Limsup = Liminf = L implies Lim = L, using neighborhoods

  • Thread starter moxy
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In summary, the author is trying to solve a problem and is stuck. He has defined limsup and liminf, and knows that sup(f(U)) ≥ L for any U in Nt, but is stuck trying to show that inf(f(U1)) ≤ L < L + ε for any ε > 0.
  • #1
moxy
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Homework Statement



I am to prove that

[itex]\lim_{x\rightarrow t}{f(x)} = L \Longleftrightarrow \limsup{f(t)} = \liminf{f(t)} = L[/itex]
where I is an interval in ℝ, and when [itex]f:I → ℝ[/itex], then [itex]\forall t \in I[/itex] we let [itex]N_t[/itex] denote the set of all neighborhoods of t relative to I, and we define:

[itex]\limsup{f(t)} := \inf_{U \in N_t} \sup{\{f(s) | s \in U\}}[/itex]

[itex]\liminf{f(t)} := \sup_{U \in N_t} \inf{\{f(s) | s \in U\}}[/itex]

The Attempt at a Solution



I'm working on [itex]\lim_{x\rightarrow t}{f(x)} = L \Longleftarrow \limsup{f(t)} = \liminf{f(t)} = L[/itex] at the moment.

So, I've assumed [itex]\limsup{f(t)} = \liminf{f(t)} = L[/itex].

That is,
[itex]\inf_{U \in N_t} \sup{\{f(s) | s \in U\}} := \inf_{U \in N_t} \sup{\{f(U)\}} = L[/itex] and
[itex]\sup_{U \in N_t} \inf{\{f(s) | s \in U\}} := \sup_{U \in N_t} \inf{\{f(U)\}} = L[/itex]

I know that sup(f(U)) ≥ L for any U in Nt, so there exists some U1 in Nt such that sup(f(U1)) - ε > L for any ε > 0.
==> sup(f(U1)) > L + ε for all ε > 0

Also, inf(f(U)) ≤ L for any U in Nt, so clearly inf(f(U1)) ≤ L < L + ε for any ε > 0.

sup(f(U1)) > L + ε
inf(f(U1)) < L + εI don't think that last step is actually getting me anywhere...I'm completely stuck. Any suggestions would be greatly appreciated. My prof recently introduced liminf, limsup, and neighborhoods, and the new terminology is really messing with my head.
 
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  • #2
When I look at this problem, I just want to use the theorem that for some sequence (tn), lim (tn) exists if and only if limsup (tn) = liminf (tn), and if we know the value of one of these, we know the value of all of them. But I assume that this can't be the proof, considering the book went to the trouble of defining the limsup and liminf in such special ways. Unless the info given is what I need to construct a sequence (tn) that converges to L.

(I'm very confused...I've been working on this problem for over five hours now.)
 
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  • #3
Maybe I'm going about it the wrong way. Is it possible to show that f is continuous at t, then I can say that limx-->t f(x) = f(t) = L? Then I can use the limsup = liminf => lim exists theorem for functions (instead of sequences, like I hinted at in my last post), and it seems that the if and only if statement that I'm trying to prove would follow immediately.
 
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1. What is the definition of Limsup, Liminf, Lim, and neighborhoods?

Limsup, or limit superior, is the largest limit point that a sequence can approach. Liminf, or limit inferior, is the smallest limit point that a sequence can approach. Lim, or limit, is the point that a sequence approaches as n (the number of terms in the sequence) approaches infinity. Neighborhoods refer to a set of points that are close to a given point in a topological space.

2. How are Limsup and Liminf related to each other?

The Limsup and Liminf are related in that they both represent the extreme limit points of a sequence. The Limsup is the largest possible limit point, while the Liminf is the smallest possible limit point. They are also related in that if a sequence has a limit, then both the Limsup and Liminf will be equal to that limit.

3. Can you give an example of a sequence where Limsup = Liminf = L?

Yes, consider the sequence {1, 1/2, 1/3, 1/4, ...}. The Limsup and Liminf of this sequence are both equal to 0, which is also the limit of the sequence.

4. How does the concept of neighborhoods play a role in Limsup = Liminf = L?

The concept of neighborhoods is essential in understanding Limsup = Liminf = L. In order for a sequence to have a limit, it must eventually be contained within any given neighborhood of that limit. This means that the Limsup and Liminf must be within the neighborhood of the limit, and therefore, all three are equal.

5. What are some real-world applications of Limsup = Liminf = L using neighborhoods?

Limsup = Liminf = L is a fundamental concept in the study of limits and sequences, and it has many real-world applications. It is used in fields such as physics, engineering, and economics to model various processes and phenomena. For example, in physics, it is used to model the behavior of particles in motion, and in economics, it is used to analyze financial trends and market behavior.

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