# Subsequence of a sequence converging to its lim sup and lim inf

$Let (x_n)$ 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.

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 $$\liminf x_n$$.

• You're welcome. – RRL Sep 3 '14 at 18:51
• It's a minor point, but does this construction ensure $k_1 < k_2 < \cdots$? – David Dec 13 '17 at 4:52
• @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. – RRL Dec 13 '17 at 5:15
• 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 . – Normal Sep 28 '18 at 12:33
• @Normal: Thanks for bringing this up. – RRL Sep 28 '18 at 18:03

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 $x_n > x^* + \epsilon$.

• Yes, i had to break down and understand all steps assumed in the proof ( including this one ) to fully understand the proof – nerdy Sep 3 '14 at 12:52