A family of open bounded intervals of $\mathbf{Q}$ whose union is $\mathbf{Q}$

Question

Let $E$ be a set and $(X_i)_{i\in I}$ a family of sets such that $E\subset\bigcup_{i\in I}X_i$. Then $(X_i)_{i\in I}$ is called a covering of $E$.

Is there a family of open, bounded intervals of $\mathbf{Q}$ (i.e. intervals of the form $]x,y[$ s.t. $x,y\in\mathbf{Q}$) that covers $\mathbf{Q}$?

Edit:

Is there a family of open, bounded intervals of $\mathbf{Q}$ (i.e. intervals of the form $]x,y[$ s.t. $x,y\in\mathbf{Q}$) whose union equals $\mathbf{Q}$?

Answer 1

As $\Bbb Q$ is countable, take an enumeration of $\Bbb Q$ say $\{q_1,q_2,\dots,q_n,\dots\}$ .

Now take your $X_n=]q_n-\frac{1}{2^n},q_n+\frac{1}{2^n}[$

Comment : There were lot of examples of coverings provided in the comments but I decided to give this rather 'fancy' covering, because the OP is trying to learn and he/she should familarize himself/herself with this, as it will be very useful while doing Measure Theory in future.

Answer 2

You have many choices. The simplest one is to take the intervals $\mathopen]-n,n\mathclose[$ (I use your notation even though I find it terrible), for $n$ running through the positive integer.

How do you show that every rational number belongs to such an interval? It is not restrictive to show it for $a/b$, with $a,b>0$. Then take $n=2a$; can you prove that $2a>a/b$? Hint: this is equivalent to $2ab>a$.