6.5 Completeness
Definition 6.5.1 (Complete Space).label Let $(X, \fU)$ be a uniform space, then $X$ is complete if every Cauchy filter converges to at least one point.
Proposition 6.5.2.label Let $\seqi{X}$ be complete uniform spaces, then $\prod_{i \in I}X_{i}$ is complete.
Proof. For any Cauchy filter $\fF$ in $\prod_{i \in I}\wh X_{i}$, $\pi_{i}(\fF)$ is a Cauchy filter base by (1) of Definition 6.2.5 and Proposition 6.4.8. For each $i \in I$, $\wh X_{i}$ is complete, so there exists $x_{i} \in \wh X_{i}$ such that $\pi_{i}(\fF)$ converges to $x_{i}$. Let $x \in X$ such that $\pi_{i}(x) = x_{i}$ for all $i \in I$, then $\fF$ converges to $x$ by Proposition 5.7.2. Therefore $\prod_{i \in I}X_{i}$ is complete.$\square$
Proposition 6.5.3.label Let $X$ be a uniform space and $A \subset X$ be closed, then $A$ is complete.
Proof. Let $\fF$ be a Cauchy filter on $A$, then there exists $x \in X$ such that $\fF$ converges to $x$. Since $A$ is closed, $x \in X$ by (4) of Definition 5.5.2.$\square$
Lemma 6.5.4.label Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $X$ is complete if and only if every Cauchy filter on $A$ converges to at least one point in $X$.
Proof. Let $\fF \subset X$ be a Cauchy filter. Using Proposition 6.4.10, assume without loss of generality that $\fF$ is minimal. By Proposition 6.4.11, $\fF$ admits a base consisting of open sets. Hence $E \cap A \ne \emptyset$ for all $E \in \fF$. Let $\fF_{A} = \bracs{E \cap A| E \in \fF}$, then $\fF_{A}$ is a filter base on $A$.
By assumption, there exists $x \in X$ such that $\fF_{A}$ converges to $x$. Since $\fF$ is a Cauchy filter contained in the filter generated by $\fF_{A}$ in $X$, $x$ is also a limit point of $\fF$ by Proposition 6.4.12.$\square$
Proposition 6.5.5 ([Proposition 2.3.11, Bou13]).label Let $X$ be a uniform space, $Y$ be a complete Hausdorff uniform space, $A \subset X$ be a dense subset, and $f \in C(A; Y)$ be a continuous function, then the following are equivalent:
- (1)
There exists a continuous function $F \in C(X; Y)$ such that $F|_{A} = f$.
- (2)
$f$ is Cauchy continuous.
Proof. By Proposition 6.1.16, $Y$ is regular. Since $Y$ is complete, (2) is equivalent to (2) of Theorem 5.9.2. Therefore the proposition follows from Theorem 5.9.2.$\square$
Theorem 6.5.6 (Extension of Cauchy Continuous Functions).label Let $(X, \fU)$ be a uniform space, $(Y, \fV)$ be a complete Hausdorff uniform space, $A \subset Y$ be a dense subset, and $f \in C(A; Y)$ be Cauchy continuous, then:
- (1)
There exists a unique $F \in C(X; Y)$ such that $F|_{A} = f$.
- (2)
If $f \in UC(A; Y)$, then $F \in UC(X; Y)$.
Proof [Theorem 1.3.2, Bou13]. (1): By Proposition 6.4.8, $f(\bracs{U \cap A| U \in \cn(x)})$ is a Cauchy filter base for all $x \in X$. Thus the existence and uniqueness of the extension follows from Proposition 6.5.5.
(2): Let $V \in \fV$. Using Proposition 6.1.14, assume without loss of generality that $V$ is symmetric and closed. Since $f \in UC(A; Y)$, there exists $U \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in U$. Using Proposition 6.1.18, assume without loss of generality that $U = \overline{U}= \ol{U \cap (A \times A)}$. In which case, since $F$ is continuous,
by Proposition 5.5.3.$\square$