Regular spaces, weight and dense subsets It is known that in case a space $X$ is regular ($T_1$ + $T_3$), its weight is less than or equal to $2^{d(X)}$, where $d(X)$ is the density of $X$.
What is an example of a Hausdorff space for which that property doesn't hold?
So far I've come up with a very simple example of a space which is not regular and fails to satisfy this condition, which is not even $T_1$: for instance, $X = \{1,2,3,4\}$ with topology $\tau = \{\emptyset, \{1\}, \{1,2\}, \{1,2,3\}, X\}$ (of course if one wishes an infinite space, the set of natural numbers with similar topology can be considered), so here $2^{d(X)} = 2$ (the set $\{1\}$ is dense), whereas the weight is clearly greater.
 A: Let $\kappa$ be an infinite cardinal. Let $\mathscr{F}$ be the set of all non-principal ultrafilters on $\kappa$ (there are $2^{2^\kappa}$ many of them). 
Define a space on $X = \kappa \cup \mathscr{F}$, by making all points of $\kappa$ isolated, and for every $\mathcal{F} \in \mathscr{F}$, the sets $\{\mathcal{F}\} \cup F$, where $F \in \mathcal{F}$ are defined to be a local base at $\mathcal{F}$. It's not too hard to check that this space is Hausdorff. (Check that if two ultrafilters $\mathcal{F}_1$ and $\mathcal{F}_2$ are such that every member of the first intersects every member of the second then $\mathcal{F}_1 = \mathcal{F}_2$ from maximality.) This space is called the Katetov extension of the discrete space of size $\kappa$.
Then $\kappa$ is dense, so $d(X) = \kappa$ and $\mathscr{F}$ is a discrete subspace of size $2^{2^\kappa}$, so $w(X) = 2^{2^\kappa} > 2^{\kappa} = 2^{d(X)}$.
It seems to be still open whether for Hausdorff spaces we have $w(X) \le 2^{2^{d(X)}}$; one can prove that for Hausdorff spaces we do have $w(X) \le 2^{2^{2^{d(X)}}}$
A: Here is a construction without ultrafilters for the case $\kappa=\aleph_0$, i.e., an example of a separable Hausdorff space $X$ with $w(X)\gt\frak c:=2^{\aleph_0}$.
Let $X=\mathbb R\times\mathbb R$ and let $\tau$ be the usual Euclidean topology of the plane. Let $\mathcal S$ be the collection of all sets $S\subseteq(\mathbb R\times\mathbb R)\setminus(\mathbb Q\times\mathbb Q)$ such that $|S\cap L|\le\aleph_0$ for every straight line $L$ through the origin. Let $\tau'=\{U\setminus S:U\in\tau,S\in\mathcal S\}$. It is easy to see that $\tau'$ is a Hausdorff topology on $X$, and it is separable because $\mathbb Q\times\mathbb Q$ is dense.
Assume for a contradiction that $(0,0)$ has a neighborhood base $\mathcal B$ with $|\mathcal B|\le\frak c$. Fix an enumeration $\mathcal B=\{B_\alpha:\alpha\lt\frak c\}$; let $B_\alpha=U_\alpha\setminus S_\alpha$ where $U_\alpha\in\tau$ and $S_\alpha\in\mathcal S$. Let $\{L_\alpha:\alpha\lt\frak c\}$ be the set of all straight lines through $(0,0)$. For each $\alpha\lt\frak c$ choose a countable set $D_\alpha\subset L_\alpha\setminus[(\mathbb Q\times\mathbb Q)\cup S_\alpha]$ such that $(0,0)$ is in the Euclidean closure of $D_\alpha$. Then $B=(\mathbb R\times\mathbb R)\setminus\bigcup_{\alpha\lt\frak c}D_\alpha$ is an open neighborhood of $(0,0)$, so there is some $\alpha\lt\frak c$ such that $B_\alpha\subseteq B$. Since $U_\alpha$ is a Euclidean neighborhood of $(0,0)$, we can choose a point $(x,y)\in U_\alpha\cap D_\alpha$. Then $(x,y)\in U_\alpha\setminus S_\alpha=B_\alpha$ but $(x,y)\notin B$, a contradiction.
