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Revision difference : closure space |
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Version 2 |
| Call a set $X$ with a closure operator defined on it a \emph{closure space}. |
Call a set $X$ with a closure operator defined on it a \emph{closure space}. |
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| Every topological space is a closure space, if we define the closure operator of the space as a function that takes any subset to its closure. The converse is also true: |
Every topological space is a closure space, if we define the closure operator of the space as a function that takes any subset to its closure. The converse is also true: |
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| \begin{prop} Let $X$ be a closure space with $c$ the associated closure operator. Define a ``closed set'' of $X$ as a subset $A$ of $X$ such that $A^c=A$, and an ``open set'' of $X$ as the complement of some closed set of $X$. Then the collection $\mathcal{T}$ of all open sets of $X$ is a topology on $X$. \end{prop} |
\begin{prop} Let $X$ be a closure space with $c$ the associated closure operator. Define a ``closed set'' of $X$ as a subset $A$ of $X$ such that $A^c=A$, and an ``open set'' of $X$ as the complement of some closed set of $X$. Then the collection $\mathcal{T}$ of all open sets of $X$ is a topology on $X$. \end{prop} |
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| \begin{proof} |
\begin{proof} |
| Since $\varnothing^c=\varnothing$, $\varnothing$ is closed. Also, $X\subseteq X^c$ and $X^c\subseteq X$ imply that $X^c=X$, or $X$ is closed. If $A,B\subseteq X$ are closed, then $(A\cup B)^c=A^c\cup B^c=A\cup B$ is closed as well. Finally, suppose $A_i$ are closed. Let $B=\bigcap A_i$. For each $i$, $A_i=B\cup A_i$, so $A_i=A_i^c=(B\cup A_i)^c=B^c\cup A_i^c=B^c\cup A_i$. This means $B^c\subseteq A_i$, or $B^c\subseteq \bigcap A_i=B$. But $B\subseteq B^c$ by definition, so $B=B^c$, or that $\bigcap A_i$ is closed. |
Since $\varnothing^c=\varnothing$, $\varnothing$ is closed. Also, $X\subseteq X^c$ and $X^c\subseteq X$ imply that $X^c=X$, or $X$ is closed. If $A,B\subseteq X$ are closed, then $(A\cup B)^c=A^c\cup B^c=A\cup B$ is closed as well. Finally, suppose $A_i$ are closed. Let $B=\bigcap A_i$. For each $i$, $A_i=B\cup A_i$, so $A_i=A_i^c=(B\cup A_i)^c=B^c\cup A_i^c=B^c\cup A_i$. This means $B^c\subseteq A_i$, or $B^c\subseteq \bigcap A_i=B$. But $B\subseteq B^c$ by definition, so $B=B^c$, or that $\bigcap A_i$ is closed. |
| \end{proof} |
\end{proof} |
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| $\mathcal{T}$ so defined is called the \emph{closure topology} of $X$ with respect to the closure operator $c$. |
$\mathcal{T}$ so defined is called the \emph{closure topology} of $X$ with respect to the closure operator $c$. |
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