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Suppose $\{v_1,v_2,v_3,\ldots\}$ and $\{w_1,w_2,w_3,\ldots\}$ be two countably infinite Hamel bases of the same vector space. Must there be infinitely many values of $n$ for which $\operatorname{span}\{v_1,\ldots,v_n\}$ $= \operatorname{span}\{w_1,\ldots,w_n\}$?

Later note: Shortly after posting this I wondered why I would think such a thing, and thought of deleting the question. Then it occurred to me that maybe this is true with a simple additional hypothesis, but I'm not sure what it is. Maybe more later?

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Now I'm trying to remember what made me think this might be the case...... –  Michael Hardy Nov 28 '12 at 19:11
    
It should be easy to find two Hamel bases such that no two finite subsets have the same span actually. –  Olivier Bégassat Nov 28 '12 at 19:13
    
A vague intuition: I omitted some simple hypothesis with which we would get this conclusion. Now I need to figure out what it might have been. This wild guess seems silly three hours after I posted it. –  Michael Hardy Nov 28 '12 at 20:56
    
Did you remember the extra hypothesis? –  Olivier Bégassat Nov 29 '12 at 20:57

1 Answer 1

No. For instance, take a Hamel Basis $(e_n)_{n\in\Bbb N}$ and define $(v_n)$ in such a way that $v_1,\dots,v_n$ is mostly even $e_{2k}$ and an occasional uneven $e_{2k+1}$ (for instance you only add an uneven term once you hit $n=2^p$), and the opposite for $w_n$. If you chose things right, there'll be no number $n$ such that the spans coincide.

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