Question:
Find a topological space that is
limit point compact (or weakly countably compact) + $\neg$countably compact + $\neg$Lindelöf
Notice such a space cannot be $T_1$ (every one-point set is closed) .
Using $\pi$-base, below are spaces satisfying the first two requirements:
However, all of them are Lindelöf.
Any hint is appreciated.
Interestingly, if Lindelöf is replaced by separable (i.e. there exists a countable dense subset), there is one:
On
Elaborating on user87690's answer, let $X$ be ordered by $<$, lack a least point, and for each open $U\subseteq X$, $a\in U$ and $b\geq a$ implies $b\in U$. (For example, the the right-ordered reals.) Then the uncountable sum of $X$ does the trick. Here are its basic properties.
This space is $T_0$ as it is the sum of $T_0$ spaces, but fails $T_1$ as it contains the non-$T_1$ space $X$. The space is completely (i.e. hereditarily) normal as $X$ is.
The space is weakly countably compact as every non-empty set has a limit point, since every non-empty subset of $X$ has a limit point (e.g. any lesser point).
The space is not even weakly Lindelöf: for each countable subcollection of the open cover using each of the uncountable copies of $X$, its union is closed and fails to contain uncountably-many copies of $X$.
The space is not countably compact as it's not even pseudocompact: a function that sends each copy of $X$ to an integer within $\mathbb R$ is continuous, and with infinitely-many copies of $X$ this can be unbounded.
The counterexample 5.7 from the book by Steen a Seebach works – an uncountable topological sum of two-point indiscrete spaces. Because of the indiscrete pieces, every nonempty subset has a limit point. And because of the uncountable sum, the space is neither countably compact not Lindelöf.
A $T_0$ example may be obtained if instead of the two-point indiscrete space we sum copies of a $T_0$ space without closed points, e.g. $ω$ with the topology $\{[0, n): n ∈ ω\}$.