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Bokuan Li
2026-03-06 14:06:15 -05:00
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src/topology/uniform/completion.tex
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@@ -17,6 +17,7 @@
}
\]
Moreover, if $f \in UC(X; Y)$, then $F \in UC(\wh X; Y)$.
\end{enumerate}
Moreover,
@@ -25,9 +26,9 @@
\[
\wh V = \bracsn{(\fF, \mathfrak{G}) \in \wh X| \exists U \in \fF \cap \mathfrak{G}: U \times U \subset V}
\]
then[(5)] The family $\wh \fB = \bracsn{\wh V| V \in \fU, V \text{ symmetric}}$ forms a fundamental system of entourages for $\wh X$. In particular, $\bracs{\wh V \cap \iota(X)| V \in \fU, V \text{ symmetric}}$ is a fundamental system of entuorages for $\iota(X)$.
then the family $\wh \fB = \bracsn{\wh V| V \in \fU, V \text{ symmetric}}$ forms a fundamental system of entourages for $\wh X$. In particular, $\bracs{\wh V \cap \iota(X)| V \in \fU, V \text{ symmetric}}$ is a fundamental system of entourages for $\iota(X)$.
\item $\iota(X)$ is dense in $\wh X$.
\item[(5)] $\iota(X)$ is dense in $\wh X$.
\end{enumerate}
@@ -39,44 +40,45 @@
\[
\wh V = \bracsn{(\fF, \mathfrak{G}) \in \wh X| \exists U \in \fF \cap \mathfrak{G}: U \times U \subset V}
\]
and $\wh \fB = \bracsn{\wh V| V \in \fU, V \text{ symmetric}}$, then
\begin{enumerate}
\item[(FB1)] Let $\wh U, \wh V \in \wh \fB$. By \ref{lemma:symmetricfundamentalentourage}, 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 $\wh W \subset \wh U \cap \wh V$.
\item[(FB1)] Let $\wh U, \wh V \in \wh \fB$. By \autoref{lemma:symmetricfundamentalentourage}, 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 $\wh W \subset \wh U \cap \wh V$.
\item[(UB1)] 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 $(\fF, \fF) \in \wh U$.
\item[(UB2)] Let $\wh U \in \wh \fB$. By \ref{lemma:symmetricfundamentalentourage}, 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 \ref{lemma:small-intersect}, $E \cup H$ 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 $\wh W \circ \wh W \subset U$.
\item[(UB2)] Let $\wh U \in \wh \fB$. By \autoref{lemma:symmetricfundamentalentourage}, 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 \autoref{lemma:small-intersect}, $E \cup H$ 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 $\wh W \circ \wh W \subset U$.
\end{enumerate}
By \ref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\wh \fU \supset \wh \fB$. Moreover, $\wh \fB$ consists of symmetric entourages by construction.
By \autoref{proposition:fundamental-entourage-criterion}, 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 \ref{definition:hausdorff}. 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, Hausdorff): It is sufficient to show that $\Delta$ is closed and use (6) of \autoref{definition:hausdorff}. 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
\begin{enumerate}
\item[(FB1)] For any $F \cup G, F' \cup G' \in \fB$, $(F \cup G) \cap (F' \cup G') \supset (F \cap F') \cup (G \cap G') \in \fB$.
\item[(FB2)] By (F3), $\emptyset \not\in \fF \cup \mathfrak{G}$, so $\emptyset \not\in \fB$.
\end{enumerate}
Thus $\fB$ is a filter base by \ref{proposition:filterbasecriterion}, and the filter $\mathfrak{H}$ generated by $\fB$ is contained in $\fF$ and $\mathfrak{G}$. By \ref{proposition:goodentourages}, 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 $\mathfrak{G}$, $\fF = \mathfrak{G} = \mathfrak{H}$.
Thus $\fB$ is a filter base by \autoref{proposition:filterbasecriterion}, and the filter $\mathfrak{H}$ generated by $\fB$ is contained in $\fF$ and $\mathfrak{G}$. By \autoref{proposition:goodentourages}, 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 \ref{proposition:cauchyfilterlimit}. 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), $(x, y) \in E \times E \in U$.
(2): For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter by (1) of \autoref{proposition:cauchyfilterlimit}. 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 \ref{proposition:cauchyinterior}, 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 \ref{definition:closure}.
(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 \autoref{proposition:cauchyinterior}, 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 \autoref{definition:closure}.
(1, Complete): Using \ref{lemma:complete-dense}, it is sufficient to show that every Cauchy filter in $\iota(X)$ converges in $\wh X$.
(1, Complete): Using \autoref{lemma:complete-dense}, 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 \ref{proposition:preimage-filterbase}.
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 \autoref{proposition:preimage-filterbase}.
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 \ref{proposition:imagecauchy}.
Let $\mathfrak{G} \subset \iota^{-1}(\wh \fF)$ be a minimal Cauchy filter, then $\iota(\mathfrak{G})$ is a Cauchy filter base by \autoref{proposition:imagecauchy}.
Let $U \in \fU$. By \ref{proposition:cauchyinterior}, there exists a $U$-small open set $E \in \mathfrak{G}$. For any $x \in E$, $E \in \cn(x)$ by \ref{lemma:openneighbourhood}, 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 $\mathfrak{G}$.
Let $U \in \fU$. By \autoref{proposition:cauchyinterior}, there exists a $U$-small open set $E \in \mathfrak{G}$. For any $x \in E$, $E \in \cn(x)$ by \autoref{lemma:openneighbourhood}, 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 \ref{theorem:uniform-continuous-extension}, 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.
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 \autoref{theorem:uniform-continuous-extension}, 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.
\end{proof}
\begin{lemma}
@@ -84,7 +86,7 @@
Let $(X, \fU)$ be a Hausdorff uniform space, then the canonical map $\iota: X \to \wh X$ is an embedding.
\end{lemma}
\begin{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 \ref{definition:hausdorff}.
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 \autoref{definition:hausdorff}.
\end{proof}
\begin{definition}[Associated Hausdorff Uniform Space, {{\cite[Proposition 2.8.16]{Bourbaki}}}]
@@ -101,13 +103,14 @@
X \ar@{->}[r]_{f} \ar@{->}[u]^{i} & Y
}
\]
\end{enumerate}
known as the \textbf{Hausdorff uniform space associated with} $(X, \fU)$.
\end{definition}
\begin{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 \ref{lemma:completion-of-hausdorff}, 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:
(U): Let $(\wh Y, \iota)$ be the Hausdorff completion of $Y$. Using \autoref{lemma:completion-of-hausdorff}, 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:
\[
\xymatrix{
@@ -117,7 +120,8 @@
}
\]
Since $\ol F(X') \subset \iota(Y) = Y'$, $F = \ol F|_{X'} \in UC(X'; Y')$ is continuous. By \ref{lemma:completion-of-hausdorff}, $\iota \in UC(Y; Y')$ is a homeomorphism. Upon identifying $Y$ with $Y'$, $F \in UC(X'; Y)$ is the desired map.
Since $\ol F(X') \subset \iota(Y) = Y'$, $F = \ol F|_{X'} \in UC(X'; Y')$ is continuous. By \autoref{lemma:completion-of-hausdorff}, $\iota \in UC(Y; Y')$ is a homeomorphism. Upon identifying $Y$ with $Y'$, $F \in UC(X'; Y)$ is the desired map.
\end{proof}
\begin{proposition}
@@ -131,9 +135,10 @@
X \ar@{->}[r]_{f} \ar@{->}[u] & Y \ar@{->}[u]
}
\]
\end{proposition}
\begin{proof}
By (U) of \ref{definition:hausdorff-completion} and \ref{definition:associated-hausdorff}.
By (U) of \autoref{definition:hausdorff-completion} and \autoref{definition:associated-hausdorff}.
\end{proof}
@@ -148,6 +153,7 @@
X \ar@{->}[r]_{f_i} \ar@{->}[u] & Y \ar@{->}[u]
}
\]
\item The uniformity of $\wh X$ is the initial uniformity induced by $\seqi{F}$.
\item There exists a unique $F \in UC(X; \prod_{i \in I}\wh Y_i)$ and $\ol{F} \in UC(\wh X; \prod_{i \in I}\wh Y_i)$ such that the following diagram commutes
@@ -159,18 +165,20 @@
}
\]
Moreover, $\ol{F}(\wh X) = \overline{F(X)}$, and $\ol{F}$ is an embedding.
\end{enumerate}
In particular, by \ref{proposition:dense-product}, there is a natural isomorphism
In particular, by \autoref{proposition:dense-product}, there is a natural isomorphism
\[
\prod_{i \in I}\wh X_i \iso \wh{\prod_{i \in I}X_i}
\]
induced by extending the identity on $\pi_{i \in I}X_i$.
\end{proposition}
\begin{proof}
(1): By (U) of \ref{definition:hausdorff-completion}.
(1): By (U) of \autoref{definition:hausdorff-completion}.
(2), (3): By (U) of the product (\ref{definition:product-topology}), there exists $f \in UC(X; \prod_{i \in I}Y_i)$ such that the following diagram commutes:
(2), (3): By (U) of the \hyperref[product]{definition:product-topology}, there exists $f \in UC(X; \prod_{i \in I}Y_i)$ such that the following diagram commutes:
\[
\xymatrix{
@@ -179,7 +187,8 @@
}
\]
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 \ref{proposition:product-complete}, 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
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 \autoref{proposition:product-complete}, 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
\[
\xymatrix{
@@ -188,9 +197,10 @@
}
\]
for all $i \in I$. By \ref{lemma:completion-of-hausdorff}, each $\iota_i \in UC(Y_i; \wh Y_i)$ is an embedding, so \ref{proposition:product-topology-embedding} 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 \ref{proposition:initial-product-topology}. Thus the composition $\iota_P \circ f$ is an embedding. As $\prod_{i \in I}\wh Y_i$ is complete by \ref{proposition:product-complete} and \ref{proposition:product-hausdorff}, $\overline{\iota_P \circ f(X)} \subset \prod_{i \in I}\wh Y_i$ is a complete Hausdorff uniform space. Let $Z$ be a complete Hausdorff uniform space and $g \in UC(X; Z)$, then \ref{theorem:uniform-continuous-extension} implies that there exists a unique $G \in UC(\overline{\iota_P \circ f(X)}; Z)$ such that the following diagram commutes
for all $i \in I$. By \autoref{lemma:completion-of-hausdorff}, each $\iota_i \in UC(Y_i; \wh Y_i)$ is an embedding, so \autoref{proposition:product-topology-embedding} 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 \autoref{proposition:initial-product-topology}. Thus the composition $\iota_P \circ f$ is an embedding. As $\prod_{i \in I}\wh Y_i$ is complete by \autoref{proposition:product-complete} and \autoref{proposition:product-hausdorff}, $\overline{\iota_P \circ f(X)} \subset \prod_{i \in I}\wh Y_i$ is a complete Hausdorff uniform space. Let $Z$ be a complete Hausdorff uniform space and $g \in UC(X; Z)$, then \autoref{theorem:uniform-continuous-extension} implies that there exists a unique $G \in UC(\overline{\iota_P \circ f(X)}; Z)$ such that the following diagram commutes
\[
\xymatrix{
@@ -200,10 +210,10 @@
}
\]
Thus $(\overline{\iota_P \circ f(X)}, \iota \circ f)$ satisfies (1), (2), and (U) of the Hausdorff completion (\ref{definition:hausdorff-completion}). Therefore $\wh X$ may be identified as a subspace of $\prod_{i \in I}\wh Y_i$ as follows:
Thus $(\overline{\iota_P \circ f(X)}, \iota \circ f)$ satisfies (1), (2), and (U) of the \hyperref[Hausdorff completion]{definition:hausdorff-completion}. Therefore $\wh X$ may be identified as a subspace of $\prod_{i \in I}\wh Y_i$ as follows:
\[
% https://darknmt.github.io/res/xypic-editor/#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
\xymatrix{
\prod_{i \in I}\wh Y_i \ar@{->}[r]^{\pi_i} & \wh Y_i \\
\overline{\iota_P \circ f(X)} \ar@{^{(}->}[u] \ar@{=}[r] & \wh X \\
@@ -211,9 +221,10 @@
}
\]
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 \ref{proposition:hausdorff-uniform-factor}, there exists $\seqi{f'}$ such that the following diagram commutes
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 \autoref{proposition:hausdorff-uniform-factor}, there exists $\seqi{f'}$ such that the following diagram commutes
\[
\xymatrix{
@@ -222,5 +233,6 @@
}
\]
for all $i \in I$. By (5) of \ref{definition:hausdorff-completion}, 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 $\seqi{f'}$.
for all $i \in I$. By (5) of \autoref{definition:hausdorff-completion}, 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 $\seqi{f'}$.
\end{proof}