Example of a Hausdorff Space with a point that has different character relative to a dense subspace I am looking for a Hausdorff space $X$ that contains a dense subset $D$ and a point $x \in D$ such that $\chi(x, D)<\chi(x, X)$.
I know that $\leq$ always holds, and I also know that $X$ cannot be regular, otherwise equality will hold.
I know such space exists because I saw it in an exercise from Engelking's book.
 A: Let $Y=(\omega+1)\times\omega$ with the product topology, let $p$ be a point not in $Y$, and let $X=\{p\}\cup Y$. Let $\mathscr{U}$ be a free (i.e., non-principal) ultrafilter on $\omega$, and let $\mathscr{F}$ be the cofinite filter on $\omega$. Basic open nbhds of $p$ are sets of the form
$$B(U,F)=\{p\}\cup\big(\{\omega\}\times U\big)\cup(\omega\times F)\;,$$
where $U\in\mathscr{U}$, and $U\subseteq F\in\mathscr{F}$. It’s not hard to check that $X$ is Hausdorff.
Let $D=(\omega\times\omega)\cup\{p\}$; clearly $D$ is dense in $X$. $\{D\cap B(F,F):F\in\mathscr{F}\}$ is a countable local base at $p$ in $D$, but $\chi(p,X)>\omega$, since $\mathscr{U}$ does not have a countable filterbase.
A: We will spoil the topology of $\mathbb R$ on $0$. Let $\mathcal S$ be the set of all irrational sequences that converge to $0$. On $0$, add the neighborhoods $A\setminus S$, where $A$ is a neighborhood of $0$ from the usual topology and $S \in \mathcal S$.
Notice that $\mathbb Q$ is a dense subset in this topology and that the subspace topology of $\mathbb Q$ is the usual topology on $\mathbb Q$, therefore, $\chi(0, \mathbb Q)=\omega$. However, it's easy to verify that $\chi(0, \mathbb R)>\omega$ and this topology is Hausdorff since the usual topology is.
