MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $X$?

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.