6.4 Compact Uniform Spaces
Definition 6.4.1.label Let $(X, \fU)$ be a uniform space, then the following are equivalent:
- (1)
For every $U \in \fU$, there exists $\seqf{x_j}\subset U$ such that $X = \bigcup_{j = 1}^{n} U(x_{j})$.
- (2)
Every ultrafilter on $X$ is Cauchy.
- (3)
For every filter $\fF_{0} \subset 2^{X}$, there exists $\fF \supset \fF_{0}$ such that $\fF$ is Cauchy.
If the above holds, then $(X, \fU)$ is totally bounded.
Proof. (1) $\Rightarrow$ (2): Let $\fF \subset 2^{X}$ be an ultrafilter and $U \in \fU$, then there exists $U$-small sets $\seqf{E_j}\subset 2^{X}$ such that $X = \bigcup_{j = 1}^{n} E_{j}$. By (3) of Definition 5.2.9, there exists $1 \le j \le n$ such that $E_{j} \in \fF$, so $\fF$ is Cauchy.
(2) $\Rightarrow$ (3): By the ultrafilter lemma.
$\neg (1) \Rightarrow \neg (3)$: Let $U \in \fU$ be symmetric such that for all $X_{0} \subset X$ finite, $\bigcup_{x \in X_0}U(x) \subsetneq X$. Let $\fF_{0}$ be the filter generated by
then for any $U$-small set $E \subset X$ and $x \in E$, $E \cap [X \setminus U(x)] = \emptyset$. Therefore no filter finer than $\fF_{0}$ contains a $U$-small set.$\square$
Proposition 6.4.2.label Let $X, Y$ be uniform spaces and $f \in UC(X; Y)$. If $X$ is totally bounded, then so is $f(X)$.
Proof. Let $U$ be an entourage of $Y$, then there exists $\seqf{x_j}\subset X$ such that $\bigcup_{j = 1}^{n}[(f \times f)^{-1}(U)](x_{j}) = X$. In which case, $f(x) \subset \bigcup_{j = 1}^{n} U(f(x_{j}))$.$\square$
Proposition 6.4.3.label Let $X$ be a uniform space, then the following are equivalent:
- (1)
$X$ is compact.
- (2)
$X$ is complete and totally bounded.
In particular, $X$ is totally bounded if and only if its Hausdorff completion is compact.
Proof. (1) $\Rightarrow$ (2): By (1) of Definition 5.16.1, $X$ is totally bounded. Let $\fF$ be a Cauchy filter, then by (4) of Definition 5.16.1, $\fF$ has at least one accumulation point. Since $\fF$ is Cauchy, the accumulation points and limit points of $\fF$ coincide by (2) of Proposition 6.6.12.
(2) $\Rightarrow$ (1): By (2) of Definition 6.4.1, every ultrafilter is Cauchy, and hence converges, which satisfies (4) of Definition 5.16.1.$\square$
Proposition 6.4.4.label Let $\seqi{X}$ be totally bounded uniform spaces, then $\prod_{i \in I}X_{i}$ is totally bounded.
Proof. Let $J \subset I$ be finite and $\seqj{U}$ such that $U_{j}$ is an entourage of $X_{j}$ for each $j \in J$. Since each $X_{j}$ is totally bounded, there exists $Y_{j} \subset X_{j}$ finite such that $X_{j} = U_{j}(Y_{j})$. Let $Y = \prod_{j \in J}Y_{j}$ and $\phi_{y} \in \pi_{J}^{-1}(y)$ for each $y \in Y$, then for any $x \in X$, there exists $y \in Y$ such that $\pi_{j}(x) \in U_{j}(\pi_{j}(y))$ for each $j \in J$. Therefore
$\square$
Proposition 6.4.5.label Let $X$ be a compact uniform space, $Y$ be a uniform space, and $f \in C(X; Y)$, then $f \in UC(X; Y)$.
Proof. Let $V$ be an entourage of $Y$. For each $x \in X$, let $U_{x}$ be an entourage of $X$ such that $(f(y), f(z)) \in V$ for all $y, z \in (U_{x} \circ U_{x})(x)$. Since $X$ is compact, there exists $\seqf{x_j}\subset X$ such that $X = \bigcup_{j = 1}^{n}U_{x_j}(x_{j})$.
Let $U = \bigcap_{j = 1}^{n} U_{x_j}$, then for any $(x, y) \in U$, there exists $1 \le j \le n$ such that $x \in U_{x_j}(x_{j})$. In which case, $x, y \in (U_{x_j}\circ U_{x_j})(x_{j})$, so $(f(x), f(y)) \in V$.$\square$