# Does existence of a non-continuous linear functional depend on Axiom of Choice?

Well, it is easy to construct a non-continuous linear functional on an arbitrary infinite-dimensional vector space (assuming Choice, and taking a basis etc.).

I think it is intuitive to say that:

Every non-finite-dimensional space has a non-continuous linear functional.

depends on the axiom of choice. But what if I am not so demanding, and just want an example of a vector space where there exists a non-continuous linear functional. Do I need choice for that?

(OBS: I'm restricting myself to normed spaces)

• haha, I didn't mean it that way. I think that the strength of requiring that "EVERY..." requires the strength of choice... and I just noted a blunder in my formulation, I'll correct it – Aloizio Macedo Apr 30 '14 at 1:05
• @AsafKaragila sorry if this is a duplicate, I did a quick search and did not find anything related – Aloizio Macedo Apr 30 '14 at 2:59
• Alozio, well, reading Nate's answer I doubt it's actually a duplicate because it asks for a normed space with a discontinuous functional. Had it been a Banach space, then this would have been discussed several times before. – Asaf Karagila Apr 30 '14 at 3:12
• Well, if you discussed this several times before, I'll not ask you to do it once again... :P, but I would be glad if you could provide me the discussions (more cannot be worse). – Aloizio Macedo Apr 30 '14 at 3:15
• @AsafKaragila: I kind of thought it probably had, but didn't find a previous question immediately, and so I thought I would seize the opportunity to write it out for myself :-) – Nate Eldredge Apr 30 '14 at 6:31

You do not need the axiom of choice to prove "There exists a normed space with a discontinuous linear functional". Consider the space $c_{00}$ of all sequences of real numbers with only finitely many nonzero terms, with the supremum norm. The functional $f(x) = \sum_{n=1}^\infty n x(n)$ is discontinuous. Everything in this argument is completely explicit and constructive, and we needed no choice whatsoever.
Very roughly: It is not hard to show it suffices to consider the case of a separable Banach space $X$, which is a Polish space (separable and completely metrizable). A subset $A$ of a Polish space has the property of Baire if it can be written $A = U \Delta M$ where $U$ is open and $M$ is meager. A function between Polish spaces is Baire measurable if the preimage of every open set has the property of Baire. There is a short but clever argument known as the Pettis lemma that shows that any linear functional on $X$ which is Baire measurable must in fact be continuous. Then, there is a deep result of Shelah and Solovay which says that it is consistent with the negation of the axiom of choice that every subset of every Polish space has the property of Baire. If so, then every function on $X$ is Baire measurable, and every linear functional on $X$ is therefore continuous.