Small dense subsets in "Hausdorff" hypergraphs Let $H=(V,E)$ be a hypergraph. We call it Hausdorff if for all $x\neq y \in V$ there are $e_1,e_2\in E$ with $e_1\cap e_2 = \emptyset$ such that $x\in e_1$ and $y\in e_2$. We say that $D\subseteq V$ is dense if $D\cap e \neq \emptyset$ for all $e\in E\setminus\{\emptyset\}$.
If $V$ is an infinite set, $H=(V,E)$ is Hausdorff, and $D\subseteq V$ is dense, is it possible that $2^{|D|} < |E|$?
 A: Let H be any Hausdorff Hypergraph.  We build G, a Greater (hyper-)Graph which has a dense set of two points and is also Hausdorff.  When H has enough edges, G will satisfy the inequality of having more than $2^2$ edges.
The ground set for G has only two points more than the ground set for H. I call these points b and c.  For every nonempty edge e from H, add edges e union (singleton) b and e union (singleton) c to G.  As H is Hausdorff (and assume H has more than one point for its ground set, otherwise add singletons b and  c  to G also), G is Hausdorff also.  Further, {b,c} is dense in G. For examples satisfying your inequality, make sure H has more than two nonempty edges.
Gerhard "Greater Generality Gives Greater Good" Paseman, 2017.09.20.
A: Take $V$ to be the set of ultrafilters on $\omega$ and let $E \subset P(V)$ be the smallest subset closed under arbitrary unions containing all sets of the form $[X]= \{ p \in V: X \in p \}\subset V$ where $X \subset \omega$ is infinite and co-infinite.
Fact: The hyper-graph $(V,E)$ is Hausdorff. 
Proof: Let $p, q \in V$ be distinct, then there is some $X \in p$ with $\,\omega \backslash X \in q$. As such, $p \in [X]\in E$, $q \in [\omega \backslash X]\in E$ and $[X] \cap [\omega \backslash X] = \emptyset$.)
Fact: The set $D = \{ p \in V: (\exists s \subset \omega)(|s| < \omega \wedge s \in p)\}$ of principal ultrafilters on $\omega$ is countable and dense with respect to $(V,E)$.
Proof: For any $\mathcal E \in E$, there is some infinite and co-infinite $X \subset \omega$ with $[X] \subset \cal E$; since $X$ is non-empty we can find some $n\in X$, in which case $p = \{ Y \subset \omega: n \in Y\} \in D\cap \cal E$.)
Fact: Assuming $\sf AC$:  $2^{|D|}=\mathfrak{c} < 2^{\mathfrak{c}} \leq \vert E \vert$.
Proof: Let $\mathcal{A} = \{ A_\alpha: \alpha \in \mathfrak{c}\}\subset P(\omega)$ be an almost disjoint family of infinite subsets of $\omega$ of size $\mathfrak{c}$ and for any non-empty $I \subset \mathfrak{c}$ define $\mathcal{E}(I) = \bigcup \{ [A_\gamma]: \gamma \in I \} $ . It follows that,
$$I \neq J \implies \mathcal{E}(I) \neq \mathcal{E}(J)$$
since, without loss of generality, we can assume there is some $\gamma \in I \backslash J$, in which case any non-principal ultrafilter containing $A_\gamma$ will witness $\mathcal{E}(I) \backslash \mathcal{E}(J) \neq\emptyset$ (that such an ultrafilter exists is a consequence of $\sf AC$.) Nothing that $E$ is closed under unions, we have $\mathcal{E}(I) \in E$ and it follows that $2^{\vert D \vert} = \mathfrak{c} < \vert P(\mathfrak{c}) \vert \leq \vert E \vert$.
