Constructing a separable space that is not hereditarily separable.

Question

The construction that I had in mind was:

Let $X$ be an uncountable space. We'll assume it has just one countable dense subset $E$. Let points $p_{1},p_{2}\in X\setminus E$ such that every open set containing either of the two points contains some or all points $a_{1},a_{2}\dots a_{n}\in E$. Then the space $X-\{ a_{1},a_{2}\dots a_{n}\}$ won't have a countable dense subset, as $E-\{ a_{1},a_{2}\dots a_{n}\}$ won't have $p_{1},p_{2}$ in its closure.

Am I making too many assumptions?

Thanks for your time!

Answer 1

This may not warrant an answer, but here are couple observations.

1. You will have to be careful in your assumption of "just one countable dense subset" since adding any point to a countable dense set will result in another. (Of course, a countable discrete space really does have only one countable dense subset, but a separable, not hereditarily separable space must be uncountable.)
2. It is not clear why $E_0 = ( E \setminus \{ a_1 , \ldots , a_n \} ) \cup \{ p_1 , p_2 \}$ couldn't be a countable dense subset of $X \setminus \{ a_1 , \ldots , a_n \}$.

Answer 2

This is not so much a construction, as a vague idea, maybe. I suppose you want a separable but not hereditarily separable space?

A simple example $X$ is $\mathbb{R}$, in the included point topology with respect to $0$. So $O$ is open iff it is empty or it contains $0$. Then $\{0\}$ is dense, so $X$ is separable, but $X\setminus\{0\}$ is discrete in itself, so not separable. But $X$ is not even $T_1$.

Nicer examples: the $S \times S$, where $S$ is the Sorgenfrey line: this is separable, but has the set $D = \{(x, -x): x \in \mathbb{R} \}$ as a closed and discrete subspace. Or any uncountable product like $\mathbb{R}^{[0,1]}$, which is separable, but the set of all points with exactly one value equal to $1$ and all others to $0$, is again discrete as a subspace (and uncountable), so we have another non-separable subspace.