\section{Uniform Structures}
\label{section:uniformstructures}


\begin{definition}[Inversion]
\label{definition:inversion}
    Let $X$ be a set and $U \subset X \times X$, then the \textbf{inversion} of $U$ is the set
    \[
    U^{-1} = \bracs{(y, x)| (x, y) \in U}
    \]

    A set $U \subset X \times X$ is \textbf{symmetric} if $U = U^{-1}$.
\end{definition}

\begin{definition}[Composition]
\label{definition:composition}
    Let $X$ be a set and $U, V \subset X \times X$, then the \textbf{composition} of $U$ and $V$ is the set
    \[
    U \circ V = \bracs{(x, z) \in X \times X| \exists y \in Y: (x, y) \in U, (y, z) \in V}
    \]

\end{definition}

\begin{definition}[Slice]
\label{definition:slice}
    Let $X, Y$ be sets, $U \subset X \times Y$, and $A \subset X$, then
    \[
    U(A) = \bracs{y \in Y| (x, y) \in U, x \in A}
    \]

    is the \textbf{slice} of $U$ at $A$.
\end{definition}


\begin{definition}[Uniformity]
\label{definition:uniformity}
    Let $X$ be a set, then a non-empty family $\fU \subset X \times X$ is a \textbf{uniformity} on $X$ if:
    \begin{enumerate}
        \item[(F1)] For any $U \in \fU$ and $V \supset U$, $V \in \fU$.
        \item[(F2)] For any $U, V \in \fU$, $U \cap V \in \fU$.
        \item[(U1)] For every $U \in \fU$, $U \supset \Delta = \bracs{(x, x)| x \in X}$.
        \item[(U2)] For any $U \in \fU$, $U^{-1} \in \fU$.
        \item[(U3)] For any $U \in \fU$, there exists $V \in \fU$ such that $V \circ V \subset U$.
    \end{enumerate}

    The elements of $\fU$ are called the \textbf{entourages} of $\fU$, and the pair $(X, \fU)$ is a \textbf{uniform space}.

    For any $x, y \in X$ and $U \in \fU$, $x$ and $y$ are \textbf{$U$-close} if $(x, y) \in U$.
\end{definition}

\begin{definition}[Subspace Uniformity]
\label{definition:subspaceuniform}
    Let $(X, \fU)$ be a uniform space and $A \subset X$, then the family
    \[
    \fU_A = \bracs{U \cap (A \times A)| U \in \fU}
    \]

    forms a uniformity on $A$, known as the \textbf{subspace uniformity} induced on $A$.
\end{definition}

\begin{proposition}
\label{proposition:subspace-uniform-compatible}
    Let $(X, \fU)$ be a uniform space and $A \subset X$, then the subspace topology of $A$ coincides with the topology induced by the subspace uniformity on $A$.
\end{proposition}
\begin{proof}
    Let $x \in A$ and $U \in \fU$, then $U(x) \cap A = [U \cap (A \times A)](x)$. Thus $V \subset X$ is a neighbourhood of $x$ with respect to the subspace topology if and only if it is a neighbourhood of $x$ with respect to the topology on $A$ induced by the subspace uniformity.
\end{proof}



\begin{definition}[Fundamental System of Entourages]
\label{definition:fundamentalentourage}
    Let $(X, \fU)$ be a uniform space, then a family $\fB \subset \fU$ is a \textbf{fundamental system of entourages} for $\fU$ if for every $U \in \fU$, there exists $V \in \fB$ such that $V \subset U$.
\end{definition}

\begin{proposition}
\label{proposition:fundamental-entourage-criterion}
    Let $X$ be a set and $\fB \subset 2^{X \times X}$ be a non-empty family of sets such that
    \begin{enumerate}
        \item[(FB1)] For each $U, V \in \fB$, there exists $W \in \fB$ such that $W \subset U \cap V$.
        \item[(UB1)] For each $V \in \fB$, $\Delta \subset V$.
        \item[(UB2)] For each $V \in \fB$, there exists $W \in \fB$ with $W \subset V^{-1}$. 
        \item[(UB3)] For each $V \in \fB$, there exists $W \in \fB$ such that $W \circ W \subset V$.
    \end{enumerate}

    and let
    \[
    \fU = \bracs{U \subset X \times X| \exists V \in \fB: V \subset U}
    \]

    then $\fU$ is the unique uniformity on $X$ such that $\fB$ is a fundamental system of entourages for $\fU$. 
\end{proposition}
\begin{proof}
    (F1): By definition of $\fU$.

    (F2): For any $U, V \in \fU$, there exists $U_0, V_0 \in \fB$ such that $U_0 \subset U$ and $V_0 \subset V$. By (FB1), there exists $W \in \fB$ with $W \subset U_0 \cap V_0 \subset U \cap V$. Thus $U \cap V \in \fU$.

    (U1), (U2), (U3): For any $U \in \fU$, there exists $U_0 \in \fB$ with $U_0 \subset U$. By (UB1), $\Delta \subset U_0 \subset U$. By (UB2) and (FB1), there exists $V_0 \in \fB$ with $V_0 \subset U_0 \cap U_0^{-1} \subset U$. By (UB3), there exists $W_0 \in \fB \subset \fU$ with $W_0 \circ W_0 \subset U_0 \subset U$.
\end{proof}


\begin{lemma}
\label{lemma:symmetricfundamentalentourage}
    Let $(X, \fU)$ be a uniform space, $\fB \subset \fU$ be a fundamental system of entourages, then
    \[
    \fB_S = \bracsn{U \cap U^{-1}| U \in \fB}
    \]

    is also a fundamental system of entourages.
\end{lemma}
\begin{proof}
    By (F2), $\fB \subset \fU$. For any $U \in \fU$, there exists $V \in \fB$ such that $V \subset U$. In which case,
    \[
    U \supset V \supset V \cap V^{-1} \in \fB_S
    \]

    so $\fb_S$ is a fundamental system of entourages.
\end{proof}

\begin{definition}[Topology of a Uniform Space]
\label{definition:uniformtopology}
    Let $(X, \fU)$ be a uniform space and
    \[
    \cn: X \to 2^X \quad x \mapsto \bracsn{U(x)| U \in \fU}
    \]

    then there exists a unique topology $\topo \subset 2^X$ such that $\cn_\topo = \cn$, known as the \textbf{topology induced by the uniform structure $\fU$}.
\end{definition}
\begin{proof}[Proof {{\cite[Proposition 2.1.2]{Bourbaki}}}. ]
Using \autoref{proposition:neighbourhoodcharacteristic}, it is sufficient to show that $\cn(x)$ is non-empty for all $x \in X$, and that it satisfies (F1), (F2), (V1), and (V2). Firstly, since $\fU \ne \emptyset$, $\cn(x) \ne \emptyset$ for all $x \in X$.

(F1): Let $U \in \fU$ and $V \supset U(x)$, then
\[
W = U \cup (\bracs{x} \times V) \supset U
\]

As $\fU$ satisfies (F1), $W \in \fU$. Thus
\[
V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
\]


(F2): Let $U, V \in \fU$, then $U(x) \cap V(x) = (U \cap V)(x)$. As $\fU$ satisfies (F2), $U \cap V \in \fU$ and $(U \cap V)(x) \in \cn(x)$.

(V1): Let $U \in \fU$. By (U1), $\Delta \subset U$, so $x \in U(x)$.

(V2): Let $U \in \fU$ and $x \in X$. By (U2), there exists $W \in \fU$ such that $W \circ W \subset U$. Let $y \in W$, then for any $z \in W(y)$, $(x, z) \in V$. Hence $W(y) \subset V(x)$ for all $y \in W$, so $V(x) \in \cn(y)$ for all $y \in W$.
\end{proof}



\begin{lemma}
\label{lemma:compositiongymnastics}
    Let $(X, \fU)$ be a uniform space, $V \in \fU$ be a symmetric entourage, and $M \subset X \times X$, then the following are equivalent:
    \begin{enumerate}
        \item $(x, y) \in V \circ M \circ V$.
        \item There exists $(p, q) \in M$ such that $(x, p) \in V$ and $(q, y) \in V$.
        \item There exists $(p, q) \in M$ such that $x \in V(p)$ and $y \in V(q)$.
        \item There exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q)$.
        \item There exists $(p, q) \in M$ such that $p \in V(x)$ and $q \in V(y)$.
        \item There exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
    \end{enumerate}
\end{lemma}
\begin{proof}
    $(1) \Rightarrow (2)$: By definition of composition, there exists $q \in X$ such that $(x, q) \in M$ and $(q, y) \in V$. Similarly, there exists $p \in X$ such that $(x, p) \in V$, $(p, q) \in M$, and $(q, y) \in V$.

    $(2) \Rightarrow (3)$: By symmetry, $(p, x) \in V$. Thus $p \in V(x)$ and $q \in V(y)$.

    $(3) \Leftrightarrow (4)$, $(5) \Leftrightarrow (6)$: By definition of product.

    $(3) \Rightarrow (5)$: By symmetry, $a \in V(b)$ if and only if $b \in V(a)$.

    $(5) \Rightarrow (1)$: By symmetry, $p \in V(x)$ implies that $(x, p) \in V$, and $(q, y) \in V$.
\end{proof}


\begin{proposition}[{{\cite[Proposition 2.1.2]{Bourbaki}}}]
\label{proposition:uniformneighbourhood}
    Let $(X, \fU)$ be a uniform space, $V \in \fU$ be a symmetric entourage, and $M \subset X \times X$, then:
    \begin{enumerate}
        \item $V \circ M \circ V \in \cn(M)$.
        \item Let $\fB$ be the set of all symmetric entourages, then $\ol{M} = \bigcap_{V \in \fB}V \circ M \circ V$.
    \end{enumerate}

    with respect to the product topology on $X \times X$.
\end{proposition}
\begin{proof}
    (1): Let $(x, y) \in V \circ M \circ V$. By \autoref{lemma:compositiongymnastics}, there exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q) \in \cn(p, q)$. In particular, if $(x, y) \in M$, then this implies that $V \circ M \circ V \in \cn(x, y)$. Thus $V \circ M \circ V \in \cn(M)$.

    (2): Let $(x, y) \in X \times X$. By \autoref{lemma:compositiongymnastics}, the following are equivalent:
    \begin{enumerate}
        \item[(a)] $(x, y) \in V \circ M \circ V$ for all $V \in \fB$.
        \item[(b)] For every $V \in \fB$, there exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
    \end{enumerate}

    As $\bracsn{V(x) \times V(y)| V \in \fB}$ is a fundamental system of neighbourhoods at $(x, y)$, (b) is equivalent to $(x, y) \in \ol{M}$. Therefore

    \[
    \ol{M} = \bigcap_{V \in \fB}V \circ M \circ V
    \]

\end{proof}

\begin{proposition}[{{\cite[Corollary 2.1.1]{Bourbaki}}}]
\label{proposition:uniformclosure}
    Let $(X, \fU)$ be a uniform space and $A \subset X$, then:
    \begin{enumerate}
        \item For any symmetric entourage $V \in \fU$, $V(A) \in \cn(A)$.
        \item Let $\fB \subset \fU$ be the family of all symmetric entourages, then $\ol{A} = \bigcap_{U \in \fB}U(A)$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): For each $x \in A$, $x \in V(x) \subset V(A)$. By (F1) of $\cn(x)$, $V(A) \in \cn(x)$ for all $x \in A$. Thus $V(A) \in \cn(A)$.

    (2): Let $x \in X$, then the following are equivalent:
    \begin{enumerate}
        \item[(a)] For all $V \in \fB$, $x \in V(A)$.
        \item[(b)] For all $V \in \fB$, there exists $y \in A$ such that $x \in V(y)$.
        \item[(c)] For all $V \in \fB$, $V(x) \cap A \ne \emptyset$.
    \end{enumerate}

    Since $\bracs{V(y): V \in \fB}$ is a fundamental system of neighbourhoods at $y$ (\autoref{lemma:symmetricfundamentalentourage}), (c) is equivalent to $x \in \overline{A}$. Therefore $\ol{A} = \bigcap_{U \in \fB}U(A)$.
\end{proof}

\begin{proposition}[{{\cite[Corollary 2.1.2]{Bourbaki}}}]
\label{proposition:goodentourages}
    Let $(X, \fU)$ be a uniform space, then the following families of sets form fundamental systems of entourages for $\fU$:
    \begin{enumerate}
        \item $\mathfrak{O} = \bracs{U^o| U \in \fU}$
        \item $\mathfrak{K} = \bracsn{\overline{U}| U \in \fU}$.
    \end{enumerate}

    By \autoref{lemma:symmetricfundamentalentourage}, there exists fundamental systems of entourages for $\fU$ consisting of symmetric and open/closed sets.
\end{proposition}
\begin{proof}
    Let $U \in \fU$, then there exists a symmetric entourage $V \in \fU$ such that $V \circ V \circ V \subset U$ by (U2) and \autoref{lemma:symmetricfundamentalentourage}. By (1) of \autoref{proposition:uniformneighbourhood}, $V \circ V \circ V \in \cn(V)$. Since
    \[
    V \subset (V \circ V \circ V)^o \subset V \circ V \circ V \subset U
    \]

    the interior $(V \circ V \circ V)^o \in \fU$, and $U$ contains the interior of an entourage. Thus (1) is a fundamental system of entourages.

    On the other hand, by (2) of \autoref{proposition:uniformneighbourhood},
    \[
    V \subset \overline{V} \subset V \circ V \circ V \subset U
    \]

    So $\overline{V} \in \fU$ and is contained in $U$. Therefore (2) is also a fundamental system of entourages.
\end{proof}

\begin{lemma}
\label{lemma:openentourageneighbourhoods}
    Let $(X, \fU)$ be a uniform space, $U \in \fU$ be a symmetric, open entourage, and $M \subset X$, then $U(M)$ is open.
\end{lemma}
\begin{proof}
    Let $U \in \fV$ and $y \in U(M)$. Since $U \subset X \times X$ is open, there exists $x \in M$, $V \in \cn(x)$, and $V' \in \cn(y)$ such that $(x, y) \subset V \times V' \subset U$. In which case, $U(M) \supset V' \in \cn(y)$. Hence $U(M)$ is open.
\end{proof}

\begin{proposition}
\label{proposition:uniform-neighbourhoods}
    Let $X$ be a uniform space and $x \in X$, then the closed neighbourhoods of $x$ form a fundamental system of neighbourhoods at $x$.
\end{proposition}
\begin{proof}
    By \autoref{proposition:goodentourages} and \autoref{lemma:openentourageneighbourhoods}, the closed neighbourhoods form a fundamental system of neighbourhoods.
\end{proof}

\begin{definition}[Separated]
\label{definition:uniform-separated}
    Let $(X, \fU)$ be a uniform space, then the following are equivalent:
    \begin{enumerate}
        \item $X$ is T0.
        \item $X$ is T1.
        \item $X$ is Hausdorff.
        \item $X$ is regular.
        \item $X$ is completely regular. 
        \item $\Delta = \bigcap_{U \in \fU}U$.
    \end{enumerate}

    If the above holds, then $X$ is \textbf{separated}.
\end{definition}
\begin{proof}
    (1) $\Rightarrow$ (6): Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U(x) \in \cn(x)$ such that $y \not\in U$. In which case, $(x, y) \not\in U$ and $\Delta \supset \bigcap_{U \in \fU}U$.

    (6) $\Rightarrow$ (5): Let $E \subset X$ be closed and $x \in X \setminus E$. Since $\Delta = \bigcap_{U \in \fU}U$, there exists $U \in \fU$ such that $U(x) \subset E^c$. By \autoref{theorem:uniform-pseudometric}, there exists a pseudometric $d: X \times X \to [0, \infty)$ such that $d(x, E) > 0$. Thus the function $y \mapsto d(x, y)$ is a continuous function that separates $x$ and $E$. 
\end{proof}


\begin{proposition}
\label{proposition:subspace-entourage}
    Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $\bracsn{\overline{U}: U \in \fU_A}$ forms a fundamental system of entourages for $X$.
\end{proposition}
\begin{proof}
    Let $U \in \fU$ be an open entourage, then by (3) of \autoref{definition:dense}, $\overline{U \cap (A \times A)} = \overline{U}$ for all $U \in \fU$, so $\overline{U \cap (A \times A)}$ is an entourage. By \autoref{proposition:goodentourages}, every closed entourage of $X$ contains an element of $\bracsn{\overline{U}: U \in \fU_A}$.
\end{proof}
