Prove or disprove that if $x_n\rightarrow x$ in the metric space $X$ then $(x_n)$ is bounded.

I was told that the statement is not true hence can be disproved. However, I think the opposite.

if it is convergent then, $d(x_n,l)<ϵ$ for any $ϵ>0$. Therefore, by taking $ϵ=1$, I can define a set such as $M=\{d(x_1,l),d(x_2,l),\ldots,d(x_N1,l),1\}$. If I define an open ball with radius of $\max M$ then wouldn't this ball be the proof that the sequence is bounded?

  • 1
    $\begingroup$ You want an open ball of radius slightly larger. $\endgroup$ – edm Aug 29 '17 at 12:39
  • $\begingroup$ Well, okay let's say it is slightly larger. Then wouldn't that ball allow us to claim that the sequence is bounded? $\endgroup$ – Larx Aug 29 '17 at 12:42
  • $\begingroup$ Yes, that is enough. $\endgroup$ – Paolo Leonetti Aug 29 '17 at 12:44
  • $\begingroup$ Then why I am told that the statement is false. The T.A. gave the example of $x_n=1/(n-1)$ and said it converges to zero, yet there is no non-trivial open set $O\subseteq X$. $\endgroup$ – Larx Aug 29 '17 at 12:49
  • $\begingroup$ Are you sure that you got him right? What you say he said is basically incorrect as your argument (mutatis mutandis) shows. If he still claims that he should be able to spot a major flaw in your proof (and not just something that can be fixed). $\endgroup$ – skyking Aug 29 '17 at 12:57

You are thinking correctly, but there are two flaws:

  1. You wrote that if $(x_n)_{n\in\mathbb N}$ converges to $l$, then $d(x_n,l)<\varepsilon$ for any $\varepsilon>0$. Not quite. Given a $\varepsilon>0$, the inequality $d(x_n,l)<\varepsilon$ holds for all but finitely many $n$'s.
  2. As you have been told, $\max M$ is not good enough. You should take $1+\max M$, for instance.

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