6.1 Topology With Respect to Families of Sets
Definition 6.1.1 (Set-Open Topology). Let $T$ be a set, $\mathfrak{S}\subset 2^{T}$ be a non-empty family of sets, and $(X, \topo)$ be a topological space. For each $S \in \mathfrak{S}$ and $U \subset X$ open, let
and
then the topology generated by $\ce$ is the $\mathfrak{S}$-open topology on $T^{X}$.
If $\cb \subset \topo$ generates $\topo$, then $\ce(\mathfrak{S}, \cb)$ generates the $\mathfrak{S}$-open topology.
Definition 6.1.2 (Set Uniformity). Let $T$ be a set, $\mathfrak{S}\subset 2^{T}$ be a non-empty family of sets, and $(X, \fU)$ be a uniform space. For each $S \in \mathfrak{S}$ and $U \in \fU$, let
and
then
$\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity $\fV$ on $X^{T}$.
The topology induced by $\fV$ is finer than the $\mathfrak{S}$-topology on $T^{X}$.
If $\mathfrak{S}$ is upward-directed with respect to inclusion, then $\mathfrak{E}(\mathfrak{S}, \fU)$ is forms a fundamental system of entourages for $\fV$.
The uniformity $\fV$ is the $\mathfrak{S}$-uniformity, and the topology induced by $\fV$ is the topology of uniform convergence on the sets $\mathfrak{S}$/$\mathfrak{S}$-uniform topology on $X^{T}$.
Proof. (1): Since $\Delta \subset E(S, U)$ for all $S \in \mathfrak{S}$ and $U \in \fU$, $\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity on $X^{T}$.
(2): Let $U \subset X$ be open, then for each $x \in U$, there exists $V_{x} \in \fU$ such that $x \in V_{x}(x) \subset U$. In which case, $U = \bigcup_{x \in U}V_{x}(x)$ and $M(S, U) = \bigcup_{x \in U}M(S, V_{x})(x)$.
(3): It is sufficient to verify
For any $S, S' \in \mathfrak{S}$, there exists $T \in \mathfrak{S}$ with $T \supset S, S'$. In which case, for any $U, U' \in \fU$,
\[E(T, U \cap U') \subset E(S \cup S', U \cap U') \subset E(S, U) \cap E(S', U')\]For any $U \in \fU$, $\Delta \subset U$. Thus the diagonal in $X^{T}$ is in $E(S, U)$ for any $S \in \mathfrak{S}$.
For any $U \in \fU$, there exists $V \in \fV$ with $V \circ V \subset U$. Thus for any $S \in \mathfrak{S}$,
\[E(S, V) \circ E(S, V) \subset E(S, V \circ V) \subset E(S, U)\]
By Proposition 5.1.8, $\mathfrak{E}$ is a fundamental system of entourages for the uniformity that it generates.$\square$
Proposition 6.1.3. Let $T$ be a set, $\mathfrak{S}\subset 2^{T}$ be a non-empty family of sets, and $(X, \fU)$ be a uniform space whose uniformity is induced by the pseudometrics $\seqi{d}$. For each $i \in I$ and $S \in \mathfrak{S}$, let
then
$\bracs{d_{i, S}| i \in I, S \in \mathfrak{S}}$ is a family of pseudometrics induces the $\mathfrak{S}$-uniformity on $X^{T}$.
If $\mathfrak{S}$ is upward-directed with respect to inclusion, then
\[\bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \mathfrak{S}}\]is a fundamental system of entourages for the $\mathfrak{S}$-uniformity on $X^{T}$.
Proof. (1): Let $S \in \mathfrak{S}$ and $U \in \fU$, then there exists $r > 0$ and $J \subset I$ finite such that $\bigcap_{j \in J}E(d_{j}, r) \subset U$, so
and the uniformity induced by $\bracs{d_{i, S}| i \in I, S \in \mathfrak{S}}$ contains the $\mathfrak{S}$-uniformity.
On the other hand, for any $i \in I$ and $r > 0$, $E(d_{j}, r/2) \in \fU$ by Definition 5.3.3. Therefore $E(S, E(d_{j}, r/2)) \subset E(d_{j, S}, r)$, so the $\mathfrak{S}$-uniformity contains the induced uniformity.
(2): If $\mathfrak{S}$ is upward-directed with respect to inclusion, then by Definition 6.1.2,
Following the same steps in (1),
is a fundamental system of entourages for the $\mathfrak{S}$-uniformity.$\square$
Definition 6.1.4 (Pointwise Topology). Let $T$ be a set and $X$ be a topological space, then the following topologies on $X^{T}$ coincide:
The product topology on $X^{T}$.
The $\mathfrak{S}$-open topology, where $\mathfrak{S}= \bracs{\bracs{x}| x \in X}$.
(If $X$ is a uniform space) The $\mathfrak{F}$-uniform topology, where $\fF = \bracs{F| F \subset X \text{ finite}}$.
This topology is the topology of pointwise convergence on $X^{T}$.
Proof. (2) $=$ (3): Let $F \subset X$ finite and $U$ be an entourage, $f \in X^{T}$, then
which is open in the product topology. The converse is given by Definition 6.1.2.$\square$
Proposition 6.1.5. Let $T$ be a set, $\mathfrak{S}\subset 2^{T}$ such that $\bigcup_{S \in \mathfrak{S}}S = T$, and $(X, \fU)$ be a complete uniform space, then $X^{T}$ equipped with the $\mathfrak{S}$-uniformity is complete.
Proof. Let $\fF \subset 2^{X^T}$ be a Cauchy filter. Let $x \in T$. If there exists $S \in \mathfrak{S}$ with $x \in S$, then $\pi_{x}(\fF) \subset 2^{X}$ is a Cauchy filter. By completeness of $X$, there exists $f: T \to X$ such that $\pi_{x}(\fF) \to f(x)$ for all $x \in \bigcup_{S \in \mathfrak{S}}S$.
Let $S \in \mathfrak{S}$ be non-empty and $V_{0}, V$ be symmetric entourages of $X$ with $V_{0} \circ V_{0} \subset V$. Since $\fF$ is Cauchy, there exists an $E(S, V_{0})$-small set $F \in \fF$. Let $x \in S$, then $\pi_{x}(\fF) \to f(x)$ implies that $F_{x} = \pi_{x}^{-1}(V_{0}(f(x))) \in \fF$. Since $F \cap F_{x} \ne \emptyset$, $(f(x), g(x)) \in V_{0} \circ V_{0} \subset V$ for all $g \in F$. As this holds for all $x \in S$, $E \subset E(S, V)(f) \in \fF$. Thus $\fF \to f$.$\square$