Can a sequence in $\mathbb{R}$ have uncountably many cluster points? I was curious about this after reading this: Example of a sequence with countable many cluster points.

In that post, there is given a sequence with countably infinite cluster points. Hence I was curious about the uncountable case, if it is even possible.


Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be any enumeration of $\Bbb Q$, the set of rational numbers; this is possible, since $\Bbb Q$ is countable. Then it’s a nice exercise to prove that every real number is a cluster point of $\sigma$.

  • $\begingroup$ Of course the last part depends on your definition of a real number! $\endgroup$ – Cameron Williams Feb 2 '16 at 6:31
  • $\begingroup$ @Cameron: How so? $\endgroup$ – Brian M. Scott Feb 2 '16 at 6:33
  • $\begingroup$ Well I mean to say that some would say that's true by definition, so it's vacuously true. Not that it has to be checked. It was clumsy wording on my part. $\endgroup$ – Cameron Williams Feb 2 '16 at 6:56
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    $\begingroup$ @Cameron: Even if you define the reals as equivalence classes of Cauchy sequences of rationals, you still have to show that any fixed enumeration of $\Bbb Q$ has a subsequence converging to each given real. $\endgroup$ – Brian M. Scott Feb 2 '16 at 6:58
  • $\begingroup$ Oh I see your point. Yeah I guess it's not quite trivial but there isn't too much work to do. Just a few lines of reasoning. $\endgroup$ – Cameron Williams Feb 2 '16 at 7:21

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