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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?

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  • $\begingroup$ Yeah , I tried copy/pasting and this is how it turned out :/ $\endgroup$
    – PainKiller
    Nov 8 '14 at 18:53
  • $\begingroup$ @PainKiller I fixed the formatting of your question. Please review it to make sure the changes are correct. $\endgroup$
    – layman
    Nov 8 '14 at 18:56
  • $\begingroup$ It's correct, thanks a lot =) $\endgroup$
    – PainKiller
    Nov 8 '14 at 18:57
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    $\begingroup$ 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 $\{ x \in \mathbb{N} \colon \exists n\in \mathbb N( 3n + 2 \leq x \leq 3n + 4) \}$. $\endgroup$
    – Git Gud
    Nov 8 '14 at 18:58
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    $\begingroup$ 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 $n$. $\endgroup$
    – Git Gud
    Nov 8 '14 at 19:03
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$B_1 = \{x \in \mathbb{N} :5 \le x \le 7 \} = \{5,6,7\}$.

$B_2 = \{ x \in \mathbb{N}: 8 \le x \le 10 \} = \{8,9,10\}$.

$B_3 = \{ x \in \mathbb{N}: 11 \le x \le 13 \} = \{11, 12,13\}$. 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.

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  • $\begingroup$ Thanks man, i was thinking into it too much, thanks a lot =) $\endgroup$
    – PainKiller
    Nov 8 '14 at 19:20

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