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](https://mathoverflow.net/questions/11767/infinite-tensor-products). 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 ;-).



Limit for divergent sequences
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.


