A generalization of Lindelöf number is extent, defined as follows: $e(X)=\sup \{|D|: D \subset X, D \text{ is closed and discrete } \} + \omega$. The weight of $X$ is defined by $w(X)= \min\{|\mathcal B|: \mathcal B \text{ a base for } X \} + \omega$.
If $e(X)=w(X)> \omega$, then must there exist a closed discrete subset $D$ of $X$ such that $|D|=w(X)$?
Thanks for the help!
No, there need not be such a subset.
Let $X=\omega_\omega+1$. Points of $\omega_\omega$ are isolated, and the point $\omega_\omega$ has a local base consisting of sets of the form $[\alpha,\omega_\omega]$ for $\alpha<\omega_\omega$. Suppose that $D$ is an infinite, closed, discrete set in $X$, and let $D_0=D\setminus\{\omega_\omega\}$. Then $D_0$ is closed and discrete, $|D_0|=|D|$, and there is an $\alpha<\omega_\omega$ such that $[\alpha,\omega_\omega]\cap D_0=\varnothing$. Thus, $|D|=|D_0|\le|\alpha|<\omega_\omega$. On the other hand, for each $n\in\omega$ the set $[0,\omega_n)$ is a closed, discrete set of cardinality $\omega_n$, so $e(X)=\sup_{n\in\omega}\omega_n=\omega_\omega$. Finally, $$\big\{\{\alpha\}:\alpha<\omega_\omega\big\}\cup\big\{[\alpha,\omega_\omega]:\alpha<\omega_\omega\big\}$$ is a base for $X$ of cardinality $\omega_\omega$, and it’s clear that $w(X)=\omega_\omega$.