Let $X$ be space. A space $X$ is called right-separated if it can be well-ordered in such a way that every initial segment is open in $X$. See the related link (left-separated).
How could we show that hereditary lindelof number is the supremum of cardinalities of right-separated subspaces of $X$?
In one direction, suppose $\kappa$ is a cardinal and $X$ has a subspace $Y$ with an open cover $\mathcal U$ that has no subcover of size $<\kappa$. Define in parallel a $\kappa$-sequence of points $y_i\in Y$ and a $\kappa$-sequence of sets $U_i\in\mathcal U$ by the following induction of length $\kappa$, in which one new $y_i$ and one new $U_i$ are chosen at each stage. At any stage, there are fewer than $\kappa$ $U_i$'s chosen at previous stages, so there is a point in $Y$ not covered by those $U_i$'s; choose one and make it the next $y_j$ in your sequence. Then choose a set in $\mathcal U$ that contains this $y_j$ and make it the next $U_j$. After $\kappa$ steps, the chosen $y_i$'s form a set $S$ of size $\kappa$, and it is right separated because any initial segment, say up to but not including $y_i$, is the intersection of $S$ with $\bigcup_{j<i}U_j$.
For the other direction, suppose $X$ has a right-separated subset $S$ of size $\kappa$. We may assume that the length of the well-ordering witnessing right-separation is $\kappa$, because if it isn't we can just delete some elements from the end and retain only the first $\kappa$ points of $S$. The proper initial segments of $S$ form an open cover of $S$. If $\kappa$ is a regular cardinal, then this cover of $S$ has no subcover of size $<\kappa$, so we're done. If $\kappa$ is singular then there is a subcover of size cf$(\kappa)$, so we have to work a little harder. The previous argument can be applied to every regular cardinal $\lambda<\kappa$, in particular to every successor cardinal $<\kappa$. It gives a subset of $X$ with an open cover of size $\lambda$ with no subcover of smaller cardinality. So the hereditary Lindelöf number of $X$ is at least $\lambda$. Since these $\lambda$'s are cofinal in $\kappa$, we're again done.
