6:46 AM
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6

Is the set $(e_n)_{n>0}$ a (vector space) basis for the sequence Hilbert space $l^2$? It is a Hilbert space basis anyway. I would say no, because the sequence $\left(\frac{1}{n}\right)_{n>0}$ is in $l^2$ but it can't be written as a finite linear combination of $e_i$'s. Is that right?

47

Let $V$ be a vector space with infinite dimensions. A Hamel basis for $V$ is an ordered set of linearly independent vectors $\{ v_i \ | \ i \in I\}$ such that any $v \in V$ can be expressed as a finite linear combination of the $v_i$'s; so $\{ v_i \ | \ i \in I\}$ spans $V$ algebraically: this is...

7

I seem to have tripped on the common Hamel/Schauder confusion. If $X$ is any vector space (not necessarily finite dimension) and $B$ is a linearly independent subset that spans $X$, then $B$ is a Hamel basis for $X$. If there exists a sequence $(e_n)$ such that for every $x \in X$ there exists ...

4

Alright, so I'm reading a book on Hilbert spaces and functional analysis, and here it defines a "Hamel basis" to be a "maximal linearly independent set". I take this to mean that $S$ is a Hamel basis if $S$ is linearly independent, and there is no linearly independent set $T$ for which $|T|>|S|$....

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