5.6 The Hausdorff Completion

Proof. (1, Uniform), (4): Let $\wh X$ be the set of all minimal Cauchy filters on $X$. For each $V \in \fU$, let

and $\wh \fB = \bracsn{\wh V| V \in \fU, V \text{ symmetric}}$, then

1. Let $\wh U, \wh V \in \wh \fB$. By Lemma 5.1.9, there exists a symmetric entourage $W \in \fU$ such that $W \subset U \cap V$. In which case, for any $(\fF, \mathfrak{G}) \in \wh W$, there exists $E \in \fF \cap \mathfrak{G}$ with $E \times E \subset W \subset U \cap V$. Thus $\wh W \subset \wh U \cap \wh V$.

2. Let $\wh U \in \wh \fB$ and $\fF \in \wh X$, then since $\fF$ is Cauchy, there exists $E \in \fF$ such that $E \times E \subset U$, so $(\fF, \fF) \in \wh U$.

3. Let $\wh U \in \wh \fB$. By Lemma 5.1.9, there exists a symmetric entourage $W \in \fU$ such that $W \circ W \subset U$. For any $\fF, \mathfrak{G}, \mathfrak{H}\in \wh X$ such that $(\fF, \mathfrak{G}), (\mathfrak{G}, \mathfrak{H}) \in \wh W$, there exists $W$-small sets $E \in \fF \cap \mathfrak{G}$ and $F \in \fF \cap \mathfrak{H}$. Since $\mathfrak{G}$ is a filter, $E \cap F \ne \emptyset$ by (F2) and (F3). By Lemma 5.4.2, $E \cup H$ is $W \circ W$-small and thus $U$-small. Using (F1), $E \cup H \in \fF \cap \mathfrak{H}$, so $(\fF, \mathfrak{H}) \in \wh U$. Therefore $\wh W \circ \wh W \subset U$.

By Proposition 5.1.8, there exists a unique uniformity $\wh \fU \supset \wh \fB$. Moreover, $\wh \fB$ consists of symmetric entourages by construction.

(1, Hausdorff): It is sufficient to show that $\Delta$ is closed and use (6) of Definition 4.8.1. Let $(\fF, \mathfrak{G}) \in \ol{\Delta}$, then $(\fF, \mathfrak{G}) \in U$ for all $U \in \fU$ closed. Let $\fB = \bracs{F \cup G| F \in \fF, G \in \mathfrak{G}}$, then

1. For any $F \cup G, F' \cup G' \in \fB$, $(F \cup G) \cap (F' \cup G') \supset (F \cap F') \cup (G \cap G') \in \fB$.

2. By (F3), $\emptyset \not\in \fF \cup \mathfrak{G}$, so $\emptyset \not\in \fB$.

Thus $\fB$ is a filter base by Proposition 4.2.3, and the filter $\mathfrak{H}$ generated by $\fB$ is contained in $\fF$ and $\mathfrak{G}$. By Proposition 5.1.14, for every $U \in \fU$, there exists a $U$-small set $E \in \fF \cap \mathfrak{G}\subset \fB \subset \mathfrak{H}$. So $\mathfrak{H}\subset \fF, \mathfrak{G}$ is a Cauchy filter. By minimality of $\fF$ and $\mathfrak{G}$, $\fF = \mathfrak{G}= \mathfrak{H}$.

(2): For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter by (1) of Proposition 5.4.10. Define $\iota: X \to \wh X$ by $x \mapsto \cn(x)$. Let $\wh U \in \wh \fU$ and $(\cn(x), \cn(y)) \in \wh U$, then there exists a $U$-small set $E \in \cn(x) \cap \cn(y)$. By (V1), $(x, y) \in E \times E \in U$.

Conversely, if $(x, y) \in U$, then $E = U(x) \cap U(y) \in \cn(x) \cap \cn(y)$ by (F2), and $E$ is $U$-small by symmetry of $U$.

Thus $(\iota \times \iota)^{-1}(\wh U) = U$, $(\iota \times \iota)^{-1}: \wh \fU \to \fU$ is a bijection, and $\iota \in UC(X; \wh X)$.

(4): Let $\fF \in \wh X$ and $\wh U \in \wh \fU$. Since $\fF$ is a Cauchy filter, there exists a $U$-small set $E \in \fF$. Using Proposition 5.4.9, assume without loss of generality that $E$ is open. Let $x \in E$, then $E \in \cn(x)$, so $(\fF, \iota(x)) \in \wh \fU$, and $X$ is dense in $\wh X$ by (3) of Definition 4.5.2.

(1, Complete): Using Lemma 5.5.4, it is sufficient to show that every Cauchy filter in $\iota(X)$ converges in $\wh X$.

Let $\wh \fF \in 2^{\iota(X)}$ be a Cauchy filter, then since $\iota: X \to \iota(X)$ is surjective, $\iota^{-1}(\wh \fF)$ is a filter base by Proposition 4.2.5.

For any $U \in \fU$, there exists a $\wh U$-small set $\wh E \in \wh \fF$. Since $U = (\iota \times \iota)^{-1}(\wh U)$, $\iota^{-1}(\wh E) \subset U$, so $\iota^{-1}(\wh \fF)$ is a Cauchy filter base.

Let $\mathfrak{G}\subset \iota^{-1}(\wh \fF)$ be a minimal Cauchy filter, then $\iota(\mathfrak{G})$ is a Cauchy filter base by Proposition 5.4.6.

Let $U \in \fU$. By Proposition 5.4.9, there exists a $U$-small open set $E \in \mathfrak{G}$. For any $x \in E$, $E \in \cn(x)$ by Lemma 4.4.3, so $(\cn(x), \mathfrak{G}) = (\iota(x), \mathfrak{G}) \in \wh U$, and $\iota(E) \subset \wh U(\mathfrak{G})$. Since $E \in \mathfrak{G}$, $\wh U(\mathfrak{G}) \supset \iota(E) \in \iota(\mathfrak{G})$, so $\iota(\mathfrak{G})$ converges to $\mathfrak{G}$.

Now, given that $\mathfrak{G}\subset \iota^{-1}(\wh \fF)$, $\iota(\mathfrak{G}) \subset \iota(\iota^{-1}(\fF)$, so $\iota(\iota^{-1}(\fF))$ converges to $\mathfrak{G}$ as well. Since $\fF$ is a filter on $\iota(X)$, $\iota(\iota^{-1}(\fF)) = \fF$, thus $\fF$ is convergent.

(U): Let $(Y, \mathfrak{V})$ be a complete Hausdorff uniform space and $f \in UC(X; Y)$. For each $\fF \in \wh X$, let $F(\fF) = \lim_{x, \fF}f(x)$. For any $x \in X$, $F(\iota(x)) = \lim_{y \to x}f(y) = f(x)$, so $f = F \circ \iota$. Since $\fF$ is a Cauchy filer, $f \in UC(X; Y)$, and $Y$ is a complete Hausdorff uniform space, the limit exists and is unique. Thus $F: \iota(X) \to Y$ is well-defined.

Let $U \in \mathfrak{V}$, then there exists a symmetric entourage $V \in \fU$ such that $(f(x), f(y)) \in U$ for all $(x, y) \in V$, then for any $(\iota(x), \iota(y)) \in \wh V \cap (\iota(X) \times \iota(X))$, $(F(\iota(x)), F(\iota(y))) \in U$, so $F$ is uniformly continuous. By Theorem 5.5.6, there exists a unique $\td F \in C(\wh X; Y)$ such that $\td F|_{\iota(X)}= F|_{\iota(X)}$, and any such $\td F$ is uniformly continuous.$\square$

Proof. Since the mapping $(\iota \times \iota)^{-1}$ is a bijection between two bases of uniformities of $X$ and $\wh X$, it is sufficient to show that $\iota$ is injective. To this end, observe that for any $x, y \in X$, $\cn(x) = \cn(y)$ if and only if $x = y$ by (4) of Definition 4.8.1.$\square$

Proof. Let $(\wh X, \iota)$ be the Hausdorff completion of $X$, $X' = \iota(X)$, and $i = \iota$, then $(X', i)$ satisfies (1) and (2).

(U): Let $(\wh Y, \iota)$ be the Hausdorff completion of $Y$. Using Lemma 5.6.2, identify $Y$ as a subspace of $\wh Y$. By (U) of the Hausdorff completion, there exists a unique $\ol F \in UC(\wh X; \wh Y)$ such that the following diagram commutes:

Since $\ol F(X') \subset \iota(Y) = Y'$, $F = \ol F|_{X'}\in UC(X'; Y')$ is continuous. By Lemma 5.6.2, $\iota \in UC(Y; Y')$ is a homeomorphism. Upon identifying $Y$ with $Y'$, $F \in UC(X'; Y)$ is the desired map.$\square$

Proof. By (U) of Definition 5.6.1 and Definition 5.6.3.$\square$

Proof. (1): By (U) of Definition 5.6.1.

(2), (3): By (U) of the product, there exists $f \in UC(X; \prod_{i \in I}Y_{i})$ such that the following diagram commutes:

First suppose that $X$ and $\seqi{Y}$ are Hausdorff. For each $i \in I$, $\iota_{i} \circ \pi_{i} \in UC(\prod_{i \in I}Y_{i}; \wh Y_{i})$, so by (U) of Proposition 5.5.2, there exists a unique $\iota_{P} \in UC(\prod_{i \in I}Y_{i}; \prod_{i \in I}\wh Y_{i})$ such that the following diagram commutes

for all $i \in I$. By Lemma 5.6.2, each $\iota_{i} \in UC(Y_{i}; \wh Y_{i})$ is an embedding, so Proposition 4.7.4 implies that $\iota_{P}$ is an embedding as well.

Since $X$ has the initial topology, $f: X \to \prod_{i \in I}Y_{i}$ is an embedding by Proposition 4.7.3. Thus the composition $\iota_{P} \circ f$ is an embedding. As $\prod_{i \in I}\wh Y_{i}$ is complete by Proposition 5.5.2 and Proposition 4.8.3, $\overline{\iota_P \circ f(X)}\subset \prod_{i \in I}\wh Y_{i}$ is a complete Hausdorff uniform space. Let $Z$ be a complete Hausdorff uniform space and $g \in UC(X; Z)$, then Theorem 5.5.6 implies that there exists a unique $G \in UC(\overline{\iota_P \circ f(X)}; Z)$ such that the following diagram commutes

Thus $(\overline{\iota_P \circ f(X)}, \iota \circ f)$ satisfies (1), (2), and (U) of the Hausdorff completion. Therefore $\wh X$ may be identified as a subspace of $\prod_{i \in I}\wh Y_{i}$ as follows:

In which case, $\wh X$ must be equipped with the initial topology induced by the projection maps.

Now assume that $X$ and $Y$ are arbitrary. Let $X'$ and $\seqi{Y'}$ be the Hausdorff spaces associated with $X$ and $\seqi{Y}$, respectively. By Proposition 5.6.4, there exists $\seqi{f'}$ such that the following diagram commutes

for all $i \in I$. By (5) of Definition 5.6.1, there is a correspondence between the uniformities of $X$ and $X'$, and $Y$ and $Y'$. Thus $X'$ is equipped with the initial uniformity genereated by $\seqi{f'}$.$\square$