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Assume you have some sequence $a_k$ where $a_k$ is a real number. Show that for every $\epsilon > 0$ there is $m \ge N$ such that $|a_k - a| < \epsilon$ for all $k \ge m$ then $\limsup( a_k) = a$ and $\liminf (a_k) = a$.

I know that $\limsup (a_k) := \inf \sup (a_k)$ and $\liminf (a_k) := \sup\inf (a_k)$

I also have that for $\epsilon > 0$ there is $m \in \Bbb N$ such that $a_k < a + \epsilon$ then $\limsup (a_k) \le a + \epsilon$. I just don't know what the next step is. Could I argue that since the def of sup is the least upper bound then $\epsilon$ would go to zero? Any help would be appreciated.

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1 Answer 1

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The second sentence doesn’t make sense as written; I expect that you’re actually supposed to show that

if for every $\epsilon>0$ there is an $m\in\Bbb N$ such that $|a_k-a|<\epsilon$ for all $k\ge m$, then $\limsup_k a_k=a$ and $\liminf_k a_k=a$.

It may be helpful, at least in getting a feel for what’s going on here, to realize that the hypothesis

for every $\epsilon>0$ there is an $m\in\Bbb N$ such that $|a_k-a|<\epsilon$ for all $k\ge m$

says precisely that the sequence $\langle a_k:k\in\Bbb N\rangle$ converges to $a$. Thus, you’re being asked to show that if a sequence has a limit, that limit is also its limit superior and its limit inferior.

The first sentence of your last paragraph is missing several crucial bits and consequently doesn’t make sense as written. It should read:

If $a=\limsup_k a_k$, then for each $\epsilon>0$ there is an $m\in\Bbb N$ such that $a_k<a+\epsilon$ whenever $k\ge m$.

Unfortunately, this isn’t much help as it stands. You need something that implies that $\limsup_k a_k=a$, and this does just the opposite: it tells you something that is implied by the statement that $\limsup_k a_k=a$. Something can be done along these lines, but at this point I think that you’re probably better served by going back to the definition.

To state it more carefully, you know that $$\limsup_k a_k=\lim_{n\to\infty}\sup_{k\ge n}a_k\;.$$ To show that $\limsup_k a_k=a$, therefore, you have to show that for each $\epsilon>0$ there is an $m\in\Bbb N$ such that $$\left|a-\sup_{k\ge n}a_k\right|<\epsilon$$ whenever $n\ge m$.

So let $\epsilon>0$ be any positive number. What do we know that might be useful? We’re told that there is an $m\in\Bbb N$ such that $|a-a_k|<\epsilon$ whenever $k\ge m$. That’s at least the right kind of statement; does it actually help with what we’re trying to prove? Suppose that $n\ge m$; what can we say about $$\sup_{k\ge n}a_k\;?$$ If $k\ge n$, then $k\ge m$, so $|a-a_k|<\epsilon$, and therefore $a-\epsilon<a_k<a+\epsilon$. In other words, the entire tail $\{a_k:k\ge n\}$ of the sequence lies between $a-\epsilon$ and $a+\epsilon$. This implies that $$a-\epsilon<\sup_{k\ge n}a_k\le a+\epsilon$$ for every $n\ge m$. (You should try to figure out why I could write a strict inequality on the left but only a non-strict inequality on the right.) In other words,

$$\left|a-\sup_{k\ge n}a_k\right|\le\epsilon$$

for every $n\ge m$. This isn’t quite what I wanted, because I have $\le$ instead of $<$, but it’s good enough to ensure that $$\lim_{n\to\infty}\sup_{k\ge n}a_k=a$$ and hence that $\limsup_k a_k=a$. (You should take a moment to see why this is true.)

The proof for the $\liminf$ is very similar.

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I tried to follow your answer but couldnt understand how you could come from the $3$th step ($3$th equation) from bottom to the second step from the bottom. –  Seyhmus Güngören Sep 2 '12 at 10:45
    
@Seyhmus: For a very good reason: I typed \sum by accident instead of \sup. –  Brian M. Scott Sep 2 '12 at 18:49
    
I was going crazy to understand))) I am happy that you had a very good reason else my heart would not resist much) –  Seyhmus Güngören Sep 2 '12 at 18:54
    
@Seyhmus: I’m sorry about that. Sometimes my fingers seem to have a mind of their own. –  Brian M. Scott Sep 2 '12 at 19:04
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@rioneye: Suppose $A$ is a set of real numbers, and I know that $0<a<1$ for each $a\in A$. Then $\sup A\le 1$, but I can’t say that $\sup A<1$: $A$ might, for instance, be $\left\{1-\frac1n:n\in\Bbb Z^+\right\}$, in which case $\sup A=1$. I do know, however, that $\sup A>0$: I just pick any $a\in A$ and observe that $0<a\le\sup A$. –  Brian M. Scott Sep 3 '12 at 21:12

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