Does finite extent $e(X)$ of a space $X$ imply that the space is compact? For a space $X$, $e(X)=\sup\{|Y| : Y\;\text{is a closed and discrete subspace of}\; X\}$ is said to be the extent of $X$. If the extent $e(X)$ is finite, then what can we say about the space?
 A: Proposition. If $e(X)<\infty$ and $X$ is $T_1$ then $X$ is finite.
Proof: Suppose otherwise, and let $Y\subseteq X$ be any finite subset of $X$. Then $Y$ is closed, and for $y\in Y$ and any $z\in Y\setminus\{y\}$ there is open $U_z$ such that $y\in U_z$ but $z\notin U_z$. Then $U = \bigcap_{z\in Y\setminus\{y\}} U_z$ is a neighbourhood of $y$ such that $U\cap Y = \{y\}$, so $Y$ is discrete. It follows that $e(X) = \infty$. $\square$
Since $X$ is finite implies $e(X)<\infty$, any non-trivial example isn't $T_1$.
Not every $T_0$ space with $e(X)<\infty$ is compact:
Let $X$ be real numbers with open sets of the form $(y, \infty)$, $X$ or $\emptyset$ (this is called right-ordered reals). Then every non-empty closed set in $X$ isn't discrete, so $e(X) = 0$. But $X$ isn't compact since $\{(n, \infty): n\in\mathbb{Z}\}$ is an open cover of $X$ with no finite subcover.
Proposition. If $D\subseteq X$ is a discrete closed subspace of compact space $X$, then $|D|<\infty$.
Proof: Since a closed subspace of a compact space is compact, $D$ is compact. Since $\{\{d\}: d\in D\}$ is a cover of $D$ by open sets with no finite subcover, it follows that $D$ is finite.
Corollary. If $X$ is compact then $e(X)\leq\aleph_0$.
Corollary. If $X$ is infinite, compact and $T_1$ then $e(X) = \aleph_0$.
But we have the following result, which also has to do with closed discrete subspaces.
Proposition. Let $X$ be metrizable. Then $X$ is compact if and only if for any closed discrete subspace $D$ of $X$ we have $|D|<\infty$.
Proof: If $X$ isn't compact there is a sequence $(x_n)_n$ with no convergent subsequence. Then $F = \{x_n : n\in\mathbb{N}\}$ is an infinite closed discrete subspace of $X$.
