# Thread: rationals and union of intervals that covers it

1. ## rationals and union of intervals that covers it

I have a question, if $\displaystyle \{q_{n}\}_{n=1}^{\infty}$ is a sequence of all rational numbers, let $\displaystyle I_{n}=[q_{n}-\frac{1}{10^{n}},q_{n}+\frac{1}{10^{n}}]$

why do we have the following inclusion $\displaystyle \mahbb{Q}\subset \displaystyle\cup_{n=1}^{\infty}I_{n}$?

2. ## Re: rationals and union of intervals that covers it

Himm, if x ∈ A and y ∈ B, is it clear to you that {x, y} ⊆ A ∪ B?

3. ## Re: rationals and union of intervals that covers it

yes it is, but I still don't understand the inclusion. Each of $\displaystyle I_{n}$ is a set (closed interval) containing infinitely many rational numbers bounded by the end points of each interval. Will their union include all rationals?

4. ## Re: rationals and union of intervals that covers it

Originally Posted by rayman
Will their union include all rationals?
Yes, it will. In my example, A ∪ B covers (i.e., is a superset of) the whole set {x, y} because each element of that set is covered by (i.e., belongs to) at least one of the sets in the union: either A or B. A completely parallel statement holds about ℚ and $\displaystyle \bigcup_{n=1}^\infty I_n$.

5. ## Re: rationals and union of intervals that covers it

Ah yes of course, I got it thank you

6. ## Re: rationals and union of intervals that covers it

Originally Posted by rayman
I have a question, if $\displaystyle \{q_{n}\}_{n=1}^{\infty}$ is a sequence of all rational numbers, let $\displaystyle I_{n}=[q_{n}-\frac{1}{10^{n}},q_{n}+\frac{1}{10^{n}}]$
why do we have the following inclusion $\displaystyle \mahbb{Q}\subset \displaystyle\cup_{n=1}^{\infty}I_{n}$?
Originally Posted by rayman
yes it is, but I still don't understand the inclusion. Each of $\displaystyle I_{n}$ is a set (closed interval) containing infinitely many rational numbers bounded by the end points of each interval. Will their union include all rationals?
To prove that $\displaystyle A\subset B$ is suffices to show $\displaystyle x \in A \to x \in B$.

If $\displaystyle t\in\mathbb{Q}$ then $\displaystyle \exists j\in\mathbb{N}[q_j=t]$.
But $\displaystyle t=q_j\in I_{j}=[q_{j}-\frac{1}{10^{j}},q_{j}+\frac{1}{10^{j}}]$

Hence, every rational number is in the large union.