Let $X \neq \emptyset$ be a 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 also possible that $|E| < |X|$ if we require $E$ to be $T_1$? Is the answer the same for $X$ finite, and $X$ infinite?

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 $|E| < |X|$ ?