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If every rational $r_i$ in $[0,1]$ lies in the interior of an open interval $I_i=(a_i,b_i)$, $a_i,b_i$ real numbers, is there a finite subset of $\{I_i\}$, such that every $r_i$ lies in an interval in this set?

Does it make any difference if the $a_i$, $b_i$ are rationals? Does it hold for every dense countable set?

What are necessary conditions on subsets of $[0,1]$ for it to hold?

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If subset of intervals is finite it is necessary bounded –  no identity Feb 9 '12 at 20:23
    
Is your notation meant to imply that we've listed all the rationals in $[0,1]$ as a sequence $(q_i)$, and that also $q_i \in (a_i,b_i)$ for each $i$ (or only for each $i$, there is some $j$ with $q_i\in (a_j,b_j)$)?? Hint: what if the sum $\sum_i b_i-a_i$ converges to something very small? –  Matthew Daws Feb 9 '12 at 20:31
    
Yes. The latter. Can you expand your hint please? –  topologycalisticalist Feb 9 '12 at 20:34
    
Sorry, my hint was pretty poor. But you have 2 good answers now... –  Matthew Daws Feb 9 '12 at 20:48

4 Answers 4

up vote 3 down vote accepted

For the final question: Heine-Borel really says that closed bounded intervals of $\mathbb R$ are compact. So a subset of $[0,1]$ will have your property if and only if it is compact. As Heine-Borel says that $[0,1]$ is compact, it's a result that subsets of $[0,1]$ are compact if and only if they are closed.

A countable set with this property is $\{ 1/n : n\in\mathbb N \}\cup\{0\}$; of course this isn't dense. As Brian Scott notes, no proper dense set can be closed, and so won't have this property.

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The answer to the first question is no: consider an irrational number $r$ in $[0,1]$. Let $\left\{b_k\right\}_{k\in\mathbb{N}}$ be a sequence of irrational numbers converging monotonically to $r$ (say the sequence is increasing). Let $I_k = (b_{k-1}, b_k)$, for $k \geq 2$. Let $I_1 = (-1, b_1)$ and $I_{\infty} = (r, 2)$ (I am using $\infty$ here just as a label). Then you obtain a countable sequence of open intervals, $\left\{I_k\right\}_{k\in\mathbb{N}}\cup \left\{I_\infty\right\}$ which covers every rational in $[0, 1]$, yet no finite subset of this countable cover covers $\mathbb{Q}\cap [0,1]$.

Reason through the other questions in a similar fashion.

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The answer is no. Let $\alpha$ be an irrational in the interval $[0,1]$, say $\alpha=1/\sqrt{2}$.

Look at the set of all open intervals of the form $(-1, \alpha -\frac{1}{n})$ or $(\alpha+\frac{1}{n},2)$, where $n$ ranges over the positive integers. Every rational in $[0,1]$ is in one of these intervals. However, this set of intervals cannot be replaced by a finite subset, since a finite subset would miss all the rationals that are close enough to $\alpha$.

Remark: We can replace the intervals $(-1,\alpha-\frac{1}{n})$ by intervals of the shape $(-1, a_n)$, where $(a_n)$ is a sequence of rationals that approaches $\alpha$ from below, and deal similarly with the $(\alpha+\frac{1}{n},2)$. Thus, in your question, replacing reals by rationals does not change the answer.

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In general there need not be such a finite subcollection. Let $x\in[0,1]$ be irrational, let $\langle p_n:n\in\omega\rangle$ be a strictly increasing sequence of rationals in $[0,1]$ converging to $x$, and let $\langle q_n:n\in\omega\rangle$ be a strictly decreasing sequence of rationals in $[0,1]$ converging to $x$. For $n\in\omega$ let $I_n=(-1,p_n)$ and $J_n=(q_n,2)$, let $\mathscr{I}=\{I_n:n\in\omega\}$, and let $\mathscr{J}=\{J_n:n\in\omega\}$. Then $\mathscr{I}\cup\mathscr{J}$ is a family of open intervals with rational endpoints, every rational in $[0,1]$ is contained in a member of $\mathscr{I}\cup\mathscr{J}$, and no finite subfamily of $\mathscr{I}\cup\mathscr{J}$ covers $\mathbb{Q}\cap[0,1]$.

This construction can clearly be generalized to any dense proper subset of $[0,1]$.

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