# What is the basis of the vector space $l^\infty$?

by Prince Thomas   Last Updated July 18, 2019 06:20 AM

We know that every vector space has a Hamel basis and also every normed space need not have a Schauder basis. As the normed space $l^\infty$ is not Separable so can't have the Schauder basis, but on the other side $l^\infty$ is also a vector space so what will be the Hamel basis of $l^\infty$?

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Even for separable infinite-dimensional Banach spaces a Hamel basis is not something we can exhibit concretely. It lives out in Axiom-of-Choice land. Even more so for $\ell^\infty$.

Robert Israel
December 09, 2016 06:15 AM

An indication that you cannot expect explicit example is that this cannot be proven in ZF.

In an infinite-dimensional normed space, existence of Hamel basis implies existence of discontinuous linear functional. And it is consistent with ZF that there are no discontinuous linear functionals on $\ell_\infty$, see here: On every infinite-dimensional Banach space there exists a discontinuous linear functional. (Perhaps there is a more straightforward argument, but this is is what I was able to come up with.)

However, if you only want proof that $\ell_\infty$ has Hamel basis, it is as standard application of Zorn's lemma. In fact, every linearly independent set is contained in a basis. (And for this proof there is nothing special about $\ell_\infty$ - it works for any vector space exactly in the same way.)

Martin Sleziak
December 09, 2016 08:10 AM