# Defining a Union of a Set of Sets formula.

I have a formula for all $$n \in \mathbb N$$, Let $$B_{n} = \{ x \in \mathbb{N} \mid 3n + 2 \leq x \leq 3n + 4 \}$$. Now I need to define: $$\bigcup \limits_{n = 1} B_{n}$$, which means I have to prove : $$\bigcup \limits_{n = 1} B_{n} = \{ x \in \mathbb{N} \mid 3n + 2 \leq x \leq 3n + 4 \}$$.

I have to show that :

1. $$\bigcup \limits_{n = 1} B_{n} \subseteq \{ x \in \mathbb{N} \mid 3n + 2 \leq x \leq 3n + 4 \}$$

2. $$\{ x \in \mathbb{N} \mid 3n + 2 \leq x \leq 3n + 4 \} \subseteq \bigcup \limits_{n = 1} B_{n}$$

I have no idea how to approach that. Could anyone help please?

• Yeah , I tried copy/pasting and this is how it turned out :/ Nov 8 '14 at 18:53
• @PainKiller I fixed the formatting of your question. Please review it to make sure the changes are correct. Nov 8 '14 at 18:56
• It's correct, thanks a lot =) Nov 8 '14 at 18:57
• The 'set' $\{ x \in \mathbb{N} \mid 3n + 2 \leq x \leq 3n + 4 \}$ isn't a set at all because $n$ isn't quantified. You probably want $\{ x \in \mathbb{N} \colon \exists n\in \mathbb N( 3n + 2 \leq x \leq 3n + 4) \}$. Nov 8 '14 at 18:58
• In the definition of $B_n$, $n$ is acting as a constant and so it's fine. but $\{ x \in \mathbb{N} \colon 3n + 2 \leq x \leq 3n + 4 \}$ makes no sense. Think about it, is $5$ in this set? You can't test it because you know nothing about $n$. Nov 8 '14 at 19:03

$B_1 = \{x \in \mathbb{N} :5 \le x \le 7 \} = \{5,6,7\}$.
$B_2 = \{ x \in \mathbb{N}: 8 \le x \le 10 \} = \{8,9,10\}$.
$B_3 = \{ x \in \mathbb{N}: 11 \le x \le 13 \} = \{11, 12,13\}$. etc.
So $\cup_n B_n = \{x \in \mathbb{N} : x \ge 5 \} = \{5,6,7,\ldots\}$.
You don't have to define $\cup_n B_n$ (it is defined when all $B_n$ are). You need to determine what set it is.