Let $\left \{ X_n: n \in \mathbb{N} \right \}$ be a sequence of metric spaces such that $X_{n+1}\subseteq X_{n}$, for all $ n \in \mathbb{N}$. Define $f_n: X_{n+1} \longrightarrow X_{n}$ as $f_n(x)=x$, for all $x \in X_{n+1}$. Prove $X_{\infty} \cong \bigcap_{n \in \mathbb{N}}X_n $.
This is more like a check to see if my work is correct:
What i tried was to define \begin{align*} \alpha: X_{\infty} &\longrightarrow \bigcap_{n \in \mathbb{N}}X_n \\ x=(x,x,\cdots&) \longrightarrow x \end{align*}
I proved $\alpha$ is bijective, by proving is open i said the following:
Let $U$ be a open subset of $X_\infty$ and $(x_n) \in U $ then there exist (We proved is a basis for $X_\infty$) $\rho^{-1}(U_{n_0})$, $ n_0 \in \mathbb{N}$ such that $$ (x_n) \in \rho^{-1}(U_{n_0}) \subset U $$ Hence $\rho((x_n)) \in U_{n_0} $ , i.e, $x \in U_{n_0}$ ($\rho$ is just the projection map restricted to $X_\infty$).
Now here comes the tricky part:
One can see that $f_n$ is continous for all $n \in \mathbb{N}$, in fact is an homeomorphism (is the inclution map) so that $f^{-1}(U_{n_0})$ is an open set in $X_{n_0+1}$ and I can keep going like this, for all $N\geq n_0$, also because $U_{n_0}$ is open in $X_{n_0}$, it is also open for every $X_1\supseteq X_2\supseteq \cdots \supseteq X_{n_0-1}$
In that case I've proved that there exist open set that contains $x$ for all $X_n$ , so if $U$ is open in $X_\infty$ then $\alpha(U)$ is open.
Is this correct?
A truly rigorous verification of your proof may depend on the exact definition of inverse limit which you are using (for instance, in Engelking’s “General topology” it required to provide the bounding maps from $X^m$ to $X^n$ for any $m\ge n$). Next, you didn’t define what is $f$ (in $f^{-1}(U_{n_0})$). At last, opennes of $U_{n_0}$ in $X_{n_0}$ does not imply that $U_{n_0}$ is open in $X_n$ for $n<n_0$, because it is not given that $X_n$ is an open subspace of $X_{n_0}$. On the other hand, it seems that it suffices to have $U_{n_0}$ open in $X_{n_0}$, because then $(x_n) \in \rho^{-1}(U_{n_0}) \subset U$, so then $x\in U_{n_0}\cap \bigcap_{n\in\Bbb N} X_n$, which should be an open subset of $\bigcap_{n\in\Bbb N} X_n $ (I assumed that $f_m$’s are topological embeddings, so $X_{m+1}$ is a subspace of $X_m$ for each $m$).