Does every vector space with a weak topology contain a dense subspace which is a direct sum of real lines?

Question

Is it true that given an index set $A$ and a topological vector subspace $E$ of $\mathbb{R}^A$ (that is, $E$ carries weak topology), there is a dense topological vector subspace of $E$ which is isomorphic as a topological vector space to a direct sum $$\bigoplus_{a\in B}\mathbb{R}_a$$ of $B$-many copies of $\mathbb{R}$ (endowed with the topology inherrited from the Tychonoff product $\mathbb{R}^B$) for some index set $B$?

Answer 1

If $E$ is a normed space, then in its weak topology there is no subspace isomorphic as a topological vector space to $\mathbb R^{\mathbb N}$.

We can prove this using: in a normed space, if a sequence converges weakly then it is bounded (in norm).

So let $E$ be a normed space. Let $e_1,e_2,\cdots$ be any sequence of nonzero vectors in $E$. Then there exists a choice $(t_k)_{k \in \mathbb N}$ of scalars, so that the equence $(\sum_{k=1}^n t_k e_k)_{n \in \mathbb N}$ is unbounded, and therefore the series $\sum_{k=1}^\infty t_k e_k$ does not converge weakly.

Finally note that if $\phi : \mathbb R^{\mathbb N} \to E$ is a linear injective map that is continuous (from the product topology of $\mathbb R^{\mathbb N}$ to the weak topology of $E$), then it gives us a sequence $(e_k)$ of nonzero vectors such that $$ \sum_{k=1}^\infty t_k e_k = \phi\Big((t_1,t_2,t_3,\cdots)\Big) $$ converges weakly for all sequences $(t_k)$ of scalars.

added (no need for weak sequential completeness)
In a normed space, for any sequence $(e_k)$ there are positive scalars so that $t_k e_k$ is unbounded, and therefore $t_k e_k$ does not converge weakly. But any linear injection $$ \phi : \bigoplus_{k \in \mathbb N} \mathbb R \to E $$ that is continuous (from the topology inherrited from the Tychonoff product $\mathbb{R}^B$ to the weak topology) would give us a counterexample to that.