x \in [0,5)$, we have that $x \in \{1,x\} = F_x \subseteq \bigcup \{F_\alpha \mid \alpha \in \beta\}$. 

*Since $1 \in \{1,\alpha\} = F_\alpha$ for all $\alpha \in\beta$, we have $\{1\} \subseteq \bigcap \{F_\alpha \mid \alpha \in \beta\}$. Conversely, assume that $x \in F_\alpha$ for all $\alpha \in \beta$. You can actually take $\alpha  = 1$, so that $x \in \{1\}$ implies $x=1$, showing the reverse inclusion.
If the notation boggles you too much, you can actually write $$\bigcup \{F_\alpha \mid \alpha \in \beta\} = \bigcup_{\alpha \in \beta}F_\alpha$$and similarly for the intersection.
