0

Prove that union of each collection of nontrivial intervals of R is the union of countable subset of that collection.

If the collection contains open intervals, then the above statement is just Lindelof's Covering theorem.What if the collection has closed intervals or half open or half closed intervals? How to prove in these cases?

mathgirl
  • 149

2 Answers2

1

HINT: Let $\mathscr{A}$ be the family of intervals, $S=\bigcup\mathscr{A}$, and let $D=S\setminus\operatorname{int}S$. Since $\operatorname{int}S$ can certainly be covered by a countable subset of $\mathscr{A}$, show that $D$ is countable. You may find it useful to know that every uncountable subset of $\Bbb R$ has a two-sided accumulation point.

Brian M. Scott
  • 616,228
  • Since any point in D belongs to an interval B( with rational end points) which can be shown to be disjoint from all the other points in D , D is countable.Is this right? – mathgirl Aug 20 '20 at 08:50
  • @pooja: Unfortunately, that doesn’t work, because $D$ need not be discrete. If you started with the intervals $[-1,0]$, $\left[\frac12,1\right]$, $\left[\frac14,\frac13\right]$, $\left[\frac16,\frac15\right]$, and so on, $D$ would be ${0}\cup\left{\frac1n:n\ge 1\right}$, and every nbhd of $D$ would contain other points of $D$. But what is always true is that for each $x\in D$ there is an $\epsilon>0$ such that either $[x,x+\epsilon)$ or $(x-\epsilon,x]$ contains no other point of $D$, because $x$ is an endpoint of one of the intervals. This is enough to ensure that $D$ is countable, ... – Brian M. Scott Aug 20 '20 at 15:54
  • ... because of the result to which I linked in my answer. – Brian M. Scott Aug 20 '20 at 15:55
  • In your example only the point '0' seems to have a problem.But i get the idea now. – mathgirl Aug 20 '20 at 16:10
  • 1
    @pooja: You’re right: in that example only one point isn’t isolated. It is possible to construct examples with infinitely many such points, though. – Brian M. Scott Aug 20 '20 at 16:18
0

Every interval of non-zero length contains a rational number. Let $U$ be the union of the set $S$ of intervals and $R$ be the (countable) set of rational numbers in $U$. For $r\in R$ take the union $U_r$ of all intervals from $S$ containing $r$. Clearly $U_r$ is an interval. Its right (left) end point is a limit of a countable sequence of right (left) end points of some intervals $I_r^i$ (resp $J_r^i$) from $S$ containing $r$. One or both of the sequences can be constant. Then $U_r$ is the union of countable set of intervals $I_r^i, J_r^i, i=1,2,...$. Moreover $U$ is clearly the union of the countable set of intervals $U_r, r\in R$. Hence $U$ is the union of the countable set of intervals $I_r^i, J_r^i$ for all $r,i$.

markvs
  • 19,653
  • The last line is unreadable. – William Elliot Aug 20 '20 at 05:43
  • I had an objection to the previous version too (which I have been writing, but it is now outdated), but now the answer is just false. Consider a covering of $\Bbb R$ by which, unluckily, you just happened to choose a covering of fat $\Bbb Q$. –  Aug 20 '20 at 05:53
  • 1
    @JCAA: The union of closed intervals need not be closed. – Brian M. Scott Aug 20 '20 at 05:58
  • 1
    @JCAA Cover $\Bbb R$ by the closed intervals with non-empty interior, enumerate $\Bbb Q$ and say that your choice of intervals was $I_{q_n}=[q_n-2^{-n},q_n+2^{-n}]$ because you were unlucky. Then $\bigcup_n I_n$ has Lebesgue measure $\le4$. Therefore it does not cover $\Bbb R$. –  Aug 20 '20 at 05:59