Find Limsup and Liminf of fn & gn

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SUMMARY

The discussion focuses on finding the limit superior (limsup) of two sequences defined by indicator functions: fn and gn. The participants confirm that limsup as n approaches infinity for fn(x) equals 0 for all x, while limsup for gn(x) equals 1 if x is 1 or 2, and 0 elsewhere. The correct definitions of fn and gn are clarified, emphasizing the importance of specifying the domain and typesetting for clarity. The final interpretations of the limits are agreed upon, confirming the accuracy of the initial conclusions.

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  • Understanding of indicator functions and their notation
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  • Ability to typeset mathematical expressions using LaTeX
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  • Learn about limit superior and limit inferior in the context of sequences
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Homework Statement


Let 1A stand for an indicator function and
Let fn = 1{n} and gn={1{1} n odd, 1{1} n even.
Find limn->∞sup{fn} and limsupn->∞{gn}

Homework Equations





The Attempt at a Solution


The limits are pointwise so I found given x, then
limn->∞sup{fn} = 0
limsupn->∞{gn} = {1 for x=1 or x=2 and 0 elsewhere}
Do you agree? I just wanted to check basically :)
 
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Your definition of gn seems to suggest it is always equal to 1{1}, was this intentional?
 
oops no 1{1} for n odd, 1{2} for n even
 
aaaa202 said:
oops no 1{1} for n odd, 1{2} for n even

What is fn? Your definition means that fn is a function, with fn(x) = 1 if x = n and fn(x) = 0 for x ≠ n. However, that does not seem to be what you really mean. Just tell us in words what are fn and gn---forget about trying to (mis)use indicator functions.
 
fn(x) = {1 if x=n, 0 elsewhere} How am I misusing indicator functions?
By definition an indicator 1A = {1 x\in A, 0 else}
 
I think you have the right answer, but it could be stated better. First, you need to include the argument ##(x)## of the functions, and second, you should try to typeset it so it will be clearer what you mean. I assume you meant the following:
$$f_n(x) = 1_{\{n\}}(x) = \begin{cases}1 & \text{ if }x = n \\
0 & \text{ otherwise}\end{cases}$$
$$g_n(x) = \begin{cases}1_{\{1\}}(x) & \text{ if }n\text{ is even} \\
1_{\{2\}}(x) & \text{ if }n\text{ is odd}\end{cases}\text{ (for all }x\text{)}$$
And I interpreted your answers as:
$$\limsup_{n \rightarrow \infty} f_n(x) = 0 \text{ (for all }x\text{)}$$
and
$$\limsup_{n \rightarrow \infty} g_n(x) = \begin{cases}
1 & \text{ if }x = 1\text{ or }x = 2 \\
0 & \text{ otherwise} \end{cases}$$
You can right click on my equations to see how they are typeset. If you show your work, we can check whether your reasoning is right.
 
Last edited:
aaaa202 said:
fn(x) = {1 if x=n, 0 elsewhere} How am I misusing indicator functions?
By definition an indicator 1A = {1 x\in A, 0 else}

Yes, I know what an indicator function is; that is why I wrote what I thought you meant for fn, and it turned out to be correct---that is exactly what you meant.

It is often impossible to tell when reading some messages whether or not the OP really knows what he/she is saying; often people write one thing when they mean another. (However, you could have made everything clear by saying that fn and gn are functions, and by specifying their domain.)
 
Last edited:

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