Let $X$ be an infinite set, and let $E \subseteq {\cal P}(X)$ be a collection of subsets of $X$. We say that $E$ is $T\_1$ (with respect to $X$) if for all $x\neq y\in X$ there is $e\in E$ with $x\in e$ and $ y\notin e$.


The set $E := \{X\setminus \{y\}: y\in X\}$ is an easy example for a $T\_1$-set $E$ with $|E| = |X|$.


Is it possible that $|E| < |X|$ ?



Infinite "$T_1$"- hypergraphs
It depends on the cardinality of $X$:



> 
> The answer to your question is no if and only if $|X|$ is a strong limit cardinal.
> 
> 
> 


Recall that $|X|$ is a [strong limit cardinal](https://en.wikipedia.org/wiki/Limit_cardinal) iff $2^\lambda < |X|$ for every $\lambda < |X|$.


If $E \subseteq \mathcal P(X)$ is "$T\_1$" then all the $|X|$-many sets of the form
$$\{e \in E : x \in E\}$$
are distinct, so that we have an injection $X \rightarrow \mathcal P(E)$. Thus $2^{|E|} \geq |X|$, and we cannot have $|E| < |X|$ if $|X|$ is a strong limit.


For the other direction, suppose $|X|$ is not a strong limit cardinal and let $\lambda$ be any cardinal with $\lambda < |X|$ and $2^\lambda \geq |X|$. Identify $X$ with a subset of $2^\lambda$, and let $E$ consist of all sets of the form
$$U\_s = \{x \in X : x(\alpha) = s(\alpha) \text{ for all }\alpha \in \mathrm{dom}(s)\}$$
where $s$ is a function from a finite subset of $\lambda$ to $\{0,1\}$. In other words, identify $X$ with a subspace of the generlized Cantor space $2^\lambda$, and then observe that this space is $T\_1$ and has a basis of size $\lambda$.


