# Union of dense intervals

If $\{ q_i \}$ is dense in the real number line, and $(a_i)$ a sequence of reals with sum $S$, and $U$ is the union of all intervals $[q_i,q_i+a_i]$. How to prove or disprove there is a real number not in $U$ if $S$ is finite, and the converse if $S$ is infinite?

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For one direction, you could consider the Lebesgue measure of $U$. – Nate Eldredge Jan 30 '12 at 13:45
For the falsity of "$S$ infinite implies the union covers $\mathbb{R}$," start with some $x \notin \{ q_i : i \in \mathbb{N} \}$ and construct a sequence of positive reals $( a_i )_{i=1}^\infty$ so that the union of closed intervals given will not contain $x$. Just need to ensure that the sum diverges to $+\infty$, but this shouldn't be too difficult. (Use the fact that if $x < q_i$ then $x \notin [ q_i , q_i + a ]$ for any value of $a$, and that the family $\{ i : x < q_i \}$ is infinite.) – arjafi Jan 30 '12 at 14:19

Let $U=\bigcup [q_i, q_i+a_i]$. Suppose the $a_i$ are nonnegative and that $\sum a_i$ is finite. The length of each $[q_i, q_i+a_i]$ is $a_i$. From this, one can show that $U$ is contained in an open set whose "length" is at most $\sum a_i<\infty$. It follows that there is and $x\in\Bbb R$ with $x\notin U$. (I'm arguing informally here, since I do not know if you've studied measure. If you have, the argument is made precise by replacing "length" by "lebesque measure".)
The converse is false. Set $a_i=1$ if $q_i$ is negative. For nonnegative $q_i$ select positive $a_i$ so that $\sum\limits_{q_i\ge0 }a_i$ is finite. Then there is an $x>1$ not in $U$ and $\sum a_i$ is infinite.
Do you mean "Then there is an $x>1$ not in $U$..."? – Quinn Culver Jan 30 '12 at 16:25