If $V$ is a linear space, then a set $B$ of linearly independent vectors in $V$ that span $V$ is called a Hamel basis for $V$.
Does every infinite dimensional vector space have a Hamel basis ?
My second question is :
On page 55 in Erwin Kreyszig's Introductory functional analysis, it reads
"Hence if B is a Hamel basis
for $V$, then every nonzero $v\in V$ has a unique representation as a linear
combination of (finitely many!) elements of B with nonzero scalars as
coefficients."
I do not understand this because I think I have what seems to be a counterexample:
The space $l^{\infty}$ of all bounded sequences of real numbers has the basis $B=\{e_i, i\geq 1\}$, where
$e_i$ is the sequence with all terms zero except for the $i$-th term which equals 1.
Now, I think of the sequence $a=(a_i)_{i\geq 1}$
where $a_i=1$. We have $a\in l^{\infty}$, but it does not have a linear combination representation w.r.t $B$ that has finitely many nonzero coefficients. What did I misss?
Thanks a lot
Best Answer
A Hamel basis is defined as a maximal linearly independent set. Every vector space has a Hamel basis and this well known result is proved using Zorn's Lemma. In your example $B$ is not a Hamel basis. (And you have proved that it is not one! You know that there is an element $x$ which is not a finite linear combination of members of $B$. This implies $B \cup \{x\}$ is linearly independent so $B$ cannot be maximal linearly independent set).