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\title[ベン図からトポスへ]{ベン図からトポスへ}
\subtitle{丸で始まる，理論の空間への旅}
\author{洞龍弥}
\institute[]{数理空間トポス 新歓 2026}
\date[2026年5月23日]{2026年5月23日}

\begin{document}

% 1
\begin{frame}[plain]
  \titlepage
  \vspace{-14pt}
  \begin{center}
  {\large $\text{命題}\;\leadsto\;\text{空間}\;\leadsto\;\text{トポス}$}
  \end{center}
\end{frame}

\section{導入}

% 2
\begin{frame}{オイラーの手紙: 論理は目に見える}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.58\textwidth}
    \begin{block}{1761年，オイラー}
      丸い図形，いやむしろ\key{空間}は，\key{論理学}の神秘を目に見えるものにする．
    \end{block}
    \vspace{4pt}
    今日の出発点:
    \[
      \boxed{\text{命題は空間の中の領域である}}
    \]
  \end{column}
  \begin{column}{0.38\textwidth}
    \centering
    \begin{tikzpicture}[scale=0.9]
      \fill[lightblue] (-1,0) circle (1.2);
      \fill[lightorange,opacity=.85] (1,0) circle (1.2);
      \draw[thick,toposblue] (-1,0) circle (1.2);
      \draw[thick,toposorange] (1,0) circle (1.2);
      \node at (-1,0) {$P$};
      \node at (1,0) {$Q$};
      \node[below] at (0,-1.55) {丸，または空間};
    \end{tikzpicture}
  \end{column}
\end{columns}
\end{frame}

% 3
\begin{frame}{今日の衝撃: 完全性定理も幾何である}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.49\textwidth}
    \begin{block}{論理の言葉}
      すべてのモデルで正しいなら，証明できる．
      \[
        T\models \sigma \quad\Rightarrow\quad T\vdash \sigma
      \]
    \end{block}
  \end{column}
  \begin{column}{0.49\textwidth}
    \begin{block}{空間の言葉}
      coherent topos には\shock{十分たくさん点がある}．
      \[
        \text{点で見る} \Rightarrow \text{全体が見える}
      \]
    \end{block}
  \end{column}
\end{columns}
\vspace{6pt}
\centering
\begin{tikzpicture}[node distance=1.2cm]
  \node[idea] (logic) {証明・モデル};
  \node[idea,right=of logic] (space) {点・開集合};
  \node[idea,right=of space] (topos) {トポス};
  \draw[->,thick,toposblue] (logic) -- (space);
  \draw[->,thick,toposblue] (space) -- (topos);
\end{tikzpicture}
\end{frame}

% 4
\begin{frame}{今日の地図}
\centering
\begin{tikzpicture}[node distance=.65cm,scale=.88,transform shape]
  \node[idea,minimum width=2.15cm] (venn) {ベン図};
  \node[idea,minimum width=2.45cm,right=of venn] (ba) {ブール代数};
  \node[idea,minimum width=2.45cm,right=of ba] (stone) {Stone空間};
  \node[idea,minimum width=2.15cm,right=of stone] (topos) {Topos};
  \draw[->,very thick,toposorange] (venn) -- (ba);
  \draw[->,very thick,toposorange] (ba) -- (stone);
  \draw[->,very thick,toposorange] (stone) -- (topos);
  \node[below=0.55cm of venn] {命題は領域};
  \node[below=0.55cm of ba] {計算規則};
  \node[below=0.55cm of stone] {論理の空間};
  \node[below=0.55cm of topos] {理論の空間};
\end{tikzpicture}
\vspace{4pt}
\begin{block}{目標}
\centering
\large \key{ベン図の丸}が，\shock{トポスの入口}だったと感じる．
\end{block}
\end{frame}

\section{ベン図}

% 5
\begin{frame}{命題は空間を区切る}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.48\textwidth}
    \begin{block}{例}
      整数 $n$ について
      \[
        P(n): n\text{ は }3\text{の倍数}
      \]
      これは整数全体 $\Z$ の中の\key{領域}を決める．
    \end{block}
    \[
      \sem{P} \subseteq \Z
    \]
  \end{column}
  \begin{column}{0.48\textwidth}
    \centering
    \begin{tikzpicture}[scale=0.75]
      \coordinate (A) at (-1.25,0);
      \coordinate (B) at (1.25,0);
      \fill[gray!12] (-3.5,-2.1) rectangle (3.5,2.1);
      \fill[lightblue] (A) circle (1.45);
      \fill[lightorange,opacity=.85] (B) circle (1.45);
      \draw[thick] (-3.5,-2.1) rectangle (3.5,2.1);
      \draw[thick,toposblue] (A) circle (1.45);
      \draw[thick,toposorange] (B) circle (1.45);
      \node[above] at (0,2.15) {$\Z$};
      \node at (-1.65,.95) {$3$の倍数};
      \node at (1.65,.95) {$5$の倍数};
      \node at (0,0) {$15$の倍数};
    \end{tikzpicture}
  \end{column}
\end{columns}
\end{frame}

% 6
\begin{frame}{論理演算は塗り絵である}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.47\textwidth}
    \begin{block}{辞書}
    \centering
    \begin{tabular}{c|c}
      論理 & 図形 \\
      \hline
      $P\land Q$ & $P\cap Q$ \\
      $P\lor Q$ & $P\cup Q$ \\
      $\neg P$ & 補集合 \\
      $\top,\bot$ & 全体，空集合
    \end{tabular}
    \end{block}
  \end{column}
  \begin{column}{0.48\textwidth}
    \centering
    \begin{tikzpicture}[scale=0.85]
      \fill[gray!10] (-3,-1.7) rectangle (3,1.7);
      \begin{scope}
        \clip (-.75,0) circle (1.15);
        \fill[toposgreen!45] (.75,0) circle (1.15);
      \end{scope}
      \draw[thick,toposblue] (-.75,0) circle (1.15) node[left=.9cm] {$P$};
      \draw[thick,toposorange] (.75,0) circle (1.15) node[right=.9cm] {$Q$};
      \draw[thick] (-3,-1.7) rectangle (3,1.7);
      \node[below] at (0,-1.95) {$P\land Q$ は重なり};
    \end{tikzpicture}
  \end{column}
\end{columns}
\vspace{4pt}
\centering
\Large \geom{推論}が，\geom{領域の計算}になる．
\end{frame}

% 7
\begin{frame}{パズル1: 「ならば」はどこを禁止する？}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.46\textwidth}
    \begin{block}{問題}
      $P\Rightarrow Q$ は，ベン図では何を意味する？
    \end{block}
    \vspace{4pt}
    答えは
    \[
      \textcolor{toposred}{P\cap \neg Q=\varnothing},\qquad P\subseteq Q.
    \]
  \end{column}
  \begin{column}{0.50\textwidth}
    \centering
    \begin{tikzpicture}[scale=0.95]
      \fill[gray!10] (-3,-1.8) rectangle (3,1.8);
      \coordinate (P) at (-.9,0);
      \coordinate (Q) at (.55,0);
      \begin{scope}
        \clip (P) circle (1.15);
        \fill[lightred] (-3,-2) rectangle (3,2);
        \fill[white] (Q) circle (1.45);
      \end{scope}
      \fill[lightblue,opacity=.45] (Q) circle (1.45);
      \draw[thick,toposblue] (P) circle (1.15);
      \draw[thick,toposorange] (Q) circle (1.45);
      \draw[thick] (-3,-1.8) rectangle (3,1.8);
      \node at (-1.65,.9) {$P$};
      \node at (1.15,.95) {$Q$};
      \node[warn] at (-1.25,-1.25) {禁止領域};
    \end{tikzpicture}
  \end{column}
\end{columns}
\end{frame}

% 8
\begin{frame}{パズル2: 三段論法は入れ子である}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.43\textwidth}
    \begin{block}{条件}
      \[
        P\Rightarrow Q,\qquad Q\Rightarrow R
      \]
    \end{block}
    \begin{block}{結論}
      \[
        P\Rightarrow R
      \]
    \end{block}
  \end{column}
  \begin{column}{0.53\textwidth}
    \centering
    \begin{tikzpicture}[scale=0.9]
      \fill[lightgreen] (0,0) circle (1.9);
      \fill[lightorange] (0,0) circle (1.25);
      \fill[lightblue] (0,0) circle (.65);
      \draw[very thick,toposgreen] (0,0) circle (1.9);
      \draw[very thick,toposorange] (0,0) circle (1.25);
      \draw[very thick,toposblue] (0,0) circle (.65);
      \node at (0,0) {$P$};
      \node at (0,1.05) {$Q$};
      \node at (0,1.68) {$R$};
      \node[below] at (0,-2.25) {$P\subseteq Q\subseteq R$};
    \end{tikzpicture}
  \end{column}
\end{columns}
\centering
\Large 論理の推論が，\key{包含関係}に変わった．
\end{frame}

% 9
\begin{frame}{パズル3: ド・モルガンの法則}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.45\textwidth}
    \begin{block}{問い}
      重なりでない場所はどこ？
      \[
        \neg(P\land Q)= ?
      \]
    \end{block}
    \vspace{8pt}
    塗ってみると
    \[
      \neg(P\land Q)=\neg P\lor \neg Q.
    \]
  \end{column}
  \begin{column}{0.52\textwidth}
    \centering
    \begin{tikzpicture}[scale=0.78]
      \fill[lightred] (-3.4,-2) rectangle (3.4,2);
      \begin{scope}
        \clip (-.85,0) circle (1.25);
        \fill[white] (.85,0) circle (1.25);
      \end{scope}
      \draw[thick] (-3.4,-2) rectangle (3.4,2);
      \draw[very thick,toposblue] (-.85,0) circle (1.25);
      \draw[very thick,toposorange] (.85,0) circle (1.25);
      \node at (-1.55,1.08) {$P$};
      \node at (1.55,1.08) {$Q$};
      \node[orangebox] at (0,-1.35) {白いところだけが $P\land Q$};
    \end{tikzpicture}
  \end{column}
\end{columns}
\end{frame}

% 10
\begin{frame}{ベン図から抽象化へ}
\centering
\begin{tikzpicture}[node distance=1.1cm]
  \node[idea,minimum width=3.2cm] (shape) {丸や四角};
  \node[idea,minimum width=3.2cm,right=of shape] (calc) {$\cap,\ \cup,\ \complement$};
  \node[idea,minimum width=3.2cm,right=of calc] (alg) {ブール代数};
  \draw[->,very thick,toposorange] (shape) -- node[above]{形を忘れる} (calc);
  \draw[->,very thick,toposorange] (calc) -- node[above]{規則を残す} (alg);
\end{tikzpicture}
\vspace{16pt}
\begin{block}{ここまでの結論}
\centering
\Large 論理は，\key{領域の代数}として扱える．
\end{block}
\vspace{4pt}
\mathside{Boolean algebra は「ベン図でできる計算」の抽象化．}
\end{frame}

\section{Stone双対}

% 11
\begin{frame}{ブール代数: 論理を代数にする}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.48\textwidth}
    \begin{block}{命題の演算}
      \[
        \land,\quad \lor,\quad \neg,\quad \top,\quad \bot
      \]
      これらの計算規則だけを取り出す．
    \end{block}
  \end{column}
  \begin{column}{0.48\textwidth}
    \begin{block}{例}
      ある集合 $X$ の部分集合全体
      \[
        \mathcal{P}(X)
      \]
      はブール代数．
    \end{block}
  \end{column}
\end{columns}
\vspace{8pt}
\centering
\begin{tikzpicture}
  \node[orangebox] {\Large ベン図の「計算」だけを，紙から切り離す};
\end{tikzpicture}
\end{frame}

% 12
\begin{frame}{可能世界の空間}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.43\textwidth}
    命題変数 $p,q$ を考える．
    \vspace{4pt}

    それぞれ true/false なので，可能世界は4つ．
    \[
      00,\ 01,\ 10,\ 11
    \]
  \end{column}
  \begin{column}{0.54\textwidth}
    \centering
    \begin{tikzpicture}[scale=1]
      \draw[->,thick] (-.6,-.6) -- (3.2,-.6) node[right] {$p$};
      \draw[->,thick] (-.6,-.6) -- (-.6,2.6) node[above] {$q$};
      \node[world] at (0,0) {$00$};
      \node[world] at (2,0) {$10$};
      \node[world] at (0,2) {$01$};
      \node[world] at (2,2) {$11$};
      \draw[dashed,gray] (0,0) -- (2,0) -- (2,2) -- (0,2) -- cycle;
    \end{tikzpicture}
  \end{column}
\end{columns}
\vspace{6pt}
\centering
\Large \key{点}とは，命題たちへの真偽値割り当てである．
\end{frame}

% 13
\begin{frame}{命題は，可能世界の集合である}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.43\textwidth}
    式
    \[
      \varphi=p\land \neg q
    \]
    は，それを真にする世界の集合を決める．
    \[
      \widehat{\varphi}=\{v\mid v(\varphi)=1\}
    \]
  \end{column}
  \begin{column}{0.54\textwidth}
    \centering
    \begin{tikzpicture}[scale=1]
      \node[world,fill=gray!10] at (0,0) {$00$};
      \node[world,fill=lightred,draw=toposred,very thick] at (2,0) {$10$};
      \node[world,fill=gray!10] at (0,2) {$01$};
      \node[world,fill=gray!10] at (2,2) {$11$};
      \draw[dashed,gray] (0,0) -- (2,0) -- (2,2) -- (0,2) -- cycle;
      \node[warn] at (2,-1.0) {$p\land\neg q$ が真};
    \end{tikzpicture}
  \end{column}
\end{columns}
\vspace{4pt}
\centering
\Large ベン図の領域は，\geom{可能世界の集合}だった．
\end{frame}

% 14
\begin{frame}{Stone双対: 論理から空間を復元する}
\centering
\begin{block}{Stone双対定理}
\centering
\Large
\[
  B \cong \Clop(\Stone(B))
\]
\end{block}
\vspace{4pt}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.48\textwidth}
    \begin{block}{左辺}
      抽象的なブール代数 $B$．\par
      つまり，命題の計算規則．
    \end{block}
  \end{column}
  \begin{column}{0.48\textwidth}
    \begin{block}{右辺}
      ある空間の clopen 領域．\par
      つまり，開かつ閉じたベン図．
    \end{block}
  \end{column}
\end{columns}
\end{frame}

% 15
\begin{frame}{Stone空間の点とは何か}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.51\textwidth}
    ブール代数 $B$ の Stone 空間は
    \[
      \Stone(B)=\Hom_{\mathrm{BA}}(B,2).
    \]
    点は，すべての命題に矛盾なく true/false を入れる方法．
  \end{column}
  \begin{column}{0.45\textwidth}
    \centering
    \begin{tikzpicture}[node distance=.6cm,scale=.88,transform shape]
      \node[idea] (B) {$B$};
      \node[idea,right=.9cm of B] (two) {$2=\{0,1\}$};
      \draw[->,very thick,toposblue] (B) -- node[above] {評価} (two);
      \node[world,below=.9cm of B] (x) {$x$};
      \node[below=.2cm of x] {点 $=$ 評価};
    \end{tikzpicture}
  \end{column}
\end{columns}
\vspace{6pt}
\centering
\begin{tikzpicture}
  \node[orangebox] {式が，領域に変わる．};
\end{tikzpicture}
\end{frame}

% 16
\begin{frame}{コンパクト性: 完全性定理の影}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.49\textwidth}
    理論 $\Gamma$ のモデル全体は
    \[
      \bigcap_{\varphi\in\Gamma}\widehat{\varphi}
    \]
    という閉集合の交わり．
    \vspace{4pt}

    有限個ずつ解けるなら，コンパクト性により全部解ける．
  \end{column}
  \begin{column}{0.48\textwidth}
    \centering
    \begin{tikzpicture}[scale=.88]
      \fill[lightblue] (0,0) circle (1.8);
      \fill[lightorange,opacity=.75] (.8,.15) circle (1.55);
      \fill[lightgreen,opacity=.75] (.25,.85) circle (1.35);
      \draw[thick,toposblue] (0,0) circle (1.8);
      \draw[thick,toposorange] (.8,.15) circle (1.55);
      \draw[thick,toposgreen] (.25,.85) circle (1.35);
      \node[dot,toposred] at (.45,.45) {};
      \node[warn] at (2.0,-1.8) {点が残る};
    \end{tikzpicture}
  \end{column}
\end{columns}
\vspace{4pt}
\centering
\Large \shock{論理の定理}が，\geom{空間の性質}になる．
\end{frame}

\section{Topos}

% 17
\begin{frame}{Stone空間だけでは足りない}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.50\textwidth}
    Stone 双対の命題は clopen．\par
    つまり白黒はっきりしている．
    \vspace{8pt}

    でも普通の空間には\shock{境界}がある．
    \[
      U=(0,\infty)\subseteq \R
    \]
    開集合の世界では
    \[
      U\lor \neg U=\R\setminus\{0\}\neq\R.
    \]
  \end{column}
  \begin{column}{0.46\textwidth}
    \centering
    \begin{tikzpicture}[scale=.88]
      \draw[->,thick] (-3,0) -- (3,0) node[right] {$\R$};
      \draw[very thick,toposblue] (.15,0) -- (2.8,0);
      \draw[very thick,toposorange] (-2.8,0) -- (-.15,0);
      \fill[white,draw=toposred,very thick] (0,0) circle (3pt);
      \node[above,toposblue] at (1.5,.15) {$U$};
      \node[above,toposorange] at (-1.5,.15) {$\neg U$};
      \node[warn] at (0,-1.1) {$0$ はどちらでもない};
    \end{tikzpicture}
  \end{column}
\end{columns}
\end{frame}

% 18
\begin{frame}{Topos: ベン図ができる宇宙}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.49\textwidth}
    集合の世界では
    \[
      \Sub_{\Set}(X)\cong \Set(X,2).
    \]
    部分集合は，特徴関数で表せる．
  \end{column}
  \begin{column}{0.49\textwidth}
    トポス $\Topos$ では
    \[
      \Sub_{\Topos}(X)\cong \Topos(X,\Omega).
    \]
    $\Omega$ は真理値の対象．
  \end{column}
\end{columns}
\vspace{4pt}
\centering
\begin{tikzpicture}[node distance=.9cm,scale=.94,transform shape]
  \node[idea] (sub) {部分対象};
  \node[idea,right=of sub] (pred) {述語};
  \node[idea,right=of pred] (truth) {真理値 $\Omega$};
  \draw[<->,very thick,toposorange] (sub) -- (pred);
  \draw[->,very thick,toposorange] (pred) -- (truth);
\end{tikzpicture}
\vspace{3pt}
\begin{tikzpicture}
  \node[orangebox] {\large Topos は，\key{ベン図ができる一般化された宇宙}．};
\end{tikzpicture}
\end{frame}

% 19
\begin{frame}{理論が空間になる}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.46\textwidth}
    coherent theory $T$ から，classifying topos
    \[
      \mathcal{E}_T
    \]
    が作られる．
    \vspace{4pt}

    その\key{点}は，$T$ の集合値モデル．
  \end{column}
  \begin{column}{0.50\textwidth}
    \centering
    \begin{tikzpicture}[node distance=.7cm]
      \node[idea,minimum width=3cm,minimum height=1.2cm] (ET) {$\mathcal{E}_T$};
      \node[world,below left=.9cm and .35cm of ET] (m1) {$M_1$};
      \node[world,below=.9cm of ET] (m2) {$M_2$};
      \node[world,below right=.9cm and .35cm of ET] (m3) {$M_3$};
      \draw[->,thick,toposblue] (m1) -- (ET);
      \draw[->,thick,toposblue] (m2) -- (ET);
      \draw[->,thick,toposblue] (m3) -- (ET);
      \node[below=.05cm of m2] {モデルたち};
    \end{tikzpicture}
  \end{column}
\end{columns}
\vspace{2pt}
\centering
\begin{tikzpicture}
  \node[orangebox] {coherent topos に点が十分ある $\Rightarrow$ 完全性定理．};
\end{tikzpicture}
\end{frame}

% 20
\begin{frame}{おまけ: 量化子も幾何である}
\begin{columns}[T,onlytextwidth]
  \begin{column}{0.48\textwidth}
    射影
    \[
      \pi:X\times Y\to X
    \]
    に沿って，述語を押し出す．
    \[
      \exists y\,R(x,y)
    \]
    は「影」として見える．
    \vspace{2pt}

    \mathside{$\exists_\pi\dashv \pi^\ast\dashv \forall_\pi$}
  \end{column}
  \begin{column}{0.48\textwidth}
    \centering
    \begin{tikzpicture}[scale=.82]
      \draw[->,thick] (-2.4,-1.5) -- (2.4,-1.5) node[right] {$X$};
      \draw[->,thick] (-2.2,-1.7) -- (-2.2,1.8) node[above] {$Y$};
      \fill[lightblue] (-.3,.25) ellipse (1.35 and .75);
      \draw[very thick,toposblue] (-.3,.25) ellipse (1.35 and .75);
      \draw[dashed,toposred,very thick] (-1.65,-1.5) -- (1.05,-1.5);
      \draw[dashed,toposred] (-1.65,-1.5) -- (-1.65,.1);
      \draw[dashed,toposred] (1.05,-1.5) -- (1.05,.1);
      \node[warn] at (-.3,-2.2) {影 $=\exists$};
    \end{tikzpicture}
  \end{column}
\end{columns}
\vspace{0pt}
\centering
\begin{tikzpicture}
  \node[orangebox] {\shock{ベン図の丸}は，\key{トポスの入口}だった．};
\end{tikzpicture}
\end{frame}

\end{document}