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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 $\{I_{\lambda}\}_{\lambda\in \Lambda}$ ?

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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

1 Answer 1

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.

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