Relationship between Sup and Limsup of Sequences

[AxiomOfChoice]

So if you have a countably infinite set \{ x_n \} and consider also the sequence (x_n), what's the relationship between \sup \{ x_n \} and \limsup x_n?

 

[disregardthat]

limsup will equal sup if and only there is a subsequence converging to sup. In general sup is always larger or equal to limsup. Both are only well-defined if the sequence is bounded above.

limsup is the largest value for which there is a subsequence converging to it. In other words it's the largest limit point of the sequence. I may be mistaken, feel free to correct me if I'm wrong.

 

[AxiomOfChoice]

limsup will equal sup if and only there is a subsequence converging to sup. In general sup is always larger or equal to limsup. Both are only well-defined if the sequence is bounded above.

limsup is the largest value for which there is a subsequence converging to it. In other words it's the largest limit point of the sequence. I may be mistaken, feel free to correct me if I'm wrong.

No, this definitely makes sense. I suppose in the case of the sequence 1, 1/2, 1/3, \ldots, we have \sup\limits_n x_n = 1 (since 1 is certainly the least upper bound), but \limsup\limits_{n\to \infty} x_n = 0 (since 0 is the only limit point of this set). Thanks!

 

[disregardthat]

No, this definitely makes sense. I suppose in the case of the sequence 1, 1/2, 1/3, \ldots, we have \sup\limits_n x_n = 1 (since 1 is certainly the least upper bound), but \limsup\limits_{n\to \infty} x_n = 0 (since 0 is the only limit point of this set). Thanks!

That seems about right, and no problem.

 

[sponsoredwalk]

I found it very clarifying to introduce the idea of a superior number & an inferior number,
as is done in this book. Just have a look at the page above the one that comes up in the link.