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Quick question about basis. Assume you have a separable, reflexive Banach space $X$ with a Schauder basis $\{e_{i}\}$. If you form subspaces $X_{n}$ such that $X_{n} = \text{span}\{e_{1},\ldots,e_{n}\}$. Then can you consider each $X_{n}$ as Banach space with Hamel basis $\{e_{1},\ldots,e_{n}\}$ since any $x \in X_{n}$ can be written as a finite linear combination of linearly independent basis vectors $\{e_{1},\ldots,e_{n}\}$?

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When you use the notation $\mathrm{span}(v_1, \ldots, v_n)$, for a finite $n$, by definition you are forming the vector space from finite linear combinations of the $v_i$'s, so you immediately retrieve a Hamel basis if the $v_i$'s are linearly independent. –  Christopher A. Wong May 29 at 21:06
    
@ChristopherA.Wong Okay thanks. Do you mind if I email you a question about something else? I saw your topics of interest and think that maybe you might have insight into something else. –  Moses May 29 at 21:14
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I do not mind, I would be happy to answer your questions. –  Christopher A. Wong May 29 at 21:20

1 Answer 1

When you use the notation $\mathrm{span}(v_1, \ldots, v_n)$, for a finite $n$, by definition you are forming the vector space from finite linear combinations of the $v_i$'s, so you immediately retrieve a Hamel basis if the $v_i$'s are linearly independent. -- Christopher A. Wong

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