A topological vector space is a vector space endowed with a topology such that the sum of vectors and the multiplication by scalars are continuous.
The weight of the topology τ is the smallest cardinal α such that there is a basis B of cardinality α for τ.
In that setting, my question is: is there an infinite-dimensional topological vector space $V$ over $\mathbb R$ or $\mathbb C$ such that the weight of $V$ is larger than the cardinality of $V$?
This question is a stronger version of that other problem. For a topological vector space $V$ over $\mathbb R$ ou $\mathbb C$, if $dim(V)\geq\mathfrak c$, then the cardinality of $V$ is equal to his dimension. So, any answer for the original problem with dimension at least $\mathfrak c$ is also a solution for the present question.
Yes, there is:
We need the following
Lemma. There is a Tychonoff (in fact, paracompact) topology on $\mathfrak c$, such that its weight $w(\mathfrak c) > \mathfrak c$.
This is certainly well-known, and there are many ways to proof it. One way is the following: Instead of $\mathfrak c$, we define it on $\mathfrak c \cup \{\infty\}$: Let $\mathfrak F$ be a $\mathfrak c$-uniform ultrafilter on $\mathfrak c$. Define the topology such that each $\alpha \in \mathfrak c$ is isolated and the neighborhoods of $\infty$ are the sets $F \cup \{\infty\}$, $F \in \mathfrak F$. It is easy to see that this topology is as required. (Note that this is a subspace of the Stone-Cech compactification of the discrete space of cardinality $\mathfrak c$.)
Now consider the free topological vector space $V(\mathfrak c)$ over $\mathfrak c$ as defined here. Then $V(\mathfrak c)$ is a Hausdorff topological vector space over $\mathbb R$ with the subspace $\mathfrak c$, and $\mathfrak c$ is a vector space basis (Theorem 2.3). Hence $|V(\mathfrak c)| = \mathfrak c$. Moreover, $w(V(\mathfrak c)) \ge w(\mathfrak c) > \mathfrak c$.