# rationals and union of intervals that covers it

• Oct 9th 2013, 10:23 AM
rayman
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 $\displaystyle \mahbb{Q}\subset \displaystyle\cup_{n=1}^{\infty}I_{n}$?
• Oct 9th 2013, 10:27 AM
emakarov
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?
• Oct 9th 2013, 10:34 AM
rayman
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?
• Oct 9th 2013, 10:45 AM
emakarov
Re: rationals and union of intervals that covers it
Quote:

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 $\displaystyle \bigcup_{n=1}^\infty I_n$.
• Oct 9th 2013, 10:51 AM
rayman
Re: rationals and union of intervals that covers it
Ah yes of course, I got it:) thank you
• Oct 9th 2013, 10:55 AM
Plato
Re: rationals and union of intervals that covers it
Quote:

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 $\displaystyle \mahbb{Q}\subset \displaystyle\cup_{n=1}^{\infty}I_{n}$?

Quote:

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.