Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $E\subset R$,if $E$ can be covered by a series of interval $\{I_{\lambda}\}_{\lambda\in \Lambda}$,then E can be covered by a countable subset series which are in $\{I_{\lambda}\}_{\lambda\in \Lambda}$ ?

share|cite|improve this question
Not correct. Consider the intervals $\{[a,a]\}_{a\in R}$. They cover $R$ but don't have a countable sub-cover. – Rabee Tourky Sep 24 '12 at 4:17
you are right.Actually I should add up that the intervals are non-empty interiors. – user39843 Sep 24 '12 at 4:32

HINT:. Let $\mathscr{J}$ be the set of all intervals with rational endpoints. For each $x\in E$ there are a $J_x\in\mathscr{J}$ and a $\lambda\in\Lambda$ such that $x\in J_x\subseteq I_\lambda$. Let $\mathscr{J}_0=\{J\in\mathscr{J}:J=J_x\text{ for some }x\in E\}$.

  1. Prove that $\mathscr{J}_0$ is countable.
  2. Show that for each $J\in\mathscr{J}_0$ there is a $\lambda(J)\in\Lambda$ such that $J\subseteq\lambda(J)$.
  3. Let $\mathscr{I}_0=\{I_{\lambda(J)}:J\in\mathscr{J}_0\}$; show that $\mathscr{I}_0$ is countable and covers $E$.

Note: I’m assuming that you’re talking about intervals with non-empty interiors, i.e., those of the forms $(a,b),[a,b),(a,b]$, and $[a,b]$ with $a<b$. If you allow degenerate intervals $[a,a]$, the statement is false.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.