Compactness and directed systems of subspaces Let $X$ be a topological space and let $K$ be a subspace of $X$. It is easy to verify the claim below:

Let $\{ U_j : j \in J \}$ be a directed system of open subspaces of $X$ with the following property:
  
  
*
  
*We have $\bigcup_{j \in J} U_j = X$.
  
  
  If $K$ is a compact subspace of $X$, then there is some $j \in J$ such that $K \subseteq U_j$.

Less well known is this fact:

Let $\{ F_\alpha : \alpha < \lambda \}$ be an increasing sequence of closed subspaces of $X$ with the following properties:
  
  
*
  
*If $\gamma$ is a limit ordinal $< \lambda$, then $F_\gamma = \bigcup_{\alpha < \gamma} F_\alpha$ and has the colimit topology.
  
*$X = \bigcup_{\alpha < \lambda} F_\alpha = X$ and has the colimit topology.
  
*If $x \in X \setminus F_\alpha$, then $\overline{\{ x \}} \subseteq X \setminus F_\alpha$.
  
  
  If $K$ is a compact closed subspace of $X$, then there is some $\alpha < \lambda$ such that $K \subseteq F_\alpha$.

Question. Is there a result generalising both of the above statements?
 A: Here is a common generalization:

Let $\lambda>0$ be an ordinal and let $\{ F_\alpha : \alpha < \lambda \}$ be an increasing sequence of subspaces of $X$ with the following properties:
  
  
*
  
*If $\gamma$ is a limit ordinal $< \lambda$, then $F_\gamma = \bigcup_{\alpha < \gamma} F_\alpha$ and has the colimit topology.
  
*$X = \bigcup_{\alpha < \lambda} F_\alpha = X$ and has the colimit topology.
  
*If $x \in X \setminus F_\alpha$, then $\overline{\{ x \}} \subseteq X \setminus F_\alpha$.
  
  
  If $K$ is a compact subspace of $X$, then there is some $\alpha < \lambda$ such that $K \subseteq F_\alpha$.

This obviously contains your second result (with the minor correction that $\lambda$ can't be $0$) as a special case: we have just removed the requirement that $K$ and the $F_\alpha$ are closed.  Let's see how it contains your first result as a special case.  If $\{U_j:j\in J\}$ is a directed family of open sets with $X=\bigcup U_j$, pick a well-ordering on $J$ and write $V_j=\bigcup_{i<j} U_j$.  Since the $U_j$ are open, the sets $V_j$ automatically satisfy the three conditions above (any space is the colimit of any open cover).  If $K\subseteq X$ is compact, we conclude that $K\subseteq V_j$ for some $j$; choose the least such $j$.  Then $j$ must be either a successor or $0$ (otherwise we could repeat the argument with $\lambda=j$ to find a smaller one); if $j\neq 0$, let $K'=K\setminus U_{j-1}$.  Then $K'\subseteq V_{j-1}$ and $K'$ is compact.  Repeating the argument with $K'$ in place of $K$, we get a minimal $j'<j$ such that $K'\subseteq V_{j'}$.  Continuing by induction, we get a descending sequence $j>j'>j''>\dots$ which must terminate since $J$ is well-ordered.  When it terminates, we find that $K$ is covered by the finitely many sets $U_{j-1},U_{j'-1},U_{j''-1},\dots$.  Since the $U_j$ are directed, this means there is a single $j^*$ such that $K\subseteq U_{j^*}$.
Let us now prove the generalization.  Suppose $X$ is a space with subspaces $F_\alpha$ as above. Suppose $K\subseteq X$ is compact and is not contained in any $F_\alpha$. Passing to a closed unbounded subsequence of the $F_\alpha$, we may choose a point $x_\alpha\in K\cap F_{\alpha+1}\setminus F_\alpha$ for each $\alpha$.  For each $\beta<\lambda$, let $C_\beta=\{x_\alpha:\beta\leq\alpha\}$ and let $D_{\beta\alpha}$ denote the closure of $C_\beta\cap F_\alpha$ in $F_\alpha$.
Let us now prove by induction on $\alpha$ that $D_{\beta\alpha}\cap F_\gamma=D_{\beta\gamma}$ for all $\gamma<\alpha$.  For $\alpha$ limit, note that the induction hypothesis implies that $\bigcup_{\gamma<\alpha} D_{\beta\gamma}$ is closed in $F_\gamma$ for each $\gamma<\alpha$ (since its intersection with $F_\gamma$ is just $D_{\beta\gamma}$) and thus $\bigcup_{\gamma<\alpha} D_{\beta\gamma}$ is closed in $F_\alpha$.  It follows that $\bigcup_{\gamma<\alpha} D_{\beta\gamma}=D_{\beta\alpha}$ and thus that $D_{\beta\alpha}\cap F_\gamma=D_{\beta\gamma}$ for all $\gamma<\alpha$.
Now suppose $\alpha$ is a successor ordinal.  Let $A=\{x_\gamma:\beta\leq\gamma<\alpha-1\}$ and $B=\{x_{\alpha-1}\}$ and note that $C_\beta\cap F_\alpha=A\cup B$.  Thus $D_{\beta\alpha}=(\overline{A}\cap F_{\alpha})\cup (\overline{B}\cap F_{\alpha})$ and so $D_{\beta\alpha}\cap F_{\alpha-1}=(\overline{A}\cap F_{\alpha-1})\cup (\overline{B}\cap F_{\alpha-1})$.  Moreover, since $A\subseteq F_{\alpha-1}$, $\overline{A}\cap F_{\alpha-1}$ is just the closure of $A$ in $F_{\alpha-1}$, which is $D_{\beta(\alpha-1)}$.  Also, $\overline{B}\cap F_{\alpha-1}=\emptyset$ since $x_{\alpha-1}\not\in F_{\alpha-1}$.  Thus $D_{\beta\alpha}\cap F_{\alpha-1}=D_{\beta(\alpha-1)}$ and the desired conclusion for arbitrary $\gamma<\alpha$ follows from the induction hypothesis.
Now let $D_\beta=\overline{C_\beta}$.  By the same argument as in the limit step above, $D_\beta=\bigcup_{\alpha<\lambda} D_{\beta\alpha}$, and so $D_\beta\cap F_\alpha=D_{\beta\alpha}$ for all $\alpha<\lambda$.  In particular, $D_\beta\cap F_\beta=D_{\beta\beta}=\emptyset$ since $C_\beta\cap F_\beta=\emptyset$.
Thus $\bigcap_{\beta<\lambda} D_\beta$ does not intersect any $F_\beta$ and so is empty.  Since the $D_\beta$ are closed and descending and $K$ is compact, there must be a single $\beta$ such that $K\cap D_\beta=\emptyset$ (this is the step where we use $\lambda>0$, since we need to know that $\lambda$ is directed).  But by construction, $x_\beta\in K\cap D_\beta$.  This is a contradiction.
