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How do we show that for any $\{x_n\}$ sequence $$\lim \sup_{n \rightarrow \infty}x_n = \inf\{ a: \text{the set} \{ n: x_n >a\}\text{ is finite} \}$$

There is a variant of this question available but I did not get the answer. Would you please explain?

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    What is your definition of $\limsup_{n\to\infty}x_n$? – 2017-01-07
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    the general definition for any sequence. $\lim\sup_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty}\sup_{k\geq n}x_k$ – 2017-01-07
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    The claim $$ \text{the set} \{ n: x_n >a\}\text{ is finite} \iff \text{the set} \{ n: x_n \leq a\}\text{ is infinite} $$ is incorrect. Consider $x_n=(-1)^n$ and $a=0$. – 2017-01-08

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How do we show that for any $\{x_n\}$ sequence $$\lim \sup_{n \rightarrow \infty}x_n = \inf\{ a: \text{the set} \{ n: x_n >a\}\text{ is finite} \}$$

If $\{x_n\}$ is not bounded above, then the LHS is $\infty$ and by convention $\inf\emptyset=\infty$.

Suppose $$\limsup_{n\to\infty}x_n=x$$ for some $x\in \mathbb{R}$. You want to show that $$ x=\inf A\tag{1} $$ where $A=\{ a\in\mathbb{R}\mid \text{the set} \{ n: x_n >a\}\text{ is finite} \}$. There are two steps to show (1)

  • $x$ is a lower bound of $A$;
  • $x$ is the biggest lower bound of $A$.

Step 1.

Let $a\in A$. The goal in this step is $x\leq a$. Then by definition of $A$, $a

Step 2.

Let $\epsilon>0$. To show that $x$ is the biggest lower bound of $A$, it suffices (exercise!) to show that there exists $a\in A$, $x+\epsilon>a$, i.e., $$ \limsup_{n\to\infty}x_n+\epsilon>a. $$ Recall the definition $ x=\limsup_{n\to\infty}x_n:=\lim_{n\to\infty}\sup\{k\geq n:x_k\}. $ Hence, for large enough $n$, $$ x-\epsilon/2\leq\sup\{k\geq n:x_k\}\leq x+\epsilon/2. $$ In particular, for large enough $n$, $$ x_n\leq x+\epsilon/2. $$ Now, if we pick any $a\in (x+\epsilon/2,x+\epsilon)$, we see that $a\in A$ and $$ a

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    Thanks for the proof....just a quick questions how do you pick 'a' in $ (x + \epsilon /2, x + \epsilon)$ (I am confused as to how I know that this $'a' \in A$ since there have to be finite $x_n >a$). – 2017-01-08
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    *Any* $a\in(x+\epsilon/2,x+\epsilon)$ would work. Because for large enough $n$, $x_n\leq x+\epsilon$ and there are only possibly finitely many $x_n$ in $(x+\epsilon/2,x+\epsilon)$. – 2017-01-08
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    The key is to note that there are only finitely many $n$ such that $x_n>x+\epsilon/2$. – 2017-01-08
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Say

$$\lim \sup_{n \rightarrow \infty}x_n = L$$

Then, for $\varepsilon>0$ we eventually have $x_n \le L+\varepsilon$, i.e the set $\{n:x_n>L+\varepsilon\}$ is finite.

Also, we have an infinite subsequence $(x_{n_k})$ which converges to $L$, so for $\varepsilon>0$, the set $\{n:x_n>L-\varepsilon\}$ will also be infinite, as it contains a tail of this subsequence.

Thus:

$$(L,\infty)\subset\{a:\text{the set }\{n:x_n>a\}\text{ is finite}\}$$

$$(-\infty,L)\subset \{a:\text{the set }\{n:x_n>a\}\text{ is finite}\}^c$$

So, $L$ is equal to the desired infimum.

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    Your proof seems okay....but in the last line "So, L is equal to the desired infimum"....how do you conclude this?...Thanks.:) – 2017-01-07