Topology of uniform convergence of functions $X\to X$?

Question

Suppose that $X$ is a compact Hausdorff space and that $\phi\colon X\to X$ is a homeomorphism. What does it mean to take the closure of $\{\phi^{k}:k\in\mathbb{Z}\}$ with respect to the topology of uniform convergence? Or more generally, what does the the topology of uniform convergence look like in this particular case? Since there is no metric involved, I don't know what 'uniform' means in this context. I read something about uniform spaces sitting properly between metric spaces and topological spaces, but it seemed quite technical.

Answer 1

$X$ is compact Hausdorff so it has a unique uniformity $\mathcal{U}$. It's convergence in the uniformity on $C(X,X)$ induced by $\mathcal{U}$ that is meant.

i.e. for all open neighbourhoods $O$ of $\{(x,x): x \in X\}=:\Delta_X \subseteq X \times X$ we have a basic open set $B(f,O)$ for $f \in C(X,X)$ defined by

$$B(f,O):= \{g \in C(X,X)\mid \forall x \in X: (f(x), g(x)) \in O\}$$

and this defines a local base at $f$ in the uniform topology. So now you know $f$ is in the closure of $\{ \phi^k\mid k \in \Bbb Z\}$ iff

$$\forall O \in \mathcal{N}(\Delta_X): \exists k \in \Bbb Z: \forall x \in X: (\phi^k(x), f(x)) \in O$$

This makes it quite concrete.