Inverse limit, contininuity question to prove $ X_{\infty} \cong \bigcap_{n \in \mathbb{N}}X_n $ 
Let $\left \{ X_n: n \in \mathbb{N} \right \}$ be a  nested metric spaces sequence , i.e , $X_{n+1}\subseteq X_{n}$, for all $ n \in \mathbb{N}$. Define $f_n: X_{n+1} \longrightarrow X_{n}$ by $f_n(x)=x$, for all $x \in X_{n+1}$. 
I want to prove  $ X_{\infty} \cong \bigcap_{n \in \mathbb{N}}X_n $
  using that 
\begin{align*} \alpha: X_{\infty} &\longrightarrow \bigcap_{n \in \mathbb{N}}X_n  \\
x=(x,x,\cdots&) \longrightarrow x
\end{align*}

Can i see the open sets in $ \bigcap_{n \in \mathbb{N}}X_n$ as open sets in $X_1$ by the subspace topology and basically use this?
https://proofwiki.org/wiki/Function_to_Product_Space_is_Continuous_iff_Composition_with_Projections_are_Continuous
Any answer would be helpful, thanks.
 A: You could use that, but a direct approach can be taken. It's not so clear to me how you would post-compose with projection, as your map has domain $X_\infty$. Even if you take $\alpha^{-1}$, you will still have to prove that it's open, so regardless you wont be able to finish this off via this tactic (or so I suspect). A solution via elementary methods follows.
The elements of $X_\infty$ are precisely the sequences $(x_i)_{i \geq 1}$ such that $x_i = x_j$ for all $i,j$. In other words, $X_\infty$ if made up of constant sequences. But as $x_i \in X_i$ for all $i$, moreover the constant term must belong to all spaces $X_i$. This (modulo some computations) says that the map $\alpha$ is bijective. 
The key fact here is the following: if $U_i \subset X_i$ are open sets for each $i$, then
$$
X_\infty \cap \prod_{i \geq 1}U_i = X_\infty \cap U_1 \times \prod_{i \geq 2}X_i.
$$
We prove the non trivial inclusion: take $(x_i)_i \in X_\infty \cap U_1 \times \prod_{i \geq 2}X_i$. Since $x_j = x_1$ for all $j \geq 2$, in particular $x_j = x_1 \in U_i$ for all $j$. Thus $(x_i)_i \in X_\infty \cap \prod_{i \geq 1}U_i$.
Now, observe that if $U \subset \cap_n X_n$ is open,
$$
X_\infty \cap U \times \prod_{i \geq 2} X_i = \{(x_i)_i \in X_\infty  :  x_1 \in U\} = \{(x_i)_i \in X_\infty  :  x_i \in U\} = \alpha^{-1}(U)
$$
is open. Reciprocally, if we have a basic open set $X_\infty \cap \prod_{i \geq 1} U_i$ with finitely proper open sets, then
$$
\alpha(X_\infty \cap \prod_{i \geq 1} U_i) = \alpha(X_\infty \cap U_1 \times  \prod_{i \geq 2}X_i) = U_1
$$
is open.
A: The general approach to introduce the concept of an inverse limit is via a universal property. An inverse system $\mathbf{X} = (X_\alpha, p^\beta_\alpha,A)$ consists of a directed set $A$, objects $X_\alpha$ for each $\alpha \in A$ and morphisms $p^\beta_\alpha : X_\beta \to X_\alpha$ for each pair $(\alpha,\beta)$ with $\beta \ge  \alpha$ which are subject to the conditions $p^\alpha_\alpha = id_{X_\alpha}$ and $p^\beta_\alpha \circ p_\beta^\gamma = p_\alpha^\gamma$ for $\gamma \ge \beta \ge \alpha$. A morphism $\mathbf{f} = (f_\alpha) : X \to \mathbf{X}$ from an object $X$ to the inverse system $\mathbf{X}$ is a collection of morphisms $f_\alpha : X \to X_\alpha$, $\alpha \in A$, such that $p^\beta_\alpha \circ f_\beta = f_\alpha$ for $\beta \ge \alpha$. An inverse limit of $\mathbf{X}$ is a morphism $\mathbf{p} = (p_\alpha) : X_\infty \to \mathbf{X}$ with the following universal property: For each morphism $\mathbf{f} : X \to \mathbf{X}$ there exists a unique morphism $\phi : X \to X_\infty$ such that $\mathbf{p} \circ \phi = \mathbf{f}$. Here $\mathbf{p} \circ \phi = (p_\alpha \circ \phi)$. Note that we do not speak about the inverse limit of $\mathbf{X}$. There may be many inverse limits, but the universal property shows that any two inverse limits $\mathbf{p} : X_\infty \to \mathbf{X}$ and $\mathbf{p'} : X'_\infty \to \mathbf{X}$ are canonically isomorphic which means that there exists a unique isomorphism $h : X_\infty \to X'_\infty$ such that $\mathbf{p'} \circ h = \mathbf{p}$. 
This works in any category, but let us do it for objects = topological spaces and morphisms = continuous maps. In this situation it is well-known that each inverse system has a standard inverse limit given by
$$X_\infty = \{ (x_\alpha) \in \prod_{\alpha \in A} X_\alpha \mid p_\alpha(x_\beta) =x_\alpha \text{ for all } \beta \ge \alpha \},$$
$$p_\alpha = \text{restriction of the projection} \prod_{\alpha \in A} X_\alpha \to  X_\alpha .$$
You can easily prove this. The only fact you have to know is that the product topology is the coarsest topology such that all projections are continuous which implies that a function to the product is continuous iff all compositions with projections are continuous.
Now let us come to your question. We generalize it a little by considering a nested  sequence of topological spaces. So let us show that $X_{\infty} = \bigcap_{n \in \mathbb{N}}X_n$ and $\mathbf{i} = (i_n) : X_\infty \to \mathbf{X}$ with inclusions $i_n : X_\infty \to X_n$ has the desired universal property.
Let $\mathbf{g} = (g_n) : Y \to \mathbf{X}$ be a map defined on a space $Y$.
We have $g_1(y) = g_n(y)$ for all $y \in Y$ and all $n$. In fact, $g_1(y) = f^n_1(g_n(y)) = g_n(y)$ because $f^n_1$ is the inclusion of $X_n$ into $X_1$. 
Hence $g_1(Y) = g_n(Y) \subset X_n$ for all $n$. We conclude $g_1(Y) \subset X_{\infty}$. Therefore $g : Y \to X_\infty, g(y) = g_1(y)$ is a well-defined continuous map which satisfies $\mathbf{i} \circ g = \mathbf{g}$. Let $g' : Y \to X_\infty$ be  any map such that $\mathbf{i} \circ g' = \mathbf{g}$. Then $i_1 \circ g = i_1 \circ g'$ which implies $g = g'$ because $i_1$ is injective.
