Let $S$ be an infinite set. Let $C(S)$ be the vector space of all functions $S \to \mathbb{R}$, and let $C_c(S)$ be the subspace of functions of finite support. Is the existence of a nonzero linear functional on $C(S)/C_c(S)$ independent of ZF? Does it follow from a choice principle that is known to be strictly weaker than AC? What about $\ell^1(S)/C_c(S)$?

(A closely related question is to ask for nonzero linear functionals on $\ell^{\infty}(S)/C_0(S)$, and here I know that examples can be constructed using the Hahn-Banach theorem or the ultrafilter lemma.)

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    $\begingroup$ For you closely related question the answer is known, it's negative. If all sets of reals have the Baire property (e.g. Solovay's model) then $\ell^\infty/c_0$ is empty for $S=\Bbb N$. $\endgroup$
    – Asaf Karagila
    Feb 22, 2013 at 1:10
  • $\begingroup$ (in my previous comment empty = trivial, of course) $\endgroup$
    – Asaf Karagila
    Feb 22, 2013 at 1:16
  • $\begingroup$ For countable $S$, the existence of a non-trivial functional on the quotient seems to imply the existence of a discontinuous linear functional on a separable completely metrizable topological vector space (since $C_c(S)$ is dense). It is a consequence of the axiom of determinacy that all linear operators (at least on separable Banach spaces) are continuous. $\endgroup$
    – Martin
    Feb 22, 2013 at 1:55
  • $\begingroup$ @Martin: Actually AD is a lot more than you need in terms of consistency strength, as it requires some very large cardinals. You can have this with equiconsistency to ZFC, see Shelah's model ("Can You Take Solovay's Inaccessible Away?"). All you really need is Baire property for all the sets of reals. $\endgroup$
    – Asaf Karagila
    Feb 22, 2013 at 2:12
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    $\begingroup$ @Qiaochu: It seems that for $S=\Bbb N$ the answer is some sort of weak version of the ultrafilter lemma should be sufficient to ensure such functional, but in models of BP (all reals have Baire property) it seems that there should be no linear functionals like that (arguments of density + automatic continuity); as for uncountable sets, I am fairly certain that the answer is that for hereditarily non-amorphous Dedekind-finite sets (i.e. sets which can be split into disjoint infinite subsets, and this is hereditary), there is no such functional. But I can't seem to prove this yet. $\endgroup$
    – Asaf Karagila
    Feb 22, 2013 at 2:30

1 Answer 1


Okay, here is something that I feel is worth posting.

We say that a Banach space is a dream space if every linear operator into a normed space is continuous, in particular every linear functional of a dream space is continuous.

This means that if we quotient a dream space by a dense subspace the resulting space do not have nontrivial functionals. Perhaps the best example for dream spaces is in Solovay's model where every Banach space is a dream space. This property follows from the assertion "Every set of real numbers has the Baire property". In fact something stronger is true in these models, every Baire-measurable homomorphism between Polish groups is continuous. In such model $C(\Bbb N)/C_c(\Bbb N)$ cannot have any linear functional.

But what other models are there for dream spaces? In $\sf ZFC$ the only dream spaces are finite dimensional spaces, but it turns out that in $\sf ZF$ we can prove the following:

Theorem (Brunner [1]). A normed space $X$ is a dream space with a Hamel basis $H$, if and only if $X$ is isomorphic with $\ell_1(D)$ for some set $D$ with a Dedekind finite power set $\mathcal P(D)$.

(Recall that $D$ is Dedekind-finite if and only if it does not have a countably infinite subset. The axiom of countable choice implies that every Dedekind-finite set is finite, but with its negation we may find infinite Dedekind-finite sets.)

First we observe that any set $D$ with a Dedekind-finite power set cannot be mapped onto $\omega$ (this is a theorem by Kuratowski), so any map into $\Bbb R$ must be with a finite range (otherwise we may assume it is unbounded, and map every point to its integer part). I will adopt the [possibly contradicting] terminology strongly Dedekind finite or, sDF for such sets.

This is consistently happening when choice fails, e.g. amorphous sets are sDF. But also linearly ordered sets may be sDF. In fact if one allows atoms (non-sets) to exist, then one may construct Mostowski's linearly ordered model in which the ultrafilter lemma holds and there are such Dedekind-finite sets which can be densely ordered (and the topology is actually locally compact and strongly connected). I'm not sure about this without atoms, but it shows quite much.

Returning to Brunner's theorem, if $D$ is sDF every element in $\ell_1(D)$ must be a function which is zero almost everywhere and indeed $\{e_d\mid d\in D\}$ is a Hamel basis (as well as a Schauder basis) for $\ell_1(D)$. We have that $\ell_1(D)=C_c(D)$. Therefore there are clearly no nonzero linear functionals on the quotient which is trivial. This fits to another proof from Brunner's paper:

Lemma (Brunner [1]). If $X$ is a Banach space with a Schauderian Hamel basis $H$, then $[H]^{<\omega}$ is Dedekind finite.

Where Schauderian Hamel basis means that the coefficient functionals are continuous, and $[H]^{<\omega}$ denotes all the finite subsets of $H$.

Turning our head to $C(D)$ we have that in fact every function is bounded, so in fact this is $\ell_\infty(D)$ by similar argument as before. To see that this is indeed a Banach space (now it's far less trivial than before) we note the following thing:

Lemma. If $D$ is sDF then $\ell_\infty(D)$ is a Banach space.

Proof. Suppose that $r_n\in\ell_\infty(D)$ is a sequence of functions from $D$ to $\Bbb R$. For every $d\in D$ let $R_d\colon\omega\to\Bbb R$ defined by $R_d(n)=r_n(d)$. The function $d\mapsto R_d$ maps elements of $d$ into $\Bbb R^\omega$, a set of cardinality $\frak c$, therefore this function has a finite range. That is, there are only finitely many different $R_d$ functions. Let $d,e\in D$ such that $R_e=R_d$, then for every $n\in\omega$ we have $r_n(e)=r_n(d)$. Therefore there exists a partition of $D$ into $D_1,\ldots, D_k$ such that for all $n$ and $i$, $r_n$ is constant on $D_i$. In particular this shows that given a sequence there is a uniform bound on the size of their ranges.

Now suppose that $r_n$ is a Cauchy sequence, for every $d$ let $r(d)=\lim_{n\to\infty} r_n(d)$. Since we work with the $\max$ norm this is a well-defined function, and it is clear that if $D_1,\ldots,D_k$ is the aforementioned partition of $D$ then $r$ is constant on $D_i$ for all $i<k$. Therefore $r$ is a well-defined element of $C(D)$, and the limit of $r_n$ as wanted. $\square$

We observe two peculiar behaviors here. The first is that $C_c(D)$ is actually $\ell_p(D)$ for every $p<\infty$; and the second is that $C(D)$ is $\ell_\infty(D)$ and pointwise convergence is the same as uniform convergence.

The next step would be to prove that every linear functional comes from $\ell_1$, that in the span of the evaluation functionals. It might be possible to first show that every continuous functional is in the span of evaluation functionals, and then that every functional is continuous. However after a few days of thinking I still don't have anything to show for in this case.

We do have to assume that the Hahn-Banach theorem fails, because it is consistent to have it for $\ell_\infty(D)$ even when $D$ is sDF. But I am not certain how to proceed from this point right now.

What choice principles could prevent this? Well, the axiom of countable choice imply that every Dedekind-finite set must be finite, so all the above becomes trivial. Although as I remarked, there are models of $\sf ZF+DC+BP$ in which $C(\Bbb N)/C_c(\Bbb N)$ must have a trivial dual, while I'm not sure if the negation of $\sf BP$ is sufficient to generate a functional, some weak form of Hahn-Banach is certainly enough (recall that Hahn-Banach itself is weaker than the ultrafilter lemma).

Do note that it is possible to have a set $D$ which is sDF and still have Hahn-Banach for $\ell_\infty(D)$. I'm not sure how the direct forcing construction would go, but just as easily one can observe that Mostowski's ordered model of $\sf ZFA$ satisfies that the set of atoms is sDF and the Boolean prime ideal theorem holds, so Hahn-Banach holds. Then use the Jech-Sochor embedding theorem to ensure that $\ell_\infty(D)$ and all its functionals are transferred into a model of $\sf ZF$. I'm not sure if the Boolean prime ideal theorem will necessarily be preserved in the transfer, but it will for that particular space.


  1. Norbert Brunner, Garnir's dream spaces with Hamel bases. Archive for Mathematical Logic, Volume 26, Number 1, pp. 123-126.
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    $\begingroup$ It's already after 6am, and I think that nearly four hours is enough for tonight. The last time someone asked something like that I ended up with a masters thesis... :-) $\endgroup$
    – Asaf Karagila
    Feb 22, 2013 at 4:18

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