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'closed set'
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| Title of object: |
closed set |
| Canonical Name: |
ClosedSet |
| Type: |
Definition |
| Created on: |
2002-03-02 01:13:01 |
| Modified on: |
2006-10-22 08:39:10 |
| Classification: |
msc:54-00 |
| Defines: |
closed |
| Synonyms: |
closed set=closed subset |
Revision comment (for changes between this and next version):
Preamble:
\usepackage{bbm}
\newcommand{\Z}{\mathbbmss{Z}}
\newcommand{\C}{\mathbbmss{C}}
\newcommand{\R}{\mathbbmss{R}}
\newcommand{\Q}{\mathbbmss{Q}}
\newcommand{\mathbb}[1]{\mathbbmss{#1}} |
Content:
Let $(X,\tau)$ be a topological space. Then a subset $C\subseteq X$ is \emph{closed} if its complement $X\backslash C$ is open under the topology $\tau$.
Example:
\begin{itemize}
\item In any topological space $(X,\tau)$, the sets $X$ and $\emptyset$ are always closed.
\item Consider $\R$ with the standard topology. Then $[0,1]$ is closed since its complement $(-\infty,0) \cup (1,\infty)$ is open (being the union of two open sets).
\item Consider $\R$ with the lower limit topology. Then $[0,1)$ is closed since its complement $(-\infty,0)\cup[1,\infty)$ is open.
\end{itemize}
Closed subsets can also be characterized as follows:
A subset $C\subseteq X$ is closed if and only if $C$ contains all of its cluster points. That is, $C=C'$.
So the set $\{1,1/2,1/3,1/4,\ldots\}$ is not closed under the standard topology on $\R$ since $0$ is a cluster point not contained in the set. |
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