Prove a double inclusion of an arbitrary unions

Question

$B_n$ is defined as follows:

$$B_n = \{ x \in \mathbb{N} \;|\; 3n+1 < x \le 3n+4 \}$$

What I need to do is:

1. "Calculate" (not sure if that's the correct term, please correct me with the right one) the arbitrary unions of $B_n$: $$\bigcup_{1 \le n \in \mathbb{N}}B_n$$
2. Prove the calculation using a double inclusion (showing that each set is a subset of the other).

What I currently have:

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

2. 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.

Answer 1

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 $y$.

Answer 2

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.)

Answer 3

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 $n\geq5$.