I've been working through Massey's A Basic Course in Algebraic Topology and I've gotten stuck on the following exercise (V.8.4):
Let $X$ be a regular topological space, and $(\tilde{X}, p)$ a covering space of $X$. Prove that for any compact $C \subset \tilde{X}$, the set $\{\phi \in A(\tilde{X}, p) : \phi(C) \cap C \neq \emptyset\}$ is finite.
Here $A(\tilde{X}, p)$ denotes the group of covering space automorphisms of $(\tilde{X}, p)$.
My thoughts so far: Since C is compact, I can show that, for any $x \in X$, $p^{-1}(x) \cap C$ must be a finite set. I also know that we can cover $C$ with a finite number of open neighborhoods which are mapped homeomorphically onto their images by $p$. Furthermore, under the assumption that the conclusion is wrong, I have shown that there is an elementary neighborhood in $X$ whose preimage has infinitely many components which intersect $C$.
I think these are the basic facts which will lead to the proof, but I'm not sure what to try. In particular, I don't know how to use the fact that $X$ is regular.
This is a lemma due to S.Kinoshita, from his paper "Notes on Covering Transformation Groups", Proceeding of AMS, volume 19, 1968. Lemma itself is on page 422. The paper is freely available from the Jstor.
Edit. Here is the detailed argument.
Let $p: \tilde{X}\to X$ is a covering with the group of automorphisms $G$. First of all, since $X$ is Hausdorff, so is $\tilde{X}$.
First, some notation. Pick a point $x\in X$ let $W$ be an (open) neighborhood of $x$ as in the definition of the covering map: $p^{-1}(W)$ is a disjoint union of open sets $\tilde{W}_j, j\in J$, such that the restriction of $p$ to each $\tilde{W}_j$ is a homeomorphism to $W$. Let $U$ be another neighborhood of $x$, whose closure is contained in $W$. (Such neighborhood exists by regularity of $X$.) For each $j\in J$ define the set $$ \tilde{U}_j:=\tilde{W}_j\cap p^{-1}(U). $$ The sets $\tilde{U}_j$ are pairwise disjoint.
Suppose that for each $j\in J$ we are given a point $$ b_j\in \tilde{U}_j. $$ Define $B=\{b_j: j\in J\}$. Then:
Lemma 1. The set $B$ has no accumulation points in $\tilde{X}$.
Proof. Suppose that such accumulation point $b$ exists. Then, by continuity of $p$, its image $a=p(b)$ belongs to the closure of $U$, hence, $a\in W$. Therefore, $b\in \tilde{W}_j$ for some $j$. However, the sets $ \tilde{W}_j$ are pairwise disjoint. Contradiction. qed
Lemma 2. The action of $G$ on $\tilde{X}$ is properly discontinuous.
Proof. Suppose not. Then there exists a compact $C\subset \tilde{X}$ such that the set $$ G_C:=\{g\in G: gC\cap C\ne \emptyset\}$$ is infinite. I will index elements of $G_C$ as $g_\alpha$, where $\alpha\in A$, $A$ is an infinite set. For each $g_\alpha$ choose $x_\alpha\in C\cap g_\alpha(C)$ and set $y_\alpha=g_\alpha^{-1}(x_\alpha)\in C$. Since the group $G$ acts on $\tilde{X}$ with finite (actually, trivial) point-stabilizers, both sets $$ P=\{x_\alpha: \alpha\in A\}, Q=\{y_\alpha: \alpha\in A\} $$ cannot be finite at the same time. I will assume that $Q$ is infinite, the argument when $P$ is infinite is similar.
By compactness of $C$, the subset $Q$ has an accumulation point $c\in C$:
for every neighborhood $V$ of $c$, the intersection
$$ Q\cap V $$ is infinite. (Here we use the fact that $\tilde{X}$ is Hausdorff.)
I will now use the notation from Lemma 1. Let $W$ and $U$ be neighborhoods of $x=p(c)$ as in Lemma 1. By definition, $c$ belongs to a unique $\tilde{U}_i$, $i\in J$. Let $V:=\tilde{U}_i$. We have infinitely many $y_\alpha$'s which belong to $V$; we also have their images $x_\alpha=g_\alpha(x_\alpha)$. Each $x_\alpha$ belongs to a unique $\tilde{U}_{j(\alpha)}=g_\alpha(V)$ and for $\beta\ne\alpha$ the sets $\tilde{U}_{j(\alpha)}, \tilde{U}_{j(\beta)}$ are distinct. (In particular, the points $x_\alpha$ are pairwise distinct.)
By Lemma 1, the set $B$ of points $b_\alpha:=x_\alpha$ cannot have an accumulation point in $\tilde{X}$. However, all these points belong to a compact $C$. Contradiction. qed