Housekeeping.
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.gitignore
vendored
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.gitignore
vendored
@@ -13,3 +13,4 @@ codebook.toml
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*.log
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*.log
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*.out
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*.out
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*.pdf
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*.pdf
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./spec.db
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16
document.tex
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document.tex
@@ -1,17 +1,17 @@
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\documentclass{report}
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% \documentclass{report}
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\usepackage{amssymb, amsmath, hyperref}
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\usepackage{amssymb, amsmath, hyperref}
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\usepackage{preamble}
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\usepackage{preamble}
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\begin{document}
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\begin{document}
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\input{./src/cat/index}
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\input{./src/topology/index}
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\input{./src/fa/index}
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\input{./src/measure/index}
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\input{./src/dg/index}
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\input{./src/process/index}
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\input{./src/cat/index.tex}
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\bibliographystyle{alpha} % We choose the "plain" reference style
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\input{./src/topology/index.tex}
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\input{./src/fa/index.tex}
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\input{./src/measure/index.tex}
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\input{./src/dg/index.tex}
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\bibliographystyle{abbrv} % We choose the "plain" reference style
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\bibliography{refs} % Entries are in the refs.bib file
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\bibliography{refs} % Entries are in the refs.bib file
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18
preamble.sty
18
preamble.sty
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\NeedsTeXFormat{LaTeX2e}
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% \NeedsTeXFormat{LaTeX2e}
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\ProvidesPackage{jerrylicious}[Jerry's Tex Mess]
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% \ProvidesPackage{jerrylicious}[Jerry's Tex Mess]
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\RequirePackage{amsthm,amssymb,amsfonts,amsmath}
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\RequirePackage{amsthm,amssymb,amsfonts,amsmath}
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@@ -8,28 +8,19 @@
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% ------------- Block Environmets ---------------
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% ------------- Block Environmets ---------------
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\makeatletter
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\theoremstyle{plain}
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\newtheorem{theorem}[subsection]{Theorem}
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\newtheorem{theorem}[subsection]{Theorem}
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\newtheorem{proposition}[subsection]{Proposition}
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\newtheorem{proposition}[subsection]{Proposition}
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\newtheorem{lemma}[subsection]{Lemma}
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\newtheorem{lemma}[subsection]{Lemma}
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\theoremstyle{definition}
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\newtheorem{definition}[subsection]{Definition}
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\newtheorem{definition}[subsection]{Definition}
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\newtheorem{example}[subsection]{Example}
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\newtheorem{example}[subsection]{Example}
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% \newtheorem{exercise}[subsection]{Exercise}
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% \newtheorem{exercise}[subsection]{Exercise}
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% \newtheorem{situation}[subsection]{Situation}
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% \newtheorem{situation}[subsection]{Situation}
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\theoremstyle{remark}
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\newtheorem{rem}[subsection]{Remark}
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\newtheorem{rem}[subsection]{Remark}
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\newtheorem{remark}[subsection]{Remark}
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\newtheorem{remark}[subsection]{Remark}
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% \newtheorem{remarks}[subsection]{Remarks}
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% \newtheorem{remarks}[subsection]{Remarks}
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\numberwithin{equation}{subsection}
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\makeatother
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% ------------- References --------------
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% ------------- References --------------
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@@ -202,10 +193,6 @@
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\newcommand{\normrd}[1]{\norms{#1}{\real^d}}
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\newcommand{\normrd}[1]{\norms{#1}{\real^d}}
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\newcommand{\normrdn}[1]{\normns{#1}{\real^d}}
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\newcommand{\normrdn}[1]{\normns{#1}{\real^d}}
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% for variation
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\newcommand{\var}{\text{var}}
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% Category Shenanigans
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% Category Shenanigans
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\newcommand{\mf}[1]{\mathfrak{#1}}
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\newcommand{\mf}[1]{\mathfrak{#1}}
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\newcommand{\catc}{\mathfrak{C}}
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\newcommand{\catc}{\mathfrak{C}}
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@@ -217,7 +204,6 @@
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\newcommand{\ppi}{\frac{\partial}{\partial x^i}}
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\newcommand{\ppi}{\frac{\partial}{\partial x^i}}
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\newcommand{\ppj}{\frac{\partial}{\partial y^j}}
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\newcommand{\ppj}{\frac{\partial}{\partial y^j}}
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\newcommand{\ppip}{\ppi\bigg\vert_p}
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\newcommand{\ppip}{\ppi\bigg\vert_p}
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\renewcommand{\part}[2]{\frac{\partial{#1}}{\partial{#2}}}
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\newcommand{\vf}{\mathfrak{X}}
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\newcommand{\vf}{\mathfrak{X}}
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\newcommand{\grad}[1]{\text{grad}\paren{#1}}
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\newcommand{\grad}[1]{\text{grad}\paren{#1}}
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\newcommand{\Grad}[1]{\text{Grad}\paren{#1}}
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\newcommand{\Grad}[1]{\text{Grad}\paren{#1}}
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9
refs.bib
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refs.bib
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isbn={978-3-319-57117-1},
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isbn={978-3-319-57117-1},
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doi={10.1007/978-3-319-57117-1}
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doi={10.1007/978-3-319-57117-1}
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}
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}
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@book{Cartan,
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title={Differential Calculus},
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author={Cartan, H.},
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isbn={9780901665140},
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series={International studies in mathematics},
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url={https://books.google.ca/books?id=ZFgZAQAAIAAJ},
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year={1983},
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publisher={Hermann}
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}
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14
spec.toml
Normal file
14
spec.toml
Normal file
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database = "spec.db"
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document = "document.tex"
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siteTitle = "Garden"
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[compiler]
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compileAll = false
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redoTags = false
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indirectReferences = true
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[website]
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font = "roboto"
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primaryColour = "blue"
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neutralColour = "grey"
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searchLimit = 16
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@@ -82,7 +82,6 @@
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\label{proposition:chain-rule-sets-conditions}
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\label{proposition:chain-rule-sets-conditions}
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Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset B(E)$ and $\tau \subset B(F)$ correspond to the following families of sets on $E$ and $F$:
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Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset B(E)$ and $\tau \subset B(F)$ correspond to the following families of sets on $E$ and $F$:
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\begin{enumerate}
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\begin{enumerate}
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\item Finite sets.
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\item Compact sets.
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\item Compact sets.
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\item Bounded sets.
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\item Bounded sets.
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\end{enumerate}
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\end{enumerate}
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@@ -96,9 +95,7 @@
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\begin{proof}
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\begin{proof}
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(1): Let $A \in \sigma$ and $U \in \cn_G(0)$. Since $T$ is continuous, there exists $V \in \cn_F(0)$ such that $T(V) \subset U$. Since $r \in \mathcal{R}_\sigma(E; G)$, there exists $t > 0$ such that $r(sA)/s \in V$ for all $s \in (0, t)$. In which case, $T \circ r(sA)/s \in U$ for all $s \in (0, t)$.
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(1): Let $A \in \sigma$ and $U \in \cn_G(0)$. Since $T$ is continuous, there exists $V \in \cn_F(0)$ such that $T(V) \subset U$. Since $r \in \mathcal{R}_\sigma(E; G)$, there exists $t > 0$ such that $r(sA)/s \in V$ for all $s \in (0, t)$. In which case, $T \circ r(sA)/s \in U$ for all $s \in (0, t)$.
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(2): Firstly, for any $A \subset E$ finite, $(T + r)(A)$ is also finite, so the claim holds directly in the case of finite sets.
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(2): To show that $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$, it is sufficient to show that for every $A \in \sigma$, $\seq{t_n} \subset \real_{> 0}$ with $t_n \downto 0$ as $n \to \infty$, and $\seq{a_n} \subset A$,
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To show that $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$, it is sufficient to show that for every $A \in \sigma$, $\seq{t_n} \subset \real_{> 0}$ with $t_n \downto 0$ as $n \to \infty$, and $\seq{a_n} \subset A$,
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\[
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\[
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\limv{n} \frac{1}{t_n}s \circ (T + r)(t_na_n) = \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
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\limv{n} \frac{1}{t_n}s \circ (T + r)(t_na_n) = \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
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\]
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\]
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@@ -112,13 +109,14 @@
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\[
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\[
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\limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
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\limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
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\]
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\]
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\end{proof}
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\end{proof}
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\begin{remark}
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\begin{remark}
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\label{remark:differentiation-bornology}
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In \autoref{definition:differentiation-small}, the system $\sigma$ can be chosen based on the bornology of $E$, and the definition of small-ness depends exclusively on $\sigma$. As such, there is an apparent disconnect between differentiation and the topology of the domain.
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In \autoref{definition:differentiation-small}, the system $\sigma$ can be chosen based on the bornology of $E$, and the definition of small-ness depends exclusively on $\sigma$. As such, there is an apparent disconnect between differentiation and the topology of the domain.
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Consider for example a Hilbert space equipped with its norm and weak topology. The norm itself is differentiable with respect to both topologies, because the bounded sets coincide. Moreover, the data for differentiability needs to only come from a neighbourhood of $0$ in the norm topology. As such, a function may be differentiable even if its domain is too small to have an interior.
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Consider for example a Hilbert space equipped with its norm and weak topology. The norm itself is differentiable with respect to both topologies, because the bounded sets coincide. Moreover, the data for differentiability needs to only come from a neighbourhood of $0$ in the norm topology. As such, a function may be differentiable even if its domain is too small to have an interior.
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@@ -31,5 +31,7 @@
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\end{proof}
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\end{proof}
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\begin{remark}[{{\cite[Section I.1]{SchaeferWolff}}}]
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\begin{remark}[{{\cite[Section I.1]{SchaeferWolff}}}]
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\label{remark:hausdorff-completion-field}
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The Hausdorff completion works in general with arbitrary valuated fields. Though the completion yields a TVS over the completion of the field, the field need not to be complete.
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The Hausdorff completion works in general with arbitrary valuated fields. Though the completion yields a TVS over the completion of the field, the field need not to be complete.
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\end{remark}
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\end{remark}
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@@ -9,6 +9,5 @@
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\input{./radon.tex}
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\input{./radon.tex}
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\input{./outer.tex}
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\input{./outer.tex}
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\input{./lebesgue-stieltjes.tex}
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\input{./lebesgue-stieltjes.tex}
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\input{./radon.tex}
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\input{./riesz.tex}
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\input{./riesz.tex}
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\input{./kolmogorov.tex}
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\input{./kolmogorov.tex}
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@@ -155,6 +155,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
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\end{proof}
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\end{proof}
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\begin{remark}
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\begin{remark}
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\label{remark:pseudometric-proof}
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Unfortunately, it was hard to access sources for this proof online, so I could not provide a specific citation. As such, I followed an online PDF given by the link: https://krex.k-state.edu/server/api/core/bitstreams/1bdf2b14-3b5a-4962-a589-93ee1998950c/content
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Unfortunately, it was hard to access sources for this proof online, so I could not provide a specific citation. As such, I followed an online PDF given by the link: https://krex.k-state.edu/server/api/core/bitstreams/1bdf2b14-3b5a-4962-a589-93ee1998950c/content
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\end{remark}
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\end{remark}
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