Polished A-A and added new lines for broken enumerates.
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@@ -41,7 +41,8 @@
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\item[(U1)] For every $U \in \fU$, $U \supset \Delta = \bracs{(x, x)| x \in X}$.
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\item[(U2)] For any $U \in \fU$, $U^{-1} \in \fU$.
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\item[(U3)] For any $U \in \fU$, there exists $V \in \fU$ such that $V \circ V \subset U$.
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\end{enumerate}
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\end\{enumerate\}
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The elements of $\fU$ are called the \textbf{entourages} of $\fU$, and the pair $(X, \fU)$ is a \textbf{uniform space}.
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For any $x, y \in X$ and $U \in \fU$, $x$ and $y$ are \textbf{$U$-close} if $(x, y) \in U$.
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@@ -79,7 +80,8 @@
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\item[(FB1)] For each $U, V \in \fB$, there exists $W \in \fB$ such that $W \subset U \cap V$.
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\item[(UB1)] For each $V \in \fB$, $\Delta \subset V$.
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\item[(UB2)] For each $V \in \fB$, there exists $W \in \fB$ such that $W \circ W \subset V$.
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\end{enumerate}
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\end\{enumerate\}
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then there exists a unique uniformity $\fU \subset 2^{X \times X}$, which is given by
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\[
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\fU = \bracs{U \subset X \times X| \exists V \in \fB: V \subset U}
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@@ -176,7 +178,8 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
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\begin{enumerate}
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\item $V \circ M \circ V \in \cn(M)$.
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\item Let $\fB$ be the set of all symmetric entourages, then $\ol{M} = \bigcap_{V \in \fB}V \circ M \circ V$.
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\end{enumerate}
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\end\{enumerate\}
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with respect to the product topology on $X \times X$.
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\end{proposition}
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\begin{proof}
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@@ -223,7 +226,8 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
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\begin{enumerate}
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\item $\mathfrak{O} = \bracs{U^o| U \in \fU}$
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\item $\mathfrak{K} = \bracsn{\overline{U}| U \in \fU}$.
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\end{enumerate}
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\end\{enumerate\}
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By \autoref{lemma:symmetricfundamentalentourage}, there exists fundamental systems of entourages for $\fU$ consisting of symmetric and open/closed sets.
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\end{proposition}
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\begin{proof}
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@@ -267,7 +271,8 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
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\item $X$ is Hausdorff.
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\item $X$ is regular.
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\item $\Delta = \bigcap_{U \in \fU}U$.
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\end{enumerate}
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\end\{enumerate\}
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If the above holds, then $X$ is \textbf{separated}.
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\end{definition}
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\begin{proof}
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