Replaced references to upward-directed families with ideals.
This commit is contained in:
Bokuan Li
@@ -3,7 +3,7 @@
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\begin{definition}[$n$-Fold Differentiability]
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\label{definition:n-differentiable-sets}
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
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Let $U \subset E$ be open, $f: U \to F$, $x_0 \in U$, and $n > 1$, then $f$ is \textbf{$n$-fold $\sigma$-differentiable at $x_0$} if
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\begin{enumerate}
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@@ -87,7 +87,7 @@
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\begin{theorem}[Symmetry of Higher Derivatives]
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\label{theorem:derivative-symmetric}
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Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed system that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
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Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset B(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
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\end{theorem}
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\begin{proof}[Proof {{\cite[Proposition 4.5.14]{Bogachev}}}. ]
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Let $\seqf{h_j} \subset E$, $E_0$ be the subspace generated by $\seqf{h_j}$, and $g = f|_{E_0 \cap U}: E_0 \cap U \to F$. Since $\sigma$ includes all bounded sets contained in finite-dimensional spaces, for any $\phi \in F^*$, the mapping $\phi \circ g: E_0 \cap U \to K$ is $n$-times Fréchet-differentiable, with
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@@ -100,7 +100,7 @@
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\begin{proposition}[Power Rule]
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\label{proposition:multilinear-derivative}
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Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a Hausdorff locally convex space, and
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Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a Hausdorff locally convex space, and
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\[
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T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
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\]
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@@ -3,7 +3,7 @@
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\begin{definition}[Small]
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\label{definition:differentiation-small}
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Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
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Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be a covering ideal, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
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\begin{enumerate}
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\item For each $A \in \sigma$, $r(th)/t^n \to 0$ uniformly on $A$.
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\item If $r_t(x) = r(tx)/t^n$, then $r_t \to 0$ as $t \to 0$ with respect to the $\sigma$-uniform topology on $F^E$.
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@@ -16,7 +16,7 @@
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\begin{proposition}
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\label{proposition:differentiation-sets}
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family contains all finite sets, $B_\sigma(E; F)$ be the space of linear operators bounded on sets in $\sigma$, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, then $(B_\sigma(E; F), \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $B_\sigma(E; F)$ be the space of linear operators bounded on sets in $\sigma$, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, then $(B_\sigma(E; F), \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
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\end{proposition}
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\begin{proof}
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Let $T \in B_\sigma(E; F)$ and suppose that there exists $V \in \cn_E(0)$ circled and $r \in \mathcal{R}_\sigma(E; F)$ such that $T|_V = r|_V$. For any $x \in V$, $\bracs{x} \in \sigma$, so $T(x) = \lim_{t \downto 0}T(tx)/t = 0$ as $F$ is separated.
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@@ -25,7 +25,7 @@
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\begin{definition}[Derivative]
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\label{definition:derivative-sets}
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in L(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in L(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
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\[
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f(x_0 + h) = f(x_0) + Th + r(h)
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\]
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@@ -37,7 +37,7 @@
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\begin{definition}[Differentiable]
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\label{definition:differentiable-sets}
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
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\end{definition}
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\begin{definition}
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@@ -48,7 +48,7 @@
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\begin{proposition}[Chain Rule]
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\label{proposition:chain-rule-sets}
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Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be upward-directed families that contain all finite sets. If:
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Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be covering ideals. If:
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\begin{enumerate}
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\item[(a)] For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
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\item[(b)] For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
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@@ -128,7 +128,7 @@
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\begin{proposition}
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\label{proposition:derivative-sets-real}
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Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be an upward-directed system that contains finite sets, then
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Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be a covering ideal, then
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\begin{enumerate}
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\item $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{B(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
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\item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, $f$ is differentiable at $x_0$ if and only if
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@@ -3,18 +3,23 @@
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\begin{proposition}
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\label{proposition:lc-spaces-linear-map}
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Let $T$ be a set, $E$ be a locally convex space defined by the seminorms $\seqi{[\cdot]}$, and $\mathfrak{S} \subset 2^T$ be an upward-directed family. For each $i \in I$ and $S \in \mathfrak{S}$, let
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Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $E$ be a locally convex space over $K$. For each $S \in \sigma$ and continuous seminorm $\rho: E \to [0, \infty)$, let
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\[
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[\cdot]_{S, i}: E^T \to [0, \infty) \quad f \mapsto \sup_{x \in S}[f(x)]_{S, i}
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\rho_S: E^T \to [0, \infty] \quad f \mapsto \sup_{x \in S}\rho(f(x))
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\]
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then the $\mathfrak{S}$-uniform topology on $E^T$ is defined by the seminorms
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then the the $\sigma$-uniform topology on $E^T$ is defined by distances of the form
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\[
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\bracs{[\cdot]_{S, i}|S \in \mathfrak{S}, i \in I}
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d_{S, \rho}: E^T \times E^T \to [0, \infty] \quad (f, g) \mapsto \rho_S(f - g)
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\]
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and hence locally convex.
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In particular, if $\cf \subset E^T$ is a subspace such that
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\begin{enumerate}
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\item[(B)] For each $f \in \cf$ and $S \in \sigma$, $f(S) \subset E$ is bounded.
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\end{enumerate}
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then the $\sigma$-uniform topology on $\cf$ is induced by seminorms of the form $\rho_S$, where $\rho$ is a continuous seminorm on $E$, and $S \in \sigma$. In which case, the $\sigma$-uniform topology on $\cf$ is locally convex.
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\end{proposition}
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\begin{proof}
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By \autoref{proposition:set-uniform-pseudometric}.
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By \autoref{proposition:tvs-set-uniformity} and \autoref{proposition:set-uniform-pseudometric}.
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\end{proof}
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@@ -3,9 +3,9 @@
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\begin{proposition}[{{\cite[III.3.1]{SchaeferWolff}}}]
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\label{proposition:tvs-set-uniformity}
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Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be an upward-directed system of sets, $F$ be a TVS over $K \in \RC$, then
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Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $F$ be a TVS over $K \in \RC$, then
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\begin{enumerate}
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\item The $\mathfrak{S}$-uniformity on $F^T$ (\autoref{definition:set-uniform}) is translation invariant.
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\item The \hyperref[$\sigma$-uniformity]{definition:set-uniform} on $F^T$ is translation invariant.
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\item The composition defined by
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\[
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T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
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@@ -13,20 +13,21 @@
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is continuous.
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\end{enumerate}
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For any vector subspace $\cf \subset F^T$, the following are equivalent:
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\begin{enumerate}
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\item[(3)] The $\mathfrak{S}$-uniform topology on $\cf \subset F^E$ is a vector space topology.
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\item[(4)] For each $S \in \mathfrak{S}$ and $f \in \cf$, $f(S) \subset E$ is bounded.
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\item[(T)] The $\sigma$-uniform topology on $\cf \subset F^E$ is a vector space topology.
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\item[(B)] For each $f \in \cf$ and $S \in \sigma$, $f(S) \subset E$ is bounded.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. Using \autoref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
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(1): Let $U \subset F \times F$ be an entourage and $S \in \sigma$. Using \autoref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
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(2): Let $f, g, f', g' \in \cf$, $S \in \mathfrak{S}$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
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(2): Let $f, g, f', g' \in \cf$, $S \in \sigma$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
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(3) $\Rightarrow$ (4): Let $f \in \cf$, $S \in \mathfrak{S}$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\mathfrak{S}$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
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(T) $\Rightarrow$ (B): Let $f \in \cf$, $S \in \sigma$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\sigma$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
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(4) $\Rightarrow$ (3): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \mathfrak{S}$, then for any $x \in X$,
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(T) $\Rightarrow$ (B): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
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\begin{align*}
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\lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
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&= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))
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@@ -83,18 +84,18 @@
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\begin{definition}[Space of Bounded Linear Maps]
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\label{definition:bounded-linear-map-space}
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Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system, and $k \in \nat$. The space $B_{\mathfrak{S}}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \mathfrak{S}$, equipped with the $\bracsn{S^k| S \in \mathfrak{S}}$-uniform topology.
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Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $k \in \nat$. The space $B_{\sigma}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform topology.
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Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\mathfrak{S}}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$.
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Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\sigma}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$.
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\end{definition}
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\begin{proposition}
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\label{proposition:multilinear-identify}
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Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system that contains all singletons, and $k \in \natp$, then
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Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
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\begin{enumerate}
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\item The map
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\[
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I: B_{\mathfrak{S}}^k(E; B_{\mathfrak{S}}(E; F)) \to B^{k+1}_{\mathfrak{S}}(E; F)
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I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
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\]
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defined by
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@@ -105,7 +106,7 @@
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is an isomorphism.
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\item The map
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\[
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I: \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} \to B^k_{\mathfrak{S}}(E; F)
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I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
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\]
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defined by
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@@ -117,33 +118,33 @@
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\end{enumerate}
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which allows the identification
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\[
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\underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} = B^k_{\mathfrak{S}}(E; F)
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\underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
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\]
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under the map $I$ in (2).
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\end{proposition}
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\begin{proof}
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(1): To see that $I$ is surjective, let $T \in B_{\mathfrak{S}}^{k+1}(E; F)$ and
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(1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
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\[
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I^{-1}T: E \to B_{\mathfrak{S}}(E; F) \quad x \mapsto T(x, \cdot)
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I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
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\]
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Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \mathfrak{S}$. Since $\mathfrak{S}$ is upward-directed and contains all singletons, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
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Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
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\[
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T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
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\]
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by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\mathfrak{S}}(E; F)$.
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by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
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In addition, for any $S_1 \in \mathfrak{S}$ and entourage $E(S_2, U)$ of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F))$.
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In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
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It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \mathfrak{S}$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
|
||||
It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
|
||||
|
||||
On the other hand, let $S_1 \in \mathfrak{S}$ and $E(S_2, U)$ be an entourage of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$. Let $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
|
||||
On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
|
||||
|
||||
(2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
|
||||
\[
|
||||
\underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F)) = B^{k+1}_{\mathfrak{S}}(E; F)
|
||||
\underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F)
|
||||
\]
|
||||
|
||||
Thus (2) holds for all $k \in \natp$.
|
||||
|
||||
52
src/topology/functions/ideal.tex
Normal file
52
src/topology/functions/ideal.tex
Normal file
@@ -0,0 +1,52 @@
|
||||
\section{Ideals}
|
||||
\label{section:set-ideal}
|
||||
|
||||
|
||||
\begin{definition}[Ideal]
|
||||
\label{definition:set-ideal}
|
||||
Let $X$ be a set and $\sigma \subset 2^X$, then $\sigma$ is an \textbf{ideal} over $X$ if:
|
||||
\begin{enumerate}[label=(I\arabic*)]
|
||||
\item For any $E \in \sigma$ and $F \subset E$, $F \in \sigma$.
|
||||
\item For any $E, F \in \sigma$, $E \cup F \in \sigma$.
|
||||
\end{enumerate}
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Covering]
|
||||
\label{definition:set-ideal-covering}
|
||||
Let $X$ be a set and $\sigma \subset 2^X$, then $\sigma$ \textbf{covers} $X$/is \textbf{covering} if $\bigcup_{E \in \sigma}E = X$.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Fundamental]
|
||||
\label{definition:set-ideal-fundamental}
|
||||
Let $X$ be a set, $\sigma \subset 2^X$ be an ideal, and $\tau \subset \sigma$, then $\tau$ is \textbf{fundamental} with respect to $\sigma$ if for any $E \in \sigma$, there exists $F \in \tau$ such that $E \subset F$.
|
||||
\end{definition}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:set-ideal-fundamental-criterion}
|
||||
Let $X$ be a set and $\tau \subset 2^X$, then the following are equivalent:
|
||||
\begin{enumerate}
|
||||
\item For any $E, F \in \tau$, there exists $G \in \tau$ such that $E \cup F \subset G$.
|
||||
\item There exists an ideal $\sigma \subset 2^X$ such that $\tau$ is fundamental with respect to $\sigma$.
|
||||
\end{enumerate}
|
||||
|
||||
If the above holds, then the ideal $\sigma$ in (2) is the \textbf{ideal generated by} $\tau$.
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
(1) $\Rightarrow$ (2): Let
|
||||
\[
|
||||
\sigma = \bracs{F \subset X| \exists E \in \tau: F \subset E}
|
||||
\]
|
||||
|
||||
then $\sigma$ satisfies (I1) by definition. For any $E, F \in \sigma$, there exists $E_0, F_0 \in \tau$ such that $E \subset E_0$ and $F \subset F_0$. By assumption, there exists $G \in \tau$ such that
|
||||
\[
|
||||
E \cup F \subset E_0 \cup F_0 \subset G
|
||||
\]
|
||||
|
||||
so $\sigma$ satisfies (I2), and is an ideal.
|
||||
|
||||
(2) $\Rightarrow$ (1): Let $E, F \in \tau$, then $E \cup F \in \sigma$. Since $\tau$ is fundamental, there exists $G \in \tau$ such that $E \cup F \subset G$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
|
||||
|
||||
@@ -1,5 +1,6 @@
|
||||
\chapter{Function Spaces}
|
||||
\label{chap:function-spaces}
|
||||
|
||||
\input{./ideal.tex}
|
||||
\input{./set-systems.tex}
|
||||
\input{./uniform.tex}
|
||||
|
||||
@@ -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}
|
||||
|
||||
@@ -17,7 +17,7 @@
|
||||
$U \circ V$ & Composition of $U, V \subset X \times X$. & \autoref{definition:composition} \\
|
||||
$U(A)$ & Slice of $U \subset X \times Y$ at $A \subset X$: $\{y \mid \exists x \in A,\, (x,y) \in U\}$. & \autoref{definition:slice} \\
|
||||
$E(S, U)$ & Entourage of the form $\{(f,g) \in X^T \mid (f(x),g(x)) \in U\ \forall x \in S\}$. & \autoref{definition:set-uniform} \\
|
||||
$\mathfrak{E}(\mathfrak{S}, \mathfrak{U})$ & $\mathfrak{S}$-uniformity, generated by $\{E(S,U) \mid S \in \mathfrak{S},\ U \in \mathfrak{U}\}$. & \autoref{definition:set-uniform} \\
|
||||
$\mathfrak{E}(\sigma, \mathfrak{U})$ & $\sigma$-uniformity, generated by $\{E(S,U) \mid S \in \sigma,\ U \in \mathfrak{U}\}$. & \autoref{definition:set-uniform} \\
|
||||
|
||||
% Function Spaces
|
||||
$\mathrm{supp}(f)$ & Support of $f$. & \autoref{definition:support} \\
|
||||
|
||||
Reference in New Issue
Block a user