Let $\{(X_{n},d_{n})\}_{n\in\mathbb{N}}$ be a sequence of complete metric spaces. If $\{f_{n}\colon X_{n}\to X_{n-1}\}_{n\in\mathbb{N}}$ is a sequence of functions continuous such that \begin{equation} \overline{f_{n}(X_{n})}=X_{n-1},\,\forall\,n\in\mathbb{N} \end{equation} show that $\overline{\bigcap_{n\in\mathbb{N}}{(f_{1}\circ\cdots\circ f_{n})(X_{n})}}=X_{0}.$
My attempt: Let $F_{n}=(f_{1}\circ\cdots\circ f_{n})(X_{n})$ be. Let's see what \begin{equation*} \overline{\textstyle\bigcap_{n\in\mathbb{N}}{F_{n}}}=X_{0}.\tag{*} \end{equation*} Note that \begin{align*} F_{n+1}&=(f_{1}\circ\cdots\circ f_{n}\circ f_{n+1})(X_{n+1})\\ &=(f_{1}\circ\cdots\circ f_{n})( f_{n+1}(X_{n+1}))\\ &\subseteq{(f_{1}\circ\cdots\circ f_{n})( \overline{f_{n+1}(X_{n+1})})}\\ &=(f_{1}\circ\cdots\circ f_{n})( X_{n})\\ &=F_{n}. \end{align*} \begin{align*} F_{n}&=(f_{1}\circ\cdots\circ f_{n})( X_{n})\\ &=(f_{1}\circ\cdots\circ f_{n})( \overline{f_{n+1}(X_{n+1})})\\ &\subseteq{\overline{(f_{1}\circ\cdots\circ f_{n})(f_{n+1}(X_{n+1}))}}\\ &=\overline{(f_{1}\circ\cdots\circ f_{n+1})( X_{n+1})}\\ &=\overline{F_{n+1}}. \end{align*} Therefore, we have \begin{equation} F_{n+1}\subseteq{F_{n}}\subseteq{X_{0}}\subseteq{\overline{F_{n}}=\overline{F_{n+1}}},\,\forall\, n\in\mathbb{N}.\tag{1} \end{equation} By $ (1), $ it follows that \begin{equation*} \overline{\textstyle\bigcap_{n\in\mathbb{N}}{F_{n}}}\subseteq{X_{0}}. \end{equation*}
Let $x\in X_{0}.$ By $(1),$ exists a sequence $(F_{n}(x_{n}))_{n\in\mathbb{N}}\subseteq{X_{0}}$ such that $d_{0}(x,F_{n}(x_{n}))<1/n.$ Now a question arises: is the sequence $ (F_{n} (x_{n})) _{n \in \mathbb {N}}$ a Cauchy sequence at $X_{0} $? If that's true, then it must converge because $ X_{0} $ is a complete metric space. Let's say $ F_{n} (X_{n}) \to y \in X_{0}. $ Like $ d_{0} (x, F_{n} (x_{n})) <1 / n, $ then $ x = y$ and it follows that $ x \in \overline {\bigcap_ {n \in \mathbb {N}} {F_ {n}}}. $
I have doubts with the previous argument. I don't know if that's enough to justify that $ X_{0} \subseteq {\overline{\textstyle\bigcap_{n\in\mathbb{N}}{F_{n}}}}. $
Another question: If for every $ n \in \mathbb {N}, $ exists a open set $ U_{n} \subseteq {X_{n}} $ such that $ \overline {f_{n} ({U_{n}})} = X_{n-1}, $ then is the previous result still valid?
Hope you can help me with some suggestions please.
