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Let $A$ be an uncountable subset of the real numbers, I think the following is true:

  1. There is an injective sequence $a:N\to A$ such that $\sum_{n=1}^\infty a_n$ diverges.

This might also be true (not quite sure though):

  1. There is an injective sequence $a:N\to A$ such that $\lim_{n\to \infty}a_n$ doesn't exist.

I tried constructing the sequences or finding a set $A$ for which 1. or 2. fails but I got nothing.

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  • $\begingroup$ For 1, try to find some $e > 0$ such that $A$ has uncountably many members outside $(-e, e)$. For 2, just pick $a \neq b$ in $A$ and consider the sequence $a, b, a, b, \dots$. $\endgroup$ – user203787 Jan 29 '15 at 0:08
  • $\begingroup$ @OohAah It has to be injective $\endgroup$ – Zero Jan 29 '15 at 0:09
  • $\begingroup$ For 2, pick two disjoint closed intervals in which $A$ has infinitely many points and then interleave two sequences from these intervals. $\endgroup$ – user203787 Jan 29 '15 at 0:12
  • $\begingroup$ @OohAah I see it now, I just have to prove that the set of limit points of $A$ is uncountable. $\endgroup$ – Zero Jan 29 '15 at 0:18
  • $\begingroup$ That would be great. You can also try to show that the set of non limit points of $A$ is countable. $\endgroup$ – user203787 Jan 29 '15 at 0:20
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Since $A$ is uncountable, either the subset of all its elements that are greater than zero is uncountable or the subset of all elements less than zero is uncountable. Thus, we can AWLOG that all elements of $A$ are positive.

We claim that there exists an $\epsilon > 0$ such that $A \setminus B_{\epsilon}$ is uncountable where $B_{\epsilon} = \{x \in \mathbb{R} | x < \epsilon\}$. Suppose that this is false. Then, in particular each $A \setminus B_{\epsilon_n}$ is at most countable where $\epsilon_n = 1 / n$. But for any element $x$ of $A$, $x \not\in B_{\epsilon_n}$ for sufficiently large $n$. Therefore, $A$ is a countable union of at most countable sets and is itself at most countable. The claim follows.

Once you have the claim, a divergent sequence can be easily constructed.

Edit: It seems OooAaa also suggested this above while I was writing this answer.

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