If $f_{n}$ has a dense image, then $\bigcap (f_{1}\circ\cdots\circ f_{n})(X_{n})$ is dense Let $\{(X_{n}, d_{n})\}_{n\in\mathbb{N}}$ be a sequence of complete metric spaces and $\{f_{n}: X_{n}\to X_{n − 1}\}_{n\in \mathbb{N}}$ a sequence of continuous functions. If $f_{n}$ has a dense image for each $n\in \mathbb{N}$, show that $\bigcap_{n\in \mathbb{N}}(f_{1}\circ f_{2}\circ \cdots \circ f_{n}) (X_{n})$ is dense at $X_{0}$.
My attemps:
if $f_{n}$ has a dense image then $\overline{f_{n}(X_{n})}=X_{n-1}$ I want to show that
$$\overline{\bigcap_{n\in \mathbb{N}}(f_{1} \circ f_{2} \circ \cdots \circ f_{n})(X_{n})} = X_{0}. $$
Let $F_{n}=(f_{1}\circ f_{2} \circ \dots ◦ f_{n})(X_{n})$. By induction, I see that $F_{n+1} ⊆ F_{n} ⊆ X_{0}$. Indeed, if $n=1$, then
$$F_{2}=(f_{1}\circ f_{2})(X_{2})=f_{1}(f_{2}(X_{2}))\subseteq f_{1}(\overline{f_{2}(X_{2}}))=f_{1}(X_{1})=F_{1}$$
if we assume that $F_{k+1}\subseteq F_{k}\subset X_{0}$. Then,
$$F_{k+2} = (f_{1} \circ f_{2} \circ \cdots \circ f_{k+1} \circ f_{k+2})(X_{k+2}) = (f_{1} \circ f_{2} \circ \cdots \circ f_{k+1})(f_{k+2}(X_{k+2}))\subseteq (f_{1} \circ f_{2} \circ \cdots \circ f_{k+1})(\overline{f_{k+2}(X_{k+2})}) = (f_{1} \circ f_{2} \circ \cdots \circ f_{k+1})(X_{k+1})= F_{k+1}\subseteq F_{k}\subseteq X_{0}$$
from the above we have
$$\cdots\subseteq F_{n+1}\subseteq F_{n} \subseteq \cdots \subset F_{2}\subseteq F_{1}$$
then
$$\cdots\subseteq \overline{F_{n+1}}\subseteq \overline{F_{n}} \subseteq \cdots \subset \overline{F_{2}}\subseteq \overline{F_{1}}=X_{0}$$
then
$$\overline{\bigcap_{n\in \mathbb{N}}F_{n} }\subseteq \bigcap_{n\in \mathbb{N}}\overline{F_{n}}\subseteq X_{0} .$$
is my reasoning correct?.I've been thinking that Biare's theorem could be used, but I don't know if it's possible since $F_{n}'s$ don't necessarily open.
If it is, how could the other containment prove?. I appreciate any help
 A: I think I found a solution but its rather technical.
The following is more of a sketch than a solution with full details.
Let $x_0\in X$ and $\varepsilon>0$ be given.
In a first step, one has to
construct points $y_{k,m}$ for $m,k\in\Bbb N_0$ with $k\leq m$
such that
$$
\begin{aligned}
y_{k,m} &\in X_k
\\
y_{0,0} &= x_0
\\
f_k(y_{k,m}) &= y_{k-1,m}
\quad\forall m\geq k\geq 1,
\\
d_k(y_{k,m},y_{k,m+1}) &< \varepsilon 2^{-m-1}
\quad\forall m\geq k\geq 0.
\end{aligned}
$$
(This requires continuity of these functions
and density of the images, and can be done recursively in $m$.
This step requires some technical details that I skipped here.)
Then, one can show that $\{y_{k,m}\}_{m\geq k}$ is a Cauchy sequence
for each $k$.
Thus, there exist points $z_k$ with
$$
\lim_{m\to\infty} y_{k,m} = z_k \quad\forall k\geq0.
$$
Due to continuity of $f_k$, one can show that
$f_k(z_k)=z_{k-1}$ holds.
Therefore, we have
$$
z_0\in F_k \quad\forall k\geq 1
\qquad\text{and therefore}\qquad
z_0 \in F_\infty:=\bigcap_{k\in\Bbb N} F_k.
$$
Finally, one can also show that
$$d_0(z_0,x_0)<\varepsilon$$
holds. Since $\varepsilon>0$ was arbitrary, this shows that
$F_\infty$ is dense in $X_0$.
A: Here is a slight reformulation of @supinf's answer, elucidating the skipped parts:

Setting. For $0 \leq i \leq j$, define the function $f^{i\leftarrow j} $ by the composition
$$ X_i \stackrel{f^{i\leftarrow j}}{\longleftarrow} X_{j}
\quad = \quad
X_i \stackrel{f_{i+1}}{\longleftarrow} X_{i+1} \stackrel{f_{i+2}}{\longleftarrow} \cdots  \stackrel{f_j}{\longleftarrow} X_j, $$
i.e.,
\begin{align*}
f^{i\leftarrow j}
:= \begin{cases}
f_{i+1} \circ \dots \circ f_{j}, & \text{if $i < j$}, \\
\operatorname{id}_{X_j}, & \text{if $i = j$}.
\end{cases}
\end{align*}
In particular, note that $f^{j \leftarrow j+1} = f_{j+1}$ and $f^{i\leftarrow j} \circ f^{j \leftarrow k} = f^{i \leftarrow k}$. Using this, OP's question boils down to showing that
$$ \bigcap_{j\geq 0} f^{0\leftarrow j}(X_j) $$
is dense in $X_0$. We establish by showing that in fact the following subset
$$ X^{0\leftarrow} := \biggl\{ z_0 \in X_0 : \text{$\exists$ $\{z_j\}_{j\geq 0} \in \prod_{j\geq 0}X_j$ s.t. $z_j = f_{j+1}(z_{j+1})$ for all $j \geq 0$} \biggr\} $$
is dense in $X_0$.
(Note. It is clear that $X^{0\leftarrow} \subseteq \bigcap_{j\geq 0} f^{0\leftarrow j}(X_j)$. What is not clear to me is whether they coincide or not.)
Proof. Fix any open ball $B(x_0,\epsilon) \subseteq X_0$. Then we construct a sequence $\{x_n\}_{n\geq 0}$ recursively as follows:

Suppose $ x_n \in X_n $ has been chosen. Since $f_{n+1}(X_{n+1})$ is dense in $X_{n}$, we can find a sequence $\{y_k\}_{k\geq 1}$ in $X_{n+1}$ such that
$$\lim_{k\to\infty} f_{n+1}(y_k) = x_{n}.$$
Then for each $j = 0, \dots, n$, by continuity of $f^{j\leftarrow n}$ and $d_j(\cdot, \cdot)$, we have
$$ \lim_{k\to\infty} d_j\bigl( f^{j\leftarrow n}(x_n), f^{j\leftarrow n}(f_{n+1}(y_k)) \bigr) = 0. $$
From this observation, we can pick $x_{n+1} \in X_{n+1}$ so that
$$ d_j\bigl( f^{j \leftarrow n}(x_n), f^{j \leftarrow n+1}(x_{n+1}) \bigr) < \epsilon 2^{-n-1} $$
for each $0 \leq j \leq n$.

Schematically, $x_n$'s are chosen as:

Now we make several observations:

*

*By the construction, for each $j \geq 0$, the sequence $\{ f^{j\leftarrow n}(x_n) \}_{n = j}^{\infty}$ is Cauchy in $X_j$, and so,
$$ z_j := \lim_{n\to\infty} f^{j\leftarrow n}(x_n) $$
exists in $X_j$.


*For each $j \geq 0$, we have
$$ z_j
= \lim_{n\to\infty} f^{j\leftarrow n}(x_n)
= \lim_{n\to\infty} f_{j+1}(f^{(j+1)\leftarrow n}(x_n))
= f_{j+1}(z_{j+1}). $$
In particular, $z_0 \in X^{0\leftarrow}$.


*We have
\begin{align*}
d_0(x_0, z_0)
&= \lim_{N\to\infty} d_0(x_0, f^{0\leftarrow N}(x_N)) \\
&\leq \lim_{N\to\infty} \sum_{n = 0}^{N-1} d_0(f^{0\leftarrow n}(x_n), f^{0\leftarrow n+1}(x_{n+1})) \\
&< \sum_{n = 0}^{\infty} \epsilon 2^{-n-1} \\
&= \epsilon,
\end{align*}
and so, $z_0 \in B(x_0, \epsilon)$.
This proves that $B(x_0, \epsilon) \cap X^{0\leftarrow} \neq \varnothing$ for any open ball $B(x_0, \epsilon)$ in $X_0$, establishing the desired claim.
