Let $X$ be a topological space. Suppose there are closed subsets $X=:F_{k}\supseteq F_{k-1}\supseteq\cdots\supseteq F_{1}\supseteq F_{0}:=\emptyset$. Is it true that

$\overline{\bigcup_{j=1}^{k}\operatorname{Int}(F_{j}\setminus F_{j-1})}=X$?

If $k=1$, the result is trivial. For $k>1$, we have $$ \overline{\bigcup_{j=1}^{k}\operatorname{Int}(F_{j}\setminus F_{j-1})}=\overline{\operatorname{Int}(X\setminus F_{k-1})}\cup\overline{\bigcup_{j=1}^{k-1}\operatorname{Int}(F_{j}\setminus F_{j-1})}=\overline{X\setminus F_{k-1}}\cup\overline{\bigcup_{j=1}^{k-1}\operatorname{Int}(F_{j}\setminus F_{j-1})}. $$

When $k=2$, the RHS simplifies to $$ \overline{X\setminus F_{1}}\cup\overline{\operatorname{Int}(F_{1})}=\overline{X\setminus\operatorname{bd}(F_{1})}=X. $$

I'm not sure if it's possible to carry out a similar reduction when $k>2$; or, if this result is indeed true. Any help is appreciated. Thank you.

  • 1
    $\begingroup$ This is true, even if the sets do not form a chain. It follows by induction on $k$ using the observation that $\mathrm{int}(A\setminus B)\subseteq\overline{\mathrm{int}(A\setminus C)\cup\mathrm{int}(C\setminus B)}$ for closed $C$. $\endgroup$ – Emil Jeřábek Mar 3 '20 at 15:54
  • $\begingroup$ I see. Thank you, this was very helpful! $\endgroup$ – ervx Mar 3 '20 at 17:15