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Given that $(a_n)$ is a bounded sequence. If $b_n=\sup\{a_n,a_{n+1},a_{n+2},a_{n+3},\ldots\}$ prove that $\lim\sup(a_n)$ is a limit of $b_n$.

The lim sup is defined as l.u.b. of $$A = \{x\in R| \ x < a_n\ \ \text{for an infinite number of values of}\ \ n\}$$

So far I have proven that the points in $b_n$ are the upper bounds of $A$, and I want to use the property that $b_n$ is a decreasing and bounded sequence to say that $l.u.b.A$ is the limit of $b_n$.

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Let's write the sequence as $(x_n)$ (instead of $(a_n)$) and say $\alpha \leq x_n \leq \beta$ for all $n \in \mathbb N$.

Considering $X_n=\{x_n, x_{n+1}, x_{n+2},\ldots\}$ and writing $b_n = \sup X_n$ and $a_n = \inf X_n$ we have that

$$\underbrace{\alpha \leq a_1 \leq a_2 \leq \ldots \leq a_n}_{\color{green}{\text{monotonic non- decreasing}}} \leq \ldots \leq \underbrace{b_n \leq \ldots \leq b_2 \leq b_1 \leq \beta}_{\color{blue}{\text{monotonic non-increasing}}}$$

Thefore the sequences are bounded and monotonic, thus convergent then

$$a = \lim a_n = \sup a_n = \sup_{n}\inf X_n\\b=\lim b_n = \inf b_n = \inf_{n} \sup X_n$$

Edit:

Example: Let the sequence $(x_n)$ be given by $x_{2n-1} = -\frac{1}{n}$ and $x_{2n} = 1 + \frac{1}{n}$.

We have that $\inf X_{2n-2} = \inf X_{2n-1} = -\frac{1}{n}$ and $\sup X_{2n-1} = \sup X_{2n} = 1 + \frac{1}{n}$, then

$$\lim \inf x_n = 0 \ \ \text{and} \ \ \lim \sup x_n = 1$$

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  • $\begingroup$ By $x_n$, what sequence do you mean? $\endgroup$ – user20391 Feb 4 '15 at 12:34
  • $\begingroup$ Just to make it easier to use the symbols. $\endgroup$ – Aaron Maroja Feb 4 '15 at 12:35
  • $\begingroup$ are you referring to the sequence $a_n$? $\endgroup$ – user20391 Feb 4 '15 at 12:35
  • $\begingroup$ That's right, is the same bounded sequence, just changed to $(x_n)$ $\endgroup$ – Aaron Maroja Feb 4 '15 at 12:36
  • $\begingroup$ Could you explain the part where you said $b = limb_n$ in slightly more detail? $\endgroup$ – user20391 Feb 4 '15 at 12:37

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