For the lower-limit line, $\chi =d=L=c={\mathrm{\aleph }}_0$ and $w=2^{{\mathrm{\aleph }}_0}$ under choice

Assume Choice and let $S$ be the lower-limit line. At $x$, the intervals $[x,x+1\mathrm{/}n)$ form a countable local base. No finite family is a local base: the intersection of finitely many neighbourhoods is still a neighbourhood and contains some $y>x$, whereas $[x,y)$ cannot contain any member of a finite local base contained in that intersection. Hence $\chi (S)={\mathrm{\aleph }}_0$.

The rationals are countable and meet every nonempty half-open interval, so $d(S)\le {\mathrm{\aleph }}_0$; no finite set is dense, since a short half-open interval can avoid it. The published lower-limit-line lemma gives Lindelöfness, while the cover $\{[-n,n):n\ge 1\}$ has no finite subcover, so $d(S)=L(S)={\mathrm{\aleph }}_0$. Density bounds cellularity above, and the disjoint family $\{[n,n+1):n\in \mathbb{Z}\}$ bounds it below, giving $c(S)={\mathrm{\aleph }}_0$.

All half-open intervals form a basis of cardinality at most $\mathrm{\mid }\mathbb{R}{\mathrm{\mid }}^2=\mathrm{\mid }\mathbb{R}\mathrm{\mid }$. Conversely, well order any basis and assign to each $x$ its first member $B_x$ with $x\in B_x\subseteq [x,x+1)$; if $x<y$, then $B_y$ cannot contain $x$, so $B_x\mathrm{\ne }{B}_y$. Thus $w(S)=\mathrm{\mid }\mathbb{R}\mathrm{\mid }=2^{{\mathrm{\aleph }}_0}$.