Discontinuous linear functional in $\mathbb{R^{\infty}}$ Let $\mathbb{R^{\infty}}=\{x=(x_n)_{n=1}^\infty:x_n\in\mathbb{R}, \forall n\in\mathbb{N}\}$, with the topology given by the set of neibourhoods of zero of the form  $\mathcal{U}(n_1,n_2,...,n_m;\epsilon)=\{x\in\mathbb{R^{\infty}}:|x_{n_j}|<\epsilon, \forall j=1,...,m\}$.
I am trying to find a linear functional $f:\mathbb{R^{\infty}}\longrightarrow\mathbb{R}$ which is not continuous for this topology.
I have tried to use the following result:
In a topological vector space $E$, a linear functional $f:\mathbb{E}\longrightarrow\mathbb{R}$ is continuous if, and only if, $\exists  \mathcal{U}$ neighbourhood of zero such that $f(\mathcal{U})$ is bounded in $\mathbb{R}$.
Everything I have tried has been useless because I have been unable to find a functional that is linear and for which $f(\mathcal{U})$ is unbounded for all $\mathcal{U}=\mathcal{U}(n_1,n_2,...,n_m;\epsilon)$.
Any hints would be very helpful. Thanks in advance.
 A: There exists a (Hamel) basis for $\mathbb R^{\infty}$ which includes the sequence $e_1,e_2,...$ (where $e_n$ has $n-th$ coordinate $1$ and all other coordinates $0$). Define $f(e_n)=1$ for all $n$ and $f(x)=0$ for all the basis elements not in $\{e_1,e_2,...\}$. You can extend $f$ to $\mathbb R^{\infty}$ by linearity and it is trivial to check that $f$ is not continuous. [Just use the fact that $f(e_n) = 1$ for all $n$ so we cannot have $|f(x)|<1$ on any basic neighborhood of $0$.
A: Why can you not "find" such an example?  Because the Axiom of Choice is required.  (It is consistent with ZF that there is no such functional.)
example
Let vectors $\mathbf e_n \in \mathbb R^\infty$ be defined by
$$
\mathbf e_n = (0,0,\dots,0,1,0,0,\dots)
$$
with $1$ for coordinate number $n$, the other coordinates all zero.
Let
$$
\mathbf u = (1,1,1,1,\dots)
$$
all coordinates $1$.
Note that the set $\mathcal A_1 := \{\mathbf u, \mathbf e_1, \mathbf e_2,\dots\}$ is linearly independent.  Extend it to a Hamel basis $\mathcal A$ for all of $\mathbb R^\infty$.  (Axiom of Choice used!)
This means: each $\mathbf x \in \mathbb R^\infty$ can be written
uniquely as
$$
\mathbf x = \sum_{\mathbf a \in \mathcal A} c_{\mathbf a} \mathbf a
$$
where $c_{\mathbf a} \in \mathbb R$ and only finitely are nonzero.
Define a linear functional $F : \mathbb R^\infty \to \mathbb R$ by
$$
F\left(\sum_{\mathbf a \in \mathcal A} c_{\mathbf a} \mathbf a\right) = c_{\mathbf u}
$$
I claim the functional $F$ is not continuous.   Indeed
$$
\lim_{n \to \infty}
\sum_{k=1}^n  \mathbf e_k = \mathbf u
$$
in the (product) topology of $\mathbb R^\infty$, but
$$
F\left(\sum_{k=1}^n{e_k}\right) = 0
\quad\text{for }n=1,2,3,\dots
\\
F(\mathbf u) = 1 .
$$
