Could you please help me understand this question:

Suppose $E \subset [0,1]$ and

for each sequence $(a_n)$ , $a_n \in E$ and there are no duplicate members at $a_n$ , the series $\sum_{n=1}^\infty a_n$ converges.

Prove $E$ is countable set.

I tried this way:

If $x_n$ is sequence of all members of E than $a_n$ are sub sequences of $x_n$.

Series $\sum_{n=1}^\infty x_n$ converges, therefore $x_n\to 0$. and series $\sum_{n=1}^\infty a_n$ also converges, therefore $a_n\to 0$.

So we have sequence converging to zero. What next ?


  • 1
    $\begingroup$ Your wording is rather confusing : Do you mean "$E\subset [0,1]$ is a set such that, for any countable set $(a_n) \subset E$, one has $\sum a_n < \infty$". And you want to show that $E$ is countable, is that right? $\endgroup$ Nov 25, 2013 at 6:13
  • $\begingroup$ $(a_n)$ is sequence, if $x\in a_n$ then $x\in E$ $\endgroup$
    – user97484
    Nov 25, 2013 at 6:17

1 Answer 1


Note that the condition of $\sum a_n$ converging is much stronger than the condition of $I$ being countable. In fact the convergence of $\sum a_n$ implies that for every $\varepsilon$, the set $I \cap [\varepsilon, 1]$ is finite. Do you see why? Use this fact to enumerate $I$.

  • $\begingroup$ No, I does not understand why this set is finite. Could you explain please. $\endgroup$
    – user97484
    Nov 25, 2013 at 6:20
  • $\begingroup$ @user97484, If it was infinite, then you could make a sequence $a_n \in I \cap [\varepsilon, 1]$, satisfying $a_i \neq a_j$ when $i \neq j$, such that $\lim \sum a_n < \lim \sum \varepsilon = \lim n\varepsilon = \infty$. $\endgroup$ Nov 25, 2013 at 8:40
  • $\begingroup$ @KarolisJuodelė : I think you mean $\lim \sum a_n > \lim \sum \epsilon$ $\endgroup$ Nov 25, 2013 at 8:44
  • $\begingroup$ @PrahladVaidyanathan, yes, I did. Thanks. $\endgroup$ Nov 25, 2013 at 17:20

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