I Limsup(a+b) = limsup(a) + limsup(b)

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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.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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