Inverse limit of compact Hausdorff spaces is nonempty and compact

Question

Suppose that $\left\{ X_n \right\}_{n=1}^{\infty}$ is a sequence of compact Hausdorff spaces and for each $n, f_n : X_{n+1} \rightarrow X_n$ is a continuous function (not necessarily onto). Show that:

$X = \lim_{\leftarrow} \left\{ X_n, f_n \right\}_{n=1}^{\infty} \neq \emptyset$

Furthermore, show that $X$ is compact.

I have seen a proof for a general inverse limit system, with $D$ being its directed set. However, I assume the proof differs in the problem I've stated. (here I guess our $D = \Bbb{N}$).

Does anyone know of a proof for this that can be found online? Or perhaps can give an outline for the proof?

Answer 1

As jgon has pointed out if is fairly simple to show that $X$ is compact using Tychonoff's theorem.

To show it is non-empty we prove the following.

For every $n\in \mathbb N$ and $x_n\in X_n$ there exists a $x\in X$ such that $\pi_n x=x_n$.

Proof: For convenience let us denote, for any $k\leq j$, $$f_{kj}:=f_{j-1}\circ f_{j-2}\circ\dots\circ f_{k+1}\circ f_k:X_j\to X_k$$ Fix an arbitrary $n$, $x_n$. Define for any $j\geq n$, $$L_j=\left\{(\tilde x)_{\ell\in \mathbb N}\in\prod_{\ell \in \mathbb N}X_\ell:\pi_n(\tilde x)=x_n,~f_{kj}\tilde x_j=\tilde x_k~\text{for all}~k\leq j\right\}.$$ We can see that each $L_j$ is closed in the product space by showing that every point in the complement is an interior point by using the continuity of the $f_k$'s. Thus each $L_j$ is compact. Furthermore, by construction, for each $(\tilde x)\in \bigcap_{j\in \mathbb N}L_j$ we have that $\pi_n \tilde x=x_n$, and for every $k\leq j$ we have $f_{kj}\tilde x_j=\tilde x_k$. We thus just need to show that $\bigcap_{j\in \mathbb N}L_j\neq \emptyset$. We do this by utilizing the fact that in a compact space if a collection of closed sets satisfies the finite intersection property then the intersection of all the sets in the collection is nonempty. Now for any finite $\{j_1,\dots j_m\}\subset \mathbb N$ there exists some $j\in \mathbb N$ such that $j_k\leq j$ for all $k\in \{1,\dots,m\}$. Finally $\emptyset\neq L_{j}\subset \bigcap_{k=1}^mL_{j_k}$, so we can conclude that every $\{L_j\}_{j\in \mathbb N}$ is a collection of closed sets with the finite intersection property, so by the above $\bigcap_{j\in \mathbb N}L_j\neq \emptyset$.