Ascending chain of compact subspaces Suppose $X$ is Hausdorff and fix a directed set $I$ (or less generally $I=\mathbb{N}$). I will say that a family $\{K_i\}_{i\in I}$ has $*$ property iff


*

*Each $K_i$ is a nonempty, compact subspace of $X$

*$X=\bigcup K_i$

*$K_i\subseteq K_j$ whenever $i\leq j$


I will additionally say that $\{K_i\}$ has $**$ property if the following holds: any subset $F\subseteq X$ is closed if and only if $F\cap K_i$ is closed in $K_i$ for every $i$. I.e. the topology on $X$ is coherent with $\{K_i\}$.
So assume that $X$ is such that there is a family that satisfies both $*$ and $**$. Does it follow that every family that satisfies $*$ also satisfies $**$ (note that for the same underlying index set $I$)? In other words if the topology on $X$ is coherent with some ascending compact chain then is it coherent with every ascending compact chain? Any hint is appreciated.
 A: No, this is very rarely true.  For a rather trivial example, let $X$ be any compact Hausdorff space that is not discrete, and let $I$ be the collection of all nonempty finite subsets of $X$.  If you take $K_i=X$ for all $i$ then $*$ and $**$ both hold, but if you take $K_i=i$ then $*$ holds but not $**$.
For a less trivial example, let $X=\mathbb{R}$ and $I=\mathbb{N}$. Then $X$ satisfies $*$ and $**$ with respect to $K_n=[-n,n]$, and satisfies $*$ but not $**$ with respect to $K_n=[-n,n]\setminus(0,1/n)$.

Here's a nontrivial example where it is true.  Let $X=\omega_1$ with its order topology and $I=\omega_1$.  Taking $K_\alpha=\alpha+1$, then $*$ and $**$ are both satisfied.  And if $(K_\alpha)$ is any family satisfying $*$, then $**$ automatically holds.  Indeed, $X$ is first-countable and in particular countably generated, and if $C\subset X$ is any countable subset, then $C\subseteq K_\alpha$ for some $\alpha$.
A: Every space has (*)-property in its general form. Let $I$ be the set of all finite subsets of $X$ under inclusion; this is a directed set. For $i \in I$ jsut use $K_i = i$, that same finite set, which is always compact. 3. is obvious and 2. as well, as we at least ahve all singletons.
So (*) is too general, any space without (**) (a non-$k$ space) is a counterexample).
So maybe for linear orders it holds, but it's dubious, not for directed sets.
