Let $X$ be a topological vector space with $\mathrm{dim}\ X=\infty$ where the neighbourhood filter of $0$ has a countable basis. Show that $X'\neq X^*$
$X'$ is the topological dual and $X^*$ is the algebraic dual.
What I tried: We need to find $f$ such that $f \in X^*$ but $f \notin X'$.
Let $I$ be the index set of the basis of $X$ and $(b_i)_{i \in \mathbb N}$ the countable basis of the neighbourhood filter of $0$.
Define $f(b_i)=0$ for $i \in \mathbb N$ and $f(b_i)=i$ for $i \in I\setminus \mathbb N$.
So $f$ is unbounded and therefore not continuous. Which implies that $f \notin X'$