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Version 4 |
| 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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| \textbf{Remark}. A closure space can be more generally defined as a set $X$ together with an operator $\cl:P(X)\to P(X)$ such that $\cl$ satisfies all of the Kuratowski's closure axioms where the equal sign ``$=$'' is replaced with set inclusion ``$\subseteq$'', and the preservation of $\varnothing$ is no longer assumed. Alternatively, a closure space in this general sense is a set $X$ and an operator $\cl$ on $P(X)$ such that |
\textbf{Remark}. A closure space can be more generally defined as a set $X$ together with an operator $\cl:P(X)\to P(X)$ such that $\cl$ satisfies all of the Kuratowski's closure axioms where the equal sign ``$=$'' is replaced with set inclusion ``$\subseteq$'', and the preservation of $\varnothing$ is no longer assumed. In this sense, a closure space is more general than a topological space. |
| \begin{itemize} |
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| \item $A\subseteq \cl(A)$, |
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| \item $\cl(\cl(A))\subseteq \cl(A)$, and |
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| \item $\cl$ is order-preserving, i.e., if $A\subseteq B$, then $\cl(A)\subseteq \cl(B)$. |
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| \end{itemize} |
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| In this sense, a closure space is more general than a topological space. |
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