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$ Let (x_n)$ be a bounded real number sequence and $ (x_n )_{n≥k} $be a subsequence of $x_n$ which only takes the values of the sequence starting from the k−th term.

Let {$x_n $} and {$x_n$ }$_{n≥k}$ denote a subset of R that contains all values of the corresponding sequences $(x_n )$ and $(x_n )_{n≥k}.$

Define
$α_k$= sup{$x_n$}$_{n≥k}$
$\beta_k$= inf{$x_n$}$_{n≥k}$
$x^∗$=lim sup⁡ $x_n$ = inf{$\alpha_k$}
$x_*$=lim inf⁡ $x_n$ = sup{$\beta_k$}

Using the above definitions, how do i prove that for every ( bounded? ) sequence $(x_n)$ there is a subsequence that converges to lim sup $x_n$ and another subsequence that converges to lim inf $x_n$ ?
Also, what's the proof or reasoning of the fact that all subsequences of $(x_n)$ can only converge to values in the following interval [lim inf $x_n$ , lim sup $x_n$].

After proving that, the important Bolzano-Weierstrass Theorem would follow imediately.

I couldn't find any math resources which uses the lim inf, lim sup treatise ( not touching on topology ) and develops the proofs of many important theorems ( such as Squeeze theorem, Bolzano-Weierstrass, Limits and unequalities Theorem ) as corollaries of that . Any recommendation ( a book, article, page, video, etc ) would be really helpful as well. Thanks a lot in advance.

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2 Answers 2

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Let

$$x^* = \limsup x_n$$

and

$$\alpha_n = \sup_{k\geq n}x_k$$

Since $x_n$ is bounded, $\alpha_n$ is bounded below and non-increasing.

Hence,

$$\lim_{n \rightarrow \infty}\sup_{k\geq n}x_k= \inf_{n}\sup_{k\geq n}x_k=x^*$$

Given $n \in \mathbf{N}$ there exists $m_n \geq n$ such that

$$x^* - 1/n < \alpha_{m_n} < x^* + 1/n.$$

Since $x^*-1/n < \sup_{k\geq m_n}x_k$, there exists $k_n \geq m_n$ such that

$$x^* - 1/n < x_{k_m} \leq\alpha_{m_n}< x^* + 1/n.$$

Hence, $|x_{k_n} - x^*| < 1/n$ where $k_n \geq n$ and the subsequence $(x_{k_n}$) converges to $x^*$.

You can make a similar argument for $\liminf x_n$.

To show that no subsequence can converge to a value greater than $\limsup x_n$, assume that a subsequence converges to $x' > x^*$. Let $\epsilon = (x'-x^*)/2$. Then there are infinitely many $x_n$ greater than $x^* + \epsilon$, a contradiction of the basic property of $x^* = \limsup x_n$.

Again, try to make similar argument for $\liminf x_n$.

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  • $\begingroup$ You're welcome. $\endgroup$
    – RRL
    Sep 3, 2014 at 18:51
  • $\begingroup$ It's a minor point, but does this construction ensure $k_1 < k_2 < \cdots$? $\endgroup$
    – David
    Dec 13, 2017 at 4:52
  • $\begingroup$ @David: Not as written above , but we can construct such a subsequence since every convergent sequence has a monotone subsequence. To construct explicitly, pick $m_1 > N$ and find $x^* \leqslant x_{k_{m_1}} < \alpha_{m_1}$. Since $\alpha_n \to x^*$ monotonically we can pick $m_2 > m_1$ and find $x^* \leqslant x_{k_{m_2}} < \alpha_{m_2} < x_{k_{m_1}}$, etc. $\endgroup$
    – RRL
    Dec 13, 2017 at 5:15
  • $\begingroup$ I have a doubt, the member $ x_{k_m} $ you chose depends on $ \epsilon $ ,I mean if I choose another $ \ epsilon$ I can get a different subsequence . $\endgroup$
    – Normal
    Sep 28, 2018 at 12:33
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    $\begingroup$ @Normal: Thanks for bringing this up. $\endgroup$
    – RRL
    Sep 28, 2018 at 18:03
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Hint: Can you show the following? For every $\epsilon > 0$, there are infinitely many $x_n$ satisfying $x_n > x^* - \epsilon$ but only finitely many such that $x_n > x^* + \epsilon$.

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  • $\begingroup$ Yes, i had to break down and understand all steps assumed in the proof ( including this one ) to fully understand the proof $\endgroup$
    – nerdy
    Sep 3, 2014 at 12:52

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