Proving Lim Sup Property for Bounded Sequences | Real Numbers HW

[barksdalemc]

I am trying to prove some things for a HW problem. Can you guys tell me if the following logic looks ok.

Let Un and Vn be bounded sequences of real numbers. If Un<=Vn for every n , show that lim sup n--->infinity Un <= lim sup n--->infinity Vn.

Here is what I wrote:

Let E1 and E2 be the set of all limits of subsequences of Un and Vn respectively. E1 and E2 are bounded and nonempty becuase Un and Vn are bounded, therefore E1 and E2 have a sup in R. Since all subsequences convege to the same limit as the sequence itself, the sup of E1 is the limit of Un and the limit of E2 is the limit of Vn. Therefore since Un<=Vn for every n the lim sup of Un<=lim supVn.

 

[barksdalemc]

I made an error obviously. All subsequences do not converge to the limit of the sequence. Any hints?

 

[nocturnal]

I made an error obviously. All subsequences do not converge to the limit of the sequence. Any hints?

Your right. its not guaranteed that all subsequences converge to the same limit (or that they converge at all), you would need to know that the sequences are also monotonic which would imply they converge (and thus their limit is equal to their lim sup and lim inf)

If $(s_n)$ is a sequence, then lim sup s_n is the limit of the sequence $(\bar{s}_N)$ whose N'th term (note the capitol N here) is $$sup\{s_n : n > N \}$$

if Un <= Vn for all n, what does this imply about sup{Un : n > N} and sup{Vn : n > N} for each N?

 

[barksdalemc]

No matter what value Un takes on at each N, it will be less than Vn so the sup Un is less than or equal to sup Vn?

 

[nocturnal]

No matter what value Un takes on at each N, it will be less than Vn so the sup Un is less than or equal to sup Vn?

The notation "sup Un" makes no sense because Un is a real number not a set of real numbers, and sup only applies to sets.

You seem confused about what the limsup of a sequence is.

If $(s_n)$ is a bounded sequence of real numbers, then we define

$$\limsup s_n = \lim_{k \rightarrow \infty}(sup \{ s_n : n > k \})$$

I used k here instead of N so as not to be confused with little n.

A way of thinking about this is that for each k we form the set {s_n : n > k}. Then define the sequence $(\bar{s}_k)$ by,
$\bar{s}_k = sup \{s_n : n > k \}$. Note that since (s_n) is a bounded sequence, {s_n : n > k} is a bounded set for each k, so the suprememum exists and is a real number. Thus $(\bar{s}_k)$ is a sequence of real numbers indexed by k. Using this notation we then have,

$$\limsup s_n = \lim_{k \rightarrow \infty} \bar{s}_k$$

Now, back to the problem. Your given that (U_n) and (V_n) are bounded sequences such that for every n, U_N <= V_n, and you need to show that limsup U_n <= limsup V_n.

Define the sequences $(\bar{U}_k)$ and $(\bar{V}_k)$ the same way we did above for (s_n). Using this notation, what we need to show is,
$$\lim_{k \rightarrow \infty} \bar{U}_k \leq \lim_{k \rightarrow \infty} \bar{V}_k$$

It is a known fact that if $(A_n)$ and $(B_n)$ are convergent sequences such that for each n, $A_n \leq B_n$, then $\limA_n \leq \limB_n$.

So if we can show that
1) $(\bar{U}_k)$ and $(\bar{V}_k)$ converge, and
2) $(\bar{U}_k) \leq (\bar{V}_k)$ for each k,
then we could use the above fact to get the desired conclusion.

Here are some hints on how to proceed:
to prove 1) show that $(\bar{U}_k)$ and $(\bar{V}_k)$ are bounded and decreasing.
Hint: if A and B are bounded sets of real numbers and $A \subseteq B$ then $supA \leq supB$.

to show 2) Try a proof by contradiction.

Does any of this make sense? If not which part(s) are confusing to you?