# A Hamel basis for $l^{\,p}$?

I am looking for an explicit example for a Hamel basis for $l^{\,p}$?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one can express any element of the vector space as a finite linear combination of these. After some trying I could not write one explicitly. A quick google search did not reveal anything useful except for the proof of uncountability of a an infinite Hamel basis. Maybe I am being a bit silly but I don't think the answer is as obvious as for a Schauder basis for the same case.

So, what is an explicit example for a Hamel basis for $l^{\,p}$??

-
You are not going to find any. This is not something one can do explicitly. – Mariano Suárez-Alvarez Sep 11 '12 at 15:18
@Mariano Interesting! Can you please provide some reference. – Abhishek Gupta Sep 11 '12 at 15:20
The fact that every vector space has a (Hamel) basis requires the Axiom of Choice (AC). Even though $\ell^p$ is a very specific space, I guess one can at least deduce a weak form of AC from an explicit basis. Choice is always an abstacle for explicitness. – Hagen von Eitzen Sep 11 '12 at 15:52
Asaf's answer in this thread on $\mathbb{R^N}$ is closely related. The first version works if you read $\ell^p$ instead of $\mathbb{R^N}$. – t.b. Sep 11 '12 at 16:02

The existence of a Hamel basis for $\ell^p$ cannot be proved without some of the axiom of choice, which in modern terms usually means that we cannot write it explicitly.

It is consistent with ZF+DC (a weak form of the axiom of choice which is sufficient to do a lot of the usual mathematics) that all sets of real numbers have Baire property, and in such model we have that every linear function from $\ell^p$ to itself is continuous.

It is also true (in ZF) that $\ell^p$ is separable for $1\leq p<\infty$. It is a known fact that continuous endomorphisms are determined completely by the countable dense set.

If there exists a Hamel basis then its cardinality is at least $\frak c$ (or rather exactly that), and therefore it has $2^\frak c$ many permutations, each extends uniquely to a linear automorphism, which is continuous.

Now, note that $\ell^p$ has size $\leq\frak c$ itself, since it is a separable metric space (and again, this is in fact $\frak c$) and therefore it has only $\frak c$ many continuous endomorphisms.

Cantor's theorem tells us that $2^\frak c\neq c$, and therefore in Shelah's model where every set of real numbers have the Baire property there is no Hamel basis for $\ell^p$.

-
It would probably be a useful idea to explain what is the relation between what you wrote and the question of whether one can make explicit a basis. Also, assuming we all know what «ZF+DC» means, or the relevance of something being true there is possibly not great. In fact, I am pretty sure that anyone who can understand what you wrote already knew the answer to the question :-) – Mariano Suárez-Alvarez Sep 11 '12 at 16:55
Mariano, due to lack of proper keyboard access this will wait for tomorrow. I agree that some improvement may be good. As for the last part I disagree completely. I knew what is ZF+DC long before I knew this answer. – Asaf Karagila Sep 11 '12 at 17:04
If you're interested, the argument can be simplified a bit: given a Hamel basis we can construct a discontinuous linear map more explicitly. Theorem ($\mathsf{ZF}+\mathsf{AC}_\omega$): If $X$ is an infinite-dimensional normed space and $B$ is a Hamel basis for $X$, then there exists a discontinuous linear functional $f : X \to \mathbb{R}$. Proof: without loss of generality suppose $B$ consists of unit vectors. Pick a countable subset $C = \{x_1, x_2, \dots\} \subset B$ (with no repetitions). Define $f$ on $B$ by $f(x_n) = n$ and $f(x) = 0$ for $x \notin B$... – Nate Eldredge Jan 21 at 19:24
Then $f$ extends to a linear functional $f : X \to \mathbb{R}$. It is discontinuous since $x_n/n \to 0$ and yet $f(x_n/n) = 1$ for all $n$. QED. If you want a map $T : X \to X$, pick your favorite nonzero $y \in X$ and let $T(x) = f(x)y$. Maybe there is a way to do it strictly in $\mathsf{ZF}$, but analysis there tends to be so weird that I don't care to try. – Nate Eldredge Jan 21 at 19:26
@Nate: That is somewhat optimal too, since Brunner proved that if $D$ is a set which cannot be mapped onto the integers, then $\ell_1(D)$ has a Hamel basis and every linear functional on it is continuous. – Asaf Karagila Jan 21 at 19:33