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Sorry for the title, but I think it's funny. Can you write down a homomorphism (of additive groups)

$\mathbb{R}^\mathbb{N} \to \mathbb{R}$,

which is nontrivial and whose kernel contains the finite sequences? For example, on the subgroup of convergent sequences, we can take the limit. The question is not if such thing exists (according to the axiom of choice, $\mathbb{R}^\mathbb{N} / \mathbb{R}^{(\mathbb{N})}$ has a basis over $\mathbb{R}$, etc.). I want to write something down1 in order to play around with this "limit for divergent sequences", which might be helpful here. Possibly all of you immediately think that this is not possible, but for which reason? Perhaps it works somehow, but it's just complicated?

1in an informal sense. I'm not interested in a discussion about mathematical logic ;-).

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  • $\begingroup$ Are you demanding that the homomorphism agree with limits when they exist? In that case, the kernel contains not only finite sequences, but those with infinite support that converge to zero. $\endgroup$
    – S. Carnahan ♦
    Commented Feb 2, 2010 at 15:51
  • $\begingroup$ no this is not imposed $\endgroup$
    – Martin Brandenburg
    Commented Feb 2, 2010 at 15:59
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    $\begingroup$ I think a logician can say this better than I can, but the precise meaning of "writing something down" seems to be at the center of your question. Therefore, it is strange to me that you don't want a discussion about mathematical logic here. $\endgroup$
    – S. Carnahan ♦
    Commented Feb 2, 2010 at 16:07
  • $\begingroup$ the rules are simple: if you are a logician and can prove that in some formal sense this hom. cannot be written down, feel free indicate a proof. if you can write something down in an informal sense, please let me know. otherwise, nothing has to be said. $\endgroup$
    – Martin Brandenburg
    Commented Feb 2, 2010 at 16:46
  • $\begingroup$ Wouldn't such a homomorphism allow you to construct a non-principal ultrafilter on N? $\endgroup$
    – Qiaochu Yuan
    Commented Feb 2, 2010 at 18:23

3 Answers 3

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Since $\mathbb{R}^\mathbb{N}$ with the product topology is a polish group and the set $F$ of finite sequences is dense in it, it follows that the only Baire measurable homomorphism $\mathbb{R}^\mathbb{N} \to \mathbb{R}$ that contains $F$ in its kernel is the trivial one. So "writing down" a nontrivial one will be pretty hard.

Now a bit of logic: It is consistent with $ZF$ that all subsets of (and hence all functions between) polish spaces are Baire measurable. So it is consistent with $ZF$ that the only homomorphism $\mathbb{R}^\mathbb{N} \to \mathbb{R}$ that contains $F$ in its kernel is the trivial one. This means that the use of (at least some of) the axiom of choice is unavoidable.

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The construction I know which comes closest to answering your question is that of a Banach limit. This is a bounded linear functional on the Banach space $\ell^{\infty}$ of bounded sequences which extends the limit of a convergent sequence and has some other nice properties. Two problems:

1) You want a functional on the set of all sequences. For this you can take a Banach limit and extend it linearly, but not in any canonical way. This brings me to

2) The construction of a Banach limit and its extension as above use the Axiom of Choice in critical ways, which you seem not to want.

I must say though that your desire to "write something down" and an unwillingness to consider the implications of that phrase make your quest somewhat quixotic. It is a generally agreed upon principle that if a certain proposition can be shown to require the Axiom of Choice in the sense of not being provable from ZF set theory, then it is futile to try to "write something down" that gives a construction. So I think you should be interested in what set-theoretic algebraists have to say about homomorphisms of additive groups of vector spaces without assuming AC. What you want to do may (I'm not saying that it has) have been shown to be impossible. Wouldn't you want to know this?

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    $\begingroup$ I strongly suspect that this is the best that can be done. Judging by the answers to my questions on duals, any such functional will restrict to an algebraic functional on l^\infty, but in some axiomatic systems, that's the continuous dual and (again, in some axiomatic systems), that's l^1. So as there are some axiomatic systems that say "It can't be done" (modulo the difference between l^0 and c_0), then in any axiomatic system you aren't going to be able to "write it down". $\endgroup$
    – Andrew Stacey
    Commented Feb 3, 2010 at 17:25
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I guess you need ultralimit :)

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  • $\begingroup$ a) ultrafilters are not concrete at all. b) R ist not compact. $\endgroup$
    – Martin Brandenburg
    Commented Feb 2, 2010 at 16:04

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