A separable, regular space which has cardinality of the continuum but is not first countable?

Question

Actually the title says it all. Is there such a topological space which is separable, regular, has cardinality of the continuum but is not first countable?

If so, is there also an example of a topological vector space having this property? Thank you!

Answer 1

Example: Let $X=\mathbb R^{s}$, where $\mathbb R$ with the usual topology and $\omega< s < \mathfrak c$. There is a result from Pondiczery, Hewitt and Marczewski

If there are not more than $\mathfrak{c}$ ( which is the continuum), separable topological spaces, then their product is still separable.

It is easily to see that $X$ is regular, separable, however it is not first countable.

Answer 2

Let $D=\{p_\xi:\xi<2^\omega\}$ be a family of $2^\omega$ free ultrafilters on $\omega$. The space $X=\omega\cup D$; clearly $|X|=2^\omega$. Points of $\omega$ are isolated, and for each $\xi<2^\omega$ the family $$\mathscr{B}_\xi=\big\{\{p_\xi\}\cup U:U\in p_\xi\big\}$$ is a local base at $p_\xi$. Then

$$\big\{\{n\}:n\in\omega\big\}\cup\bigcup_{\xi<2^\omega}\mathscr{B}_\xi$$

is a clopen base for $X$, which is clearly $T_1$, so $X$ is Tikhonov (and hence certainly $T_3$). $X$ is separable, since $\omega$ is dense in $X$. Finally, $X$ is not first countable at any point of $D$.