# Is a finite increasing chain of closed sets the closure of the union of the interiors of the relative complements? [closed]

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.

• 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 $C$. – Emil Jeřábek Mar 3 '20 at 15:54
• I see. Thank you, this was very helpful! – ervx Mar 3 '20 at 17:15