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Let $\{x_n\}_{n=0}^\infty$ be a sequence of real numbers. Let $\alpha = \liminf_{n\to\infty} x_n$. Which of the following statements are true?

a. For every $\epsilon \gt 0$, there exists a subsequence $\{x_{n_k}\}$ such that $x_{n_k} \leq \alpha + \epsilon$ for all $k\in N$.

b. For every $\epsilon \gt 0$, there exists a subsequence $\{x_{n_k}\}$ such that $x_{n_k} \leq \alpha - \epsilon$ for all $k\in N$.

c. There exists a subsequence $\{x_{n_k}\}$ such that $x_{n_k} \rightarrow \alpha$ as $k\rightarrow \infty$.

If the sequence converges then $$\liminf_{n\to\infty} x_n=\limsup_{n\to\infty} x_n$$ then $x_n \lt \alpha + \epsilon$ (only a few will be greater than $b+\epsilon$) then subsequence will also follow and a,c will be right...What if sequence diverges?

In exam I took particular examples like $-\frac {1}{n}$ and $-\frac{1}{2n}$ and such...but don't know proper explanation...

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    $\begingroup$ For b: try $x_n = \frac{1}{n}$. $\endgroup$ – Daniel Schepler Aug 4 '17 at 17:34
  • $\begingroup$ i don't want to only discard options,i want to solve it properly and by applying defination if possible. $\endgroup$ – Pranita Gupta Aug 4 '17 at 17:37
  • $\begingroup$ You want $\{x_{n_k}\}$ in place of $\{x_n\}$ in a and b. $\endgroup$ – zhw. Aug 5 '17 at 22:18
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(a) is true. Let's see why using the definition of lim inf.

Let $\epsilon > 0$. $\liminf \limits_{n \to \infty} x_{n} = a$ means $\lim \limits_{n \to \infty} g_{n} = a$, where $g_{n} := \inf \limits_{k \geq n} \{x_{k}\}$, right? So, for some point $N$ in the sequence, we have all later points (i.e., for all $n \geq N$) satisfying $g_{n} \in (a - \epsilon, a + \epsilon)$. That means $\inf \limits_{k \geq n}\{x_{k}\} \in (a - \epsilon, a + \epsilon)$ for all $n \geq N$. Ok, so let's look at $N$ first. $g_{N} \in (a - \epsilon, a + \epsilon)$ implies $\inf \limits_{k \geq N} \{x_{k}\} \in (a - \epsilon, a + \epsilon)$, and so by properties of infimum, we can find some point in the sequence $x_{n_{1}}$, with $n_{1} \geq N$. Now, since $n_{1} \geq N$, that means $g_{n_{1}} \in (a - \epsilon, a + \epsilon)$. By the same argument as above, we can find some $n_{2} > n_{1}$ so that $x_{n_{2}} \in (a - \epsilon, a + \epsilon)$. Then we can find $n_{3} > n_{2}$, and so on. In particular, we have a subsequence $x_{n_{k}}$ satisfying $x_{n_{k}} < a + \epsilon$, which is what we wanted to find, so (a) is true.


(b) is false, as Daniel Schepler noted in the comments.

To show this using the definitions, note that $a - \epsilon < a$, right? Now, if $\liminf \limits_{n \to \infty} x_{n} = a$, that means $\lim \limits_{n \to \infty} g_{n} = a$, where $g_{n} := \inf \limits_{k \geq n} \{x_{k}\}$, right? So, after some $N$, all of the $g_{n}$ are within $\frac{\epsilon}{2}$-distance of $a$ by definition of limit. In particular, for some $N$, $a - \frac{\epsilon}{2} < g_{N} = \inf \limits_{k \geq N} \{x_{k}\}$, implying that $a - \frac{\epsilon}{2} \leq x_{k}$ for all $k \geq N$. This shows that you can't find a subsequence less than $a - \epsilon$.

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    $\begingroup$ Yes ...or can we take any increasing divergent sequence or else? Do a and C implies each other? $\endgroup$ – Pranita Gupta Aug 4 '17 at 18:11
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    $\begingroup$ @Ikuyuki Yes, the fact that (a) holds implies that (c) holds. Since you can find a subsequence in which all of the terms are less than $\epsilon$-distance from $a$ (by part (a)), apply this fact to $\epsilon = \frac{1}{k}$. Then choose an element from the sequence that works for $\frac{1}{k}$. Do this for each $k$, and you will have a subsequence $x_{n_{k}}$ of $x_{n}$ which converges to $a$. It is a good exercise for you to prove this rigorously (i.e., spend time to go over the proof mentioned above in detail), but the general idea of the proof is written in this comment. $\endgroup$ – layman Aug 4 '17 at 18:33
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    $\begingroup$ @Ikuyuki I stated the definition of lim inf in my answer, but I'll state it again here. $\liminf \limits_{n \to \infty} x_{n}$ means the following: first you define the sequence $g_{n} := \inf \limits_{k \geq n} \{x_{k}\}$. Then you take the limit of $g_{n}$. So the lim inf of $x_{n}$ is defined as $\lim \limits_{n \to \infty} g_{n}$. I recommend you think about this with the following example: $x_{0} = 1, x_{1} = -1, x_{2} = 1, x_{3} = -1, x_{4} = 1, x_{5} = -1, ...$. If this is your sequence $x_{n}$, what is $g_{n}$ for each $n$? Then what is $\liminf \limits_{n \to \infty} x_{n}$? $\endgroup$ – layman Aug 4 '17 at 19:05
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    $\begingroup$ I read it somewhere ..take subsequences and there limits and then infimum of those limits. I took $\{1,1,1,1...\}$ And $ \{-1,-1,-1,...\}$ and then set of subsequential limits $E_0 = \{-1,1\}$ and it's infimum is -1 and supremum is 1.. So I think it should be -1. .. The way you said to do it I'm getting $g_1=-1$ $g_2=-1$ $g_3=-1$ ... So limit will be -1 . M I doing it right? In case sequence $\{x_n\}=\frac 1n$ is all $g_1,g_2,...$ will be 0 ,right? $\endgroup$ – Pranita Gupta Aug 5 '17 at 3:52
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    $\begingroup$ @Ikuyuki You are correct with both sequences you gave, the $-1,1,-1,1,...$ and the $\frac{1}{n}$ sequences. For me, the method I explained in the above comment is much easier for me to compute the $\liminf$. The other definition you gave is also correct, but I find it easier to compute $\liminf$'s (and $\limsup$'s) using the method I described above. $\endgroup$ – layman Aug 5 '17 at 4:43
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Fact: the set $S$ of subsequential limits of a sequence $(x_n)_n$ is a closed subset of the extended reals $[-\infty, +\infty]$. Its maximum is $\limsup_n x_n$ and its minimum is $\liminf_n x_n$.

This immediately implies that c) holds.

a) Also holds, for it it failed, there would exists a $\varepsilon_0 > 0$ such that all convergent subsequences have limits $> \alpha + \varepsilon_0$ ,this implies that $S \subseteq (\alpha+\epsilon_0 ,\infty]$ so that $\alpha$ cannot be its minimum, contradiction.

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  • $\begingroup$ Could it be $ [0, \infty)$ ? $\endgroup$ – Pranita Gupta Aug 4 '17 at 18:16
  • $\begingroup$ @Ikuyuki No, that set is not closed in the extended reals. It could be $[0,\infty]$ or $\{0,1, \infty\}$, e.g. $\endgroup$ – Henno Brandsma Aug 4 '17 at 18:17
  • $\begingroup$ What is S? Set of all limits? I am not familiar with concept of extended real numbers but it looks like if we add $\infty$ or $-\infty$ then it will become extended real numbers or any closed set will properly contain either $\infty$ or $-\infty$ means not of the form [0, $\infty$) or (-$\infty$,0] like you said..$(\alpha+\epsilon_0 ,\infty]$ this is not closed do we have to take it open here? $\endgroup$ – Pranita Gupta Aug 4 '17 at 18:37
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    $\begingroup$ @Ikuyuki $(\alpha+\varepsilon_0 , +\infty]$ is not closed, but $S$ lies in it ($S$ is indeed the set of all subsequential limits). $[-\infty, +\infty]$ is topologised as $[-1,1]$ is. A compatble metric is $d(x,y) = |\arctan(x) - \arctan(y)|$ where the values at the infinities are the assymptotic values $\pm \frac{\pi}{2}$, of course. $\endgroup$ – Henno Brandsma Aug 4 '17 at 18:47

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