If $X=(x_n)$ is a bounded sequence in $\mathbb{R}$ show that there is a subsequence of $X$ which converges to $\lim \text{inf} (x_n)$.

My attempt:

Let $\lim \text{inf} (x_n) =x$. Then by definition, there are infintely many $x_n$ such that $$x-\epsilon <x_n<x+\epsilon.$$ Which implies that $\exists$ $x_{n_k}$ such that $x_{n_k}\in \{a-\epsilon,a+\epsilon\}= a-\epsilon <x_n<a+\epsilon$... But now how can I get my subsequence to converge to $\lim \text{inf} (x_n)$?


Choose $\varepsilon_n = \frac{1}{n}$ and use your method to extract a single $x_n$ for each $\varepsilon_n$. Because there are infinitely many $x_n$ to choose from for each $\varepsilon_n$, we can make it so that we choose in order so that it actually is a subsequence.

  • $\begingroup$ Why exactly did you choose $\varepsilon_n = \frac{1}{n}?$? $\endgroup$ – Q.matin Mar 27 '13 at 7:00
  • $\begingroup$ Because it goes to $0$ and it's enumerated by an integer. I could have picked $\varepsilon_n = \frac{1}{n^2}$ if I wanted. $\endgroup$ – muzzlator Mar 27 '13 at 7:01
  • $\begingroup$ Ah, I see, thanks. I am going to try and attempt your technique. $\endgroup$ – Q.matin Mar 27 '13 at 7:05
  • $\begingroup$ If I choose $\varepsilon_n = \frac{1}{n}$ then $|x_n-x|<\epsilon_n \implies |x_{n_k}-x|< \varepsilon_n =\frac{1}{n}$. Is this correct? $\endgroup$ – Q.matin Mar 27 '13 at 7:12
  • 1
    $\begingroup$ Hmm, confusing, maybe I should read up more on subsequence. Thanks a lot, muzz! $\endgroup$ – Q.matin Mar 27 '13 at 8:09

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