5.1 Uniform Structures
Definition 5.1.1 (Inversion). Let $X$ be a set and $U \subset X \times X$, then the inversion of $U$ is the set
A set $U \subset X \times X$ is symmetric if $U = U^{-1}$.
Definition 5.1.2 (Composition). Let $X$ be a set and $U, V \subset X \times X$, then the composition of $U$ and $V$ is the set
Definition 5.1.3 (Slice). Let $X, Y$ be sets, $U \subset X \times Y$, and $A \subset X$, then
is the slice of $U$ at $A$.
Definition 5.1.4 (Uniformity). Let $X$ be a set, then a non-empty family $\fU \subset X \times X$ is a uniformity on $X$ if:
For any $U \in \fU$ and $V \supset U$, $V \in \fU$.
For any $U, V \in \fU$, $U \cap V \in \fU$.
For every $U \in \fU$, $U \supset \Delta = \bracs{(x, x)| x \in X}$.
For any $U \in \fU$, $U^{-1}\in \fU$.
For any $U \in \fU$, there exists $V \in \fU$ such that $V \circ V \subset U$.
The elements of $\fU$ are called the entourages of $\fU$, and the pair $(X, \fU)$ is a uniform space.
For any $x, y \in X$ and $U \in \fU$, $x$ and $y$ are $U$-close if $(x, y) \in U$.
Definition 5.1.5 (Subspace Uniformity). Let $(X, \fU)$ be a uniform space and $A \subset X$, then the family
forms a uniformity on $A$, known as the subspace uniformity induced on $A$.
Proposition 5.1.6. 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 $A$.
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.$\square$
Definition 5.1.7 (Fundamental System of Entourages). Let $(X, \fU)$ be a uniform space, then a family $\fB \subset \fU$ is a fundamental system of entourages for $\fU$ if for every $U \in \fU$, there exists $V \in \fB$ such that $V \subset U$.
Proposition 5.1.8. Let $X$ be a set and $\fB \subset 2^{X \times X}$ be a non-empty family of sets. If
For each $U, V \in \fB$, there exists $W \in \fB$ such that $W \subset U \cap V$.
For each $V \in \fB$, $\Delta \subset V$.
For each $V \in \fB$, there exists $W \in \fB$ such that $W \circ W \subset V$.
then there exists a unique uniformity $\fU \subset 2^{X \times X}$, which is given by
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) and (U2): 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), there exists $V_{0} \in \fB \subset \fU$ with $V_{0} \circ V_{0} \subset U_{0} \subset U$.$\square$
Lemma 5.1.9. Let $(X, \fU)$ be a uniform space, $\fB \subset \fU$ be a fundamental system of entourages, then
is also a fundamental system of entourages.
Proof. By (F2), $\fB \subset \fU$. For any $U \in \fU$, there exists $V \in \fB$ such that $V \subset U$. In which case,
so $\fb_{S}$ is a fundamental system of entourages.$\square$
Definition 5.1.10 (Topology of a Uniform Space, [Proposition 2.1.2, Bou13]). Let $(X, \fU)$ be a uniform space and
then there exists a unique topology $\topo \subset 2^{X}$ such that $\cn_{\topo} = \cn$, known as the topology induced by the uniform structure $\fU$.
Proof. Using Proposition 4.4.4, 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
As $\fU$ satisfies (F1), $W \in \fU$. Thus
(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 $y \in W$.$\square$
Lemma 5.1.11. 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:
$(x, y) \in V \circ M \circ V$.
There exists $(p, q) \in M$ such that $(x, p) \in V$ and $(q, y) \in V$.
There exists $(p, q) \in M$ such that $x \in V(p)$ and $y \in V(q)$.
There exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q)$.
There exists $(p, q) \in M$ such that $p \in V(x)$ and $q \in V(y)$.
There exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
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 $(q, y) \in V$.$\square$
Proposition 5.1.12. Let $(X, \fU)$ be a uniform space, $V \in \fU$ be a symmetric entourage, and $M \subset X \times X$, then:
$V \circ M \circ V \in \cn(M)$.
Let $\fB$ be the set of all symmetric entourages, then $\ol{M}= \bigcap_{V \in \fB}V \circ M \circ V$.
with respect to the product topology on $X \times X$.
Proof. (1): Let $(x, y) \in V \circ M \circ V$. By Lemma 5.1.11, 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 Lemma 5.1.11, the following are equivalent:
$(x, y) \in V \circ M \circ V$ for all $V \in \fB$.
For every $V \in \fB$, there exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
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
Proposition 5.1.13. Let $(X, \fU)$ be a uniform space and $A \subset X$, then:
For any symmetric entourage $V \in \fU$, $V(A) \in \cn(A)$.
Let $\fB \subset \fU$ be the family of all symmetric entourages, then $\ol{A}= \bigcap_{U \in \fB}U(A)$.
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:
For all $V \in \fB$, $x \in V(A)$.
For all $V \in \fB$, there exists $y \in A$ such that $x \in V(y)$.
For all $V \in \fB$, $V(x) \cap A \ne \emptyset$.
Since $\bracs{V(y): V \in \fB}$ is a fundamental system of neighbourhoods at $y$ (Lemma 5.1.9), (c) is equivalent to $x \in \overline{A}$. Therefore $\ol{A}= \bigcap_{U \in \fB}U(A)$.$\square$
Proposition 5.1.14. Let $(X, \fU)$ be a uniform space, then the following families of sets form fundamental systems of entourages for $\fU$:
$\mathfrak{O}= \bracs{U^o| U \in \fU}$
$\mathfrak{K}= \bracsn{\overline{U}| U \in \fU}$.
By Lemma 5.1.9, there exists fundamental systems of entourages for $\fU$ consisting of symmetric and open/closed sets.
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 Lemma 5.1.9. By (1) of Proposition 5.1.12, $V \circ V \circ V \in \cn(V)$. Since
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 Proposition 5.1.12,
So $\overline{V}\in \fU$ and is contained in $U$. Therefore (2) is also a fundamental system of entourages.$\square$
Lemma 5.1.15. Let $(X, \fU)$ be a uniform space, $U \in \fU$ be a symmetric, open entourage, and $M \subset X$, then $U(M)$ is open.
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.$\square$
Proposition 5.1.16. Let $X$ be a uniform space and $x \in X$, then the closed neighbourhoods of $x$ form a fundamental system of neighbourhoods at $x$.
Proof. By Proposition 5.1.14 and Lemma 5.1.15, the closed neighbourhoods form a fundamental system of neighbourhoods.$\square$
Definition 5.1.17 (Separated). Let $(X, \fU)$ be a uniform space, then the following are equivalent:
$X$ is T0.
$X$ is T1.
$X$ is Hausdorff.
$X$ is regular.
$\Delta = \bigcap_{U \in \fU}U$.
If the above holds, then $X$ is separated.
Proof. $(1) \Rightarrow (5)$: 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 $\Delta \supset \bigcap_{U \in \fU}U$.
$(5) \Rightarrow (2)$: By Proposition 5.1.14, $\ol \Delta \subset \bigcap_{U \in \fU}\ol U = \Delta$, so $\ol \Delta$ is closed. By (6) of Definition 4.8.1, $X$ is Hausdorff.
$(1) \Rightarrow (4)$: $X$ is T1 and satisfies (2) of Definition 4.9.1 by Proposition 5.1.16, so $X$ is regular.
$(4) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$: (T3) $\Rightarrow$ (T2) $\Rightarrow$ (T1) $\Rightarrow$ (T0).$\square$
Proposition 5.1.18. 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 $X$.
Proof. Let $U \in \fU$ be an open entourage, then by (3) of Definition 4.5.5, $\overline{U \cap (A \times A)}= \overline{U}$ for all $U \in \fU$, so $\overline{U \cap (A \times A)}$ is an entourage. By Proposition 5.1.14, every closed entourage of $X$ contains an element of $\bracsn{\overline{U}: U \in \fU_A}$.$\square$