Graduate Limits and Supremum: Is It True?

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The discussion centers on the convergence of a sequence ##a_n## in norm to a limit ##a## and the behavior of the supremum of the inner product with respect to a set ##S##. It questions whether the supremum ##\sup_{s\in S} <a,s> < +\infty## holds true given that ##\sup_{s\in S} <a_n,s> < +\infty## for all ##n \ge 0##. Participants suggest that if there exists a constant ##M \in R## such that ##\sup <a_n,s> < M## and an epsilon condition is satisfied, it implies a boundedness in the limit. The conversation emphasizes the need for clear definitions and conditions to establish the validity of the supremum statement. Overall, the thread explores the implications of convergence and boundedness in the context of functional analysis.
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We have ##a_n## converges in norm to ##a## and a set ##S## such that for all ##n\ge 0##
$$\sup_{s\in S} <a_n,s><+\infty .$$ Is it true that ##\sup_{s\in S} <a,s><+\infty##
 
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You need to use two hashes (or two dollars) for Latex delimiters.
 
Is this a homework-type of problem? There is a format for those and you must show work before we can give hints.
Suppose ##M \in R## is such that ##sup<a_n,s> \lt M##. Also, suppose ##\epsilon \gt 0## and ##m\in N## are such that ##<a_n,a> \lt \epsilon## ##\forall n\gt m##. What can you say then?
 

Thread 'About the definition of topological manifold using closed sets'

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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