\section{Topologies With Respect to Ideals}
\label{section:pointwise}

\begin{definition}[Set-Open Topology]
\label{definition:set-open}
    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
    \[
    \ce(\sigma, \topo) = \bracs{M(S, U)| S \in \sigma, U \in \topo}
    \]

    then the topology generated by $\ce$ is the \textbf{$\sigma$-open topology} on $T^X$.

    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, $\sigma \subset 2^T$ 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
    \[
    \mathfrak{E}(\sigma, \fU) = \bracs{E(S, U)| S \in \sigma, U \in \fU}
    \]

    then
    \begin{enumerate}
        \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{$\sigma$-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 $S \in \sigma$ 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 $S, S'  \in \sigma$, 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 $S \in \sigma$.
        \item[(UB2)] For any $U \in \fU$, there exists $V \in \fV$ with $V \circ V \subset U$. Thus for any $S \in \sigma$,
        \[
        E(S, V) \circ E(S, V) \subset E(S, V \circ V) \subset E(S, U)
        \]

    \end{enumerate}

    By \autoref{proposition:fundamental-entourage-criterion}, $\mathfrak{E}$ is a fundamental system of entourages for the uniformity that it generates.
\end{proof}

\begin{proposition}
\label{proposition:set-uniform-pseudometric}
    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 $\bracs{d_{i, S}| i \in I, S \in \sigma}$ is a family of pseudometrics induces the $\sigma$-uniformity on $X^T$.
        \item The family
        \[
        \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \sigma}
        \]

        is a fundamental system of entourages for the $\sigma$-uniformity on $X^T$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (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 $\bracs{d_{i, S}| i \in I, S \in \sigma}$ 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 $\sigma$-uniformity contains the induced uniformity.

    (2): By \autoref{definition:set-uniform},
    \[
    \bracs{E(S, U)| U \in \fU, S \in \sigma}
    \]

    is a fundamental system of entourages for the $\sigma$-uniformity. Following the same steps in (1),
    \[
    \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \sigma}
    \]

    is a fundamental system of entourages for the $\sigma$-uniformity.
\end{proof}



\begin{definition}[Pointwise Topology]
\label{definition:pointwise}
    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 $\sigma$-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 $X^T$.
\end{definition}
\begin{proof}
    (2) $=$ (3): Let $F \subset X$ finite and $U$ be an entourage, $f \in X^T$, then
    \[
    E(F, U)(f) = \bigcap_{x \in F}\pi_x^{-1}(U(f(x)))
    \]

    which is open in the product topology. The converse is given by \autoref{definition:set-uniform}.
\end{proof}

\begin{proposition}
\label{proposition:set-uniform-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 $S \in \sigma$ 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 $S \in \sigma$ 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}

\begin{proposition}
\label{proposition:compact-uniform-open}
    Let $X$ be a topological space, $\kappa \subset 2^X$ be the collection of all precompact sets in $X$, and $(Y, \fU)$ be a uniform space, then the $\kappa$-open topology and $\kappa$-uniform topology on $C(X; Y)$ coincide. 
\end{proposition}
\begin{proof}
    By \autoref{definition:set-uniform}, the $\kappa$-uniform topology is finer than the $\kappa$-open topology. 

    Let $K \in \kappa$ be compact, $U \in \fU$ be symmetric, and $f \in C(X; Y)$. For each $x \in K$, there exists $V_x \in \cn_X(x)$ such that $f(V_x) \subset U(f(x))$. Since $K$ is compact, there exists $\seqf{x_j} \subset X$ such that $K \subset \bigcup_{j = 1}^n V_{x_j}$. 

    Let $g \in \bigcap_{j = 1}^n M(V_{x_j} \cap K, U(f(x_j)))$ and $y \in K$, then there exists $1 \le j \le n$ such that $y \in V_{x_j}$. In which case, since $f(y), g(y) \in U(f(x_j))$, $(f(y), g(y)) \in U \circ U$. Therefore
    \[
    f \in  \bigcap_{j = 1}^n M(V_{x_j} \cap K, U(f(x_j))) \subset [E(K, U \circ U)](f)
    \]

    and the $\kappa$-open topology is finer than the $\kappa$-uniform topology.
\end{proof}