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Replaced references to upward-directed families with ideals.

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Bokuan Li
2026-05-04 17:08:01 -04:00
parent e4da295fd9
commit 60115baa41
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64
src/topology/functions/set-systems.tex
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@@ -1,58 +1,58 @@
\section{Topology With Respect to Families of Sets}
\section{Topology With Respect to Ideals}
\label{section:pointwise}
\begin{definition}[Set-Open Topology]
\label{definition:set-open}
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
Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $(X, \topo)$ be a topological space. For each $S \in \sigma$ and $U \subset X$ open, let
\[
M(S, U) = \bracs{f \in X^T| f(S) \subset U}
\]
and
\[
\ce(\mathfrak{S}, \topo) = \bracs{M(S, U)| S \in \mathfrak{S}, U \in \topo}
\ce(\sigma, \topo) = \bracs{M(S, U)| S \in \sigma, U \in \topo}
\]
then the topology generated by $\ce$ is the \textbf{$\mathfrak{S}$-open topology} on $T^X$.
then the topology generated by $\ce$ is the \textbf{$\sigma$-open topology} on $T^X$.
If $\cb \subset \topo$ generates $\topo$, then $\ce(\mathfrak{S}, \cb)$ generates the $\mathfrak{S}$-open topology.
If $\cb \subset \topo$ generates $\topo$, then $\ce(\sigma, \cb)$ generates the $\sigma$-open topology.
\end{definition}
\begin{definition}[Set Uniformity]
\label{definition:set-uniform}
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
Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $(X, \fU)$ be a uniform space. For each $S \in \sigma$ and $U \in \fU$, let
\[
E(S, U) = \bracs{(f, g) \in X^T|(f(x), g(x)) \in U \forall x \in S}
\]
and
\[
\mathfrak{E}(\mathfrak{S}, \fU) = \bracs{E(S, U)| S \in \mathfrak{S}, U \in \fU}
\mathfrak{E}(\sigma, \fU) = \bracs{E(S, U)| S \in \sigma, U \in \fU}
\]
then
\begin{enumerate}
\item $\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity $\fV$ on $X^T$.
\item The topology induced by $\fV$ is finer than the $\mathfrak{S}$-topology on $T^X$.
\item If $\mathfrak{S}$ is upward-directed with respect to inclusion, then $\mathfrak{E}(\mathfrak{S}, \fU)$ is forms a fundamental system of entourages for $\fV$.
\item $\mathfrak{E}(\sigma, \fU)$ generates a uniformity $\fV$ on $X^T$.
\item The topology induced by $\fV$ is finer than the $\sigma$-open topology on $T^X$.
\item If $\mathfrak{E}(\sigma, \fU)$ forms a fundamental system of entourages for $\fV$.
\end{enumerate}
The uniformity $\fV$ is the \textbf{$\mathfrak{S}$-uniformity}, and the topology induced by $\fV$ is the \textbf{topology of uniform convergence on the sets $\mathfrak{S}$}/\textbf{$\mathfrak{S}$-uniform topology} on $X^T$.
The uniformity $\fV$ is the \textbf{$\sigma$-uniformity}, and the topology induced by $\fV$ is the \textbf{topology of uniform convergence on the sets $\sigma$}/\textbf{$\sigma$-uniform topology} on $X^T$.
\end{definition}
\begin{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 $X^T$.
(1): Since $\Delta \subset E(S, U)$ for all $S \in \sigma$ and $U \in \fU$, $\mathfrak{E}(\sigma, \fU)$ 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
\begin{enumerate}
\item[(FB1)] For any $S, S' \in \mathfrak{S}$, there exists $T \in \mathfrak{S}$ with $T \supset S, S'$. In which case, for any $U, U' \in \fU$,
\item[(FB1)] For any $S, S' \in \sigma$, there exists $T \in \sigma$ with $T \supset S, S'$. In which case, for any $U, U' \in \fU$,
\[
E(T, U \cap U') \subset E(S \cup S', U \cap U') \subset E(S, U) \cap E(S', U')
\]
\item[(UB1)] For any $U \in \fU$, $\Delta \subset U$. Thus the diagonal in $X^T$ is in $E(S, U)$ for any $S \in \mathfrak{S}$.
\item[(UB2)] For any $U \in \fU$, there exists $V \in \fV$ with $V \circ V \subset U$. Thus for any $S \in \mathfrak{S}$,
\item[(UB1)] For any $U \in \fU$, $\Delta \subset U$. Thus the diagonal in $X^T$ is in $E(S, U)$ for any $S \in \sigma$.
\item[(UB2)] For any $U \in \fU$, there exists $V \in \fV$ with $V \circ V \subset U$. Thus for any $S \in \sigma$,
\[
E(S, V) \circ E(S, V) \subset E(S, V \circ V) \subset E(S, U)
\]
@@ -63,43 +63,43 @@
\begin{proposition}
\label{proposition:set-uniform-pseudometric}
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
Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $(X, \fU)$ be a uniform space whose uniformity is induced by the pseudometrics $\seqi{d}$. For each $i \in I$ and $S \in \sigma$, let
\[
d_{i, S}: X^T \times X^T \quad (f, g) \mapsto \sup_{x \in S}d_i(f(x), g(x))
\]
then
\begin{enumerate}
\item $\bracs{d_{i, S}| i \in I, S \in \mathfrak{S}}$ is a family of pseudometrics induces the $\mathfrak{S}$-uniformity on $X^T$.
\item If $\mathfrak{S}$ is upward-directed with respect to inclusion, then
\item $\bracs{d_{i, S}| i \in I, S \in \sigma}$ is a family of pseudometrics induces the $\sigma$-uniformity on $X^T$.
\item The family
\[
\bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \mathfrak{S}}
\bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \sigma}
\]
is a fundamental system of entourages for the $\mathfrak{S}$-uniformity on $X^T$.
is a fundamental system of entourages for the $\sigma$-uniformity on $X^T$.
\end{enumerate}
\end{proposition}
\begin{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
(1): Let $S \in \sigma$ 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
\[
\bigcap_{j \in J}E(d_{j, S}, r) \subset E\paren{S, \bigcap_{j \in J}E(d_j, r)} \subset E(S, U)
\]
and the uniformity induced by $\bracs{d_{i, S}| i \in I, S \in \mathfrak{S}}$ contains the $\mathfrak{S}$-uniformity.
and the uniformity induced by $\bracs{d_{i, S}| i \in I, S \in \sigma}$ contains the $\sigma$-uniformity.
On the other hand, for any $i \in I$ and $r > 0$, $E(d_j, r/2) \in \fU$ by \autoref{definition:pseudometric-uniformity}. Therefore $E(S, E(d_j, r/2)) \subset E(d_{j, S}, r)$, so the $\mathfrak{S}$-uniformity contains the induced uniformity.
On the other hand, for any $i \in I$ and $r > 0$, $E(d_j, r/2) \in \fU$ by \autoref{definition:pseudometric-uniformity}. Therefore $E(S, E(d_j, r/2)) \subset E(d_{j, S}, r)$, so the $\sigma$-uniformity contains the induced uniformity.
(2): If $\mathfrak{S}$ is upward-directed with respect to inclusion, then by \autoref{definition:set-uniform},
(2): By \autoref{definition:set-uniform},
\[
\bracs{E(S, U)| U \in \fU, S \in \mathfrak{S}}
\bracs{E(S, U)| U \in \fU, S \in \sigma}
\]
Following the same steps in (1),
is a fundamental system of entourages for the $\sigma$-uniformity. Following the same steps in (1),
\[
\bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \mathfrak{S}}
\bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \sigma}
\]
is a fundamental system of entourages for the $\mathfrak{S}$-uniformity.
is a fundamental system of entourages for the $\sigma$-uniformity.
\end{proof}
@@ -109,7 +109,7 @@
Let $T$ be a set and $X$ be a topological space, then the following topologies on $X^T$ coincide:
\begin{enumerate}
\item The product topology on $X^T$.
\item The $\mathfrak{S}$-open topology, where $\mathfrak{S} = \bracs{\bracs{x}| x \in X}$.
\item The $\sigma$-open topology, where $\sigma$ is the collection of all finite sets.
\item (If $X$ is a uniform space) The $\mathfrak{F}$-uniform topology, where $\fF = \bracs{F| F \subset X \text{ finite}}$.
\end{enumerate}
This topology is the \textbf{topology of pointwise convergence} on $X^T$.
@@ -125,10 +125,10 @@
\begin{proposition}
\label{proposition:set-uniform-complete}
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.
Let $T$ be a set, $\sigma \subset 2^T$ be a covering ideal, and $(X, \fU)$ be a complete uniform space, then $X^T$ equipped with the $\sigma$-uniformity is complete.
\end{proposition}
\begin{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 $\fF \subset 2^{X^T}$ be a Cauchy filter. Let $x \in T$. If there exists $S \in \sigma$ 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 $x \in \bigcup_{S \in \sigma}S$.
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 $\fF \to f$.
Let $S \in \sigma$ 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 $\fF \to f$.
\end{proof}