A particular TVS

Question

I am looking for a topological vector space $X$ satisfying in the following properties:

(1) Cardinal number of $X$ is at most of continuum.

(2) $X$ is not a hereditary Lindelof space.

(3) $X$ is a separable topological vector space.

Answer 1

Let $\mathbb{S}$ be the Sorgenfrey line and let $X = C_p( \mathbb{S})$, the space of continuous functions on $\mathbb{S}$ in the pointwise (i.e. product) topology. This has size at most $\mathfrak{c}$, as $\mathbb{S}$ is separable. All spaces $C_p(X)$ for $X$ Tychonoff are topological (locally convex) vector spaces (a quite interesting class of them and very well studied).

(For the following facts and relevant definitions, see Tkachuk A $C_p$-Theory Problem Book (Topological and function spaces))

$d(X) = d(C_p(\mathbb{S}) = iw( \mathbb{S}) = \aleph_0$ and $C_p(\mathbb{S})$ is not even normal, let alone (hereditarily) Lindelöf (this can be proved with Jones' lemma as for the Sorgenfrey plane). Or more directly, define $f_r: \mathbb{S} \to \mathbb{R}$ by $f_r(x) = 0$ if $x < r$, $f_r(x)= 1$ if $x \ge r$, for $r \in \mathbb{S}$ and note that $S=\{f_r : r \in \mathbb{S}\}$ is a discrete-in-itself subspace of $C_p(\mathbb{S})$ of size $\mathfrak{c}$ so that $C_p(\mathbb{S})$ cannot be hereditarily Lindelöf. One can also show that $S$ is closed as well and then the separability of $C_p(\mathbb{S})$ shows by Jones' lemma that it is not normal, while a Lindelöf Tychonoff space would be normal. But this is more work and $S$ suffices for your purpose.

In a way the simplest example is $\mathbb{R}^{\omega_1}$ in the product topology, which is just $C_p(D(\omega_1))$ of course, but this only consistently satisfies (1); in some models of ZFC we have $2^{\aleph_1} = 2^{\aleph_0}$ in others we have strict inequality. Without demand (1) that space would have been my go-to example.