Prove a double inclusion of an arbitrary unions $B_n$ is defined as follows:  
$$B_n = \{ x \in \mathbb{N} \;|\; 3n+1 < x \le 3n+4 \}$$
What I need to do is:


*

*"Calculate" (not sure if that's the correct term, please correct me with the right one) the arbitrary unions of $B_n$:
$$\bigcup_{1 \le n \in \mathbb{N}}B_n$$

*Prove the calculation using a double inclusion (showing that each set is a subset of the other).


What I currently have:


*

*Considering that $1 \le n$, I "calculated" the arbitrary unions of $B_n$ to be:
$$\bigcup_{1 \le n \in \mathbb{N}}B_n = \{ x \in \mathbb{N} \;|\; 4 < x \}$$

*Proving the first direction:
Let:
$$ x \in \bigcup_{1 \le n \in \mathbb{N}}B_n $$
Since $1 \le n \in \mathbb{N}$, then also $4 < x \in \mathbb{N}$. 
Thus:
$$ \bigcup_{1 \le n \in \mathbb{N}}B_n \subseteq \{ x \in \mathbb{N} \;|\; 4 < x \} $$
Now, I'm not sure if my first-direction proof is correctly written - and please correct me if it's not - but what's more of an issue here is this: 
How do I approach the opposite direction proof?
I'd be glad for any guidance.
 A: The word calculate is fine—you could also say compute or evaluate, and you're correct that the union $\bigcup_{1 \le n \in \mathbb{N}} B_n$ is equal to $\{ x \in \mathbb{N} \mid 4 < x \}$, so let's look at the proof.
Your proof of the $\subseteq$ inclusion is correct but it is lacking some details. The definition of an indexed union is
$$\bigcup_{i \in I} X_i = \{ x \mid x \in X_i \text{ for some } i \in I \}$$
and this is really the definition you should be working with directly in your proof.
A more complete proof of the $\subseteq$ inclusion would look something like this:

Let $y \in \bigcup_{1 \le n \in \mathbb{N}} B_n$. Then $y \in B_n$ for some $n \in \mathbb{N}$ with $n \ge 1$.
By definition of $B_n$, we have $3n+1 < y \le 3n+4$.
Since $n \ge 1$, we have $y > 4$, and so $y \in \{ x \in \mathbb{N} \mid 4 < x \}$.

(Notice that I used the variable $y$ to avoid overloading the variable $x$.)
For the $\supseteq$ direction, you need to let $y \in \{ x \in \mathbb{N} \mid 4 < x \}$ and derive $y \in \bigcup_{1 \le n \in \mathbb{N}} B_n$, using the definition of the sets and set operations involved. Thus your proof should look like this:

Let $y \in \{ x \in \mathbb{N} \mid 4 < x \}$.
[...here you need to find (with proof) a value of $n \in \mathbb{N}$ with $n \ge 1$ such that $y \in B_n$...]
Hence $y \in \bigcup_{1 \le n \in \mathbb{N}} B_n$.

The value of $n$ that you find will be given in terms of $y$.
A: The word 'calculate' is ok, but you could also use e.g. 'find'.
The inclusion is correct, maybe you could insert a middle step: if $x$ is in the union of $B_n$'s, it means $x\in B_n$ for some $n\ge1$, thus $4\le 3n+1<x$.
For the other direction, given an $x\in\Bbb N$ greater than $4$, how can you associate the $n$ to it, such that $x\in B_n$? (Try small examples.) 
A: By evaluating the frist $B_n$'s,we get: $B_1=\{5,6,7\}$, $B_2=\{8,9,10\}$, $B_3=\{11,12,13\}$, so we see a pattern here. Each set has three consecutive numbers and, for each $n\in\mathbb N$, the last element of $B_n$ is $3n+4$ and the first element of $B_{n+1}$ is $3(n+1)+2=3n+5$, so the first element of $B_{n+1}$ is consecutive to the last element of $B_n$. Therefore, their union covers all natural numbers $n\geq5$.
