Undergrad Limsup(a+b) = limsup(a) + limsup(b)

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The discussion centers on the relationship between the limit superior of the sum of two bounded sequences and the limit superiors of the individual sequences. It is established that while the inequality limsup(a+b) ≤ limsup(a) + limsup(b) holds, the reverse equality limsup(a+b) = limsup(a) + limsup(b) requires additional conditions. Specifically, one of the sequences must be convergent for the equality to hold true. An example using the sequence a_n = (-1)^n illustrates that limsup(-a_n) does not equal -limsup(a_n) in general. The conclusion emphasizes the necessity of convergence for the equality to be valid.
Mr Davis 97
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Let ##\{a_n\}## and ##\{b_n\}## be bounded sequences. Say that we already know that ##\displaystyle \limsup_{n\to\infty} (a_n+b_n) \le \limsup_{n\to\infty}a_n + \limsup_{n\to\infty}b_n##.

But isn't it also true then that $$\limsup_{n\to\infty} b_n = \limsup_{n\to\infty} ((a_n+b_n) +(- a_n)) \le \limsup_{n\to\infty} (a_n+b_n) + \limsup_{n\to\infty} (-a_n) = \limsup_{n\to\infty} (a_n+b_n) - \limsup_{n\to\infty} a_n,$$ and so ##\limsup_{n\to\infty} b_n + \limsup_{n\to\infty} a_n \le \limsup_{n\to\infty} (a_n+b_n)##. So we conclude that ##\displaystyle \limsup_{n\to\infty} (a_n+b_n) = \limsup_{n\to\infty}a_n + \limsup_{n\to\infty}b_n##. Is something going wrong with this argument? I think in general you need one of the sequences to be convergent for this to be true.
 
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##\limsup_{n \to \infty} (-a_n) \neq - \limsup_{n \to \infty} a_n##

Take ##a_n = (-1)^n##.
 
Math_QED said:
##\limsup_{n \to \infty} (-a_n) \neq - \limsup_{n \to \infty} a_n##

Take ##a_n = (-1)^n##.
Ah, I see. But if ##a_n## converges then ##\limsup_{n \to \infty} (-a_n) = - \limsup_{n \to \infty} a_n##.
 
Mr Davis 97 said:
Ah, I see. But if ##a_n## converges then ##\limsup_{n \to \infty} (-a_n) = - \limsup_{n \to \infty} a_n##.

Correct.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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