Invers limit function.

Question

Consider a countable collection of sets {Xn}n≥1 and functions fn : Xn+1 → Xn. We could organize these into a diagram that looks as follows
$..\rightarrow_f X3\rightarrow _f X2 \rightarrow _f X1.$

Define $lim_\leftarrow n$ Xn (the ‘inverse limit’ of the Xn’s) to be the subset of
$\Pi_{n\geq 1} Xn$ consisting of those elements $(x_1, x_2, . . .)$ for which $x_n = f_n(x_n+1)$.

(a) Equip each $Xn$ with the discrete topology, give $\Pi_{n≥1} X_n$ the product topology and give $lim_{\leftarrow n}$ $X_n$ the corresponding subspace topology.
Explain why this topology does not have to be the discrete topology.

(b) Show that $lim_{\leftarrow n}$ $X_n$ is a closed subset of $\Pi_{n≥1} X_n$ and that it is Hausdorff.

(c) Use Tychonoff’s theorem to show that if each $X_n$ is a finite set, then $lim_{\leftarrow n}$ $X_n$ is compact

(d) The inverse limit $lim_{\leftarrow n}$ $X_n$ can be infinite, despite all the $X_n$ being finite . In this case, why does the result of part (c) give another way to conclude that $lim_{\leftarrow n}$ $X_n$ does not have the discrete topology?

Answer 1

(b) We will show that $\lim X_n$ is closed by showing that its complement is open. So let $(x_1, x_2,\ldots)\not\in\lim X_n$. In particular there is $f_k$ such that $f_k(x_{k+1})\neq x_k$. Since $f_k$ is continous (which has to be assumed, otherwise (b) is not true) then there exists an open neighbourhood $U\subseteq X_{k+1}$ of $x_{k+1}$ such that $f_k(u)\neq x_k$ for any $u\in U$. Put

$$A_n=X_n \mbox{ if } n\neq k+1$$ $$A_{k+1} = U$$ What we've shown is that $\prod A_n\cap\lim X_n=\emptyset$ so $\prod A_n$ is a subset of a complement of $\lim X_n$. On the other hand $\prod A_n$ is open in $\prod X_n$ by definition. This proves that the complement of $\lim X_n$ is open and thus $\lim X_n$ is closed.

(c) By Tychonoff's theorem $\prod X_n$ is compact and since $\lim X_n$ is closed by (b) then it has to be compact.

(d) That's because a Hausdorff space is both discrete and compact if and only if it is finite.