# Inverse Limit of Compact/Hausdorff Spaces is Nonempty and Compact

#### joypav

##### Active member
Problem:

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 online?
Or perhaps can give an outline for the proof?

#### Euge

##### MHB Global Moderator
Staff member
Hi joypav ,

If $X = \emptyset$, then to each $x\in X$ corresponds an index $k$ such that $x_k \neq f_k(x_{k+1})$. Since $X_k$ is Hausdorff, $x_k$ and $f_k(x_{k+1})$ are separated by some open sets $U_k \ni x_k$ and $V_k\ni f_k(x_{k+1})$; continuity of $f_k$ allows us to find an open set $W_{k+1} \ni x_{k+1}$ such that $f_k(W_{k+1}) \subset V_k$. Set $\Sigma(k) := X_1 \times X_2 \times \cdots \times X_{k-1}\times U_k \times W_{k+1}\times X_{k+2}\cdots$, for $k\in \Bbb N$. The collection $\{\Sigma(k): k\in \Bbb N\}$ is an open cover of $\prod_n X_n$. Tychonoff's theorem ensures compactness of $\prod_n X_n$ (since each $X_n$ is compact), so there are indices $k_1 < k_2 < \cdots < k_j$ such that $\prod_n X_n = \Sigma(k_1)\cup \cdots \cup \Sigma(k_j)$. If $v_k \in V_k$, then $v = (v_1,v_2,v_3,\ldots)\in \Sigma(k_m)$ for some $m$. Thus $v_{k_m}\in U_{k_m}$, so that $v_{k_m} \in V_{k_m}\cap U_{k_m}$, a contradiction.

Since $\prod_n X_n$ is compact, to show that $X$ is compact, it suffices to show $X$ is closed in $\prod_n X_n$. Take a point $x\notin X$, and let $k\in \Bbb N$ such that $x_k \notin f_k(x_{k+1})$. Using the same notation as above, consider the set $\Sigma(k)$, which is an open neighborhood of $x$. Given $y\in \Sigma(k)$, $y_k \in U_k$ and $y_{k+1}\in W_{k+1}$. Since $f(W_{k+1})\subset V_k$ and $U_k$ is disjoint from $V_k$, then $y_k \neq f_k(y_{k+1})$. Consequently, $y\notin X$. Therefore, $X$ has open complement, i.e., $X$ is closed.