# Algebraic dual of $\mathbb{R}^{\mathbb{N}}$

As in the title, my question is: What is the algebraic dual space of $\mathbb{R}^{\mathbb{N}}$ - the space of all $\mathbb{R}$-valued sequences. I think that it should be the direct sum $\bigoplus_{\mathbb{N}} \mathbb{R}$, via the natural duality, but I cannot proof it.

It's not the dual. Denote by $\def\<#1>{\left<#1\right>}\<,> \colon \mathbf R^{\mathbf N} \times (\mathbf R^{\mathbf N})^* \to \mathbf R$ the duality. We embed $\def\s{\mathbf R^{(\mathbf N)}}\s := \bigoplus_{\mathbf N}\mathbf R$ in $(\mathbf R^{\mathbf N})^*$ via $\def\p{\mathbf R^{\mathbf N}}$ $$\<(a_n), (b_n)> := \sum_n a_n b_n, \quad a \in \p, b \in \s$$ We will show that this map is not onto. Let $c \subseteq \p$ denote the subspace of converging sequences, and let $\ell \in (\p)^*$ be a linear map such that $\<a,\ell> = \lim a_n$ for all $a \in c$. Suppose $\ell$ were represented by some $b \in \s$. Let $e^k \in c$ denote the sequence $(\delta_{kn})_n$, then $$\ell(e^k) = \lim_n \delta_{kn} = 0$$ Hence $$b_k = \<b, e^k> = \ell(e^k) = 0$$ So $b =0$, but $\ell\ne 0$. That is $\ell \in (\p)^* \setminus \s$.
• I do not believe that this is a counterexample. Why should it be possible to extend the limit operator $\ell$ to $\mathbb{R}^{\mathbb{N}}$? – Sebastian Sep 23 '15 at 13:30
• By the axiom of choice. Let $B$ a basis of $c$. Extend it to a basis $B \uplus C$ of $\mathbf R^{\mathbf N}$. Define $\ell$ on $B \uplus C$ be $\ell(b) := \lim_n b_n$, $b \in B$ and $\ell(c) := 0$, $c \in C$, extended by linearity. – martini Sep 23 '15 at 13:32