Definition:Closed Set
This page is about closed set in the context of topology. For other uses, see closed.
![]() | This page has been identified as a candidate for refactoring of advanced complexity. In particular: Reconsider approach, maybe make this a disambiguation, too? As it stands, the Closure Operator definition is awkward here Until this has been finished, please leave {{Refactor}} in the code.
New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Refactor}} from the code. |
Definition
Topology
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
$H$ is closed (in $T$) if and only if its complement $S \setminus H$ is open in $T$.
That is, $H$ is closed if and only if $\paren {S \setminus H} \in \tau$.
That is, if and only if $S \setminus H$ is an element of the topology of $T$.
Metric Space
In the context of metric spaces, the same definition applies:
$H$ is closed (in $M$) if and only if its complement $A \setminus H$ is open in $M$.
Normed Vector Space
Let $V = \struct {X, \norm {\,\cdot\,} }$ be a normed vector space.
Let $F \subset X$.
Definition 1
$F$ is closed in $V$ if and only if its complement $X \setminus F$ is open in $V$.
Definition 2
$F$ is closed (in $V$) if and only if every limit point of $F$ is also a point of $F$.
That is: if and only if $F$ contains all its limit points.
Complex Analysis
Let $S \subseteq \C$ be a subset of the complex plane.
$S$ is closed (in $\C$) if and only if every limit point of $S$ is also a point of $S$.
That is: if and only if $S$ contains all its limit points.
Real Analysis
Let $S \subseteq \R$ be a subset of the set of real numbers.
Then $S$ is closed (in $\R$) if and only if its complement $\R \setminus S$ is an open set.
Under Closure Operator
The concept of closure can be made more generally than on a topological space:
Let $S$ be a set.
Let $\cl: \powerset S \to \powerset S$ be a closure operator.
Let $T \subseteq S$ be a subset.
Definition 1
$T$ is closed (with respect to $\cl$) if and only if:
- $\map \cl T = T$
Definition 2
$T$ is closed (with respect to $\cl$) if and only if $T$ is in the image of $\cl$:
- $T \in \Img \cl$
Also see
- Results about closed sets can be found here.
Internationalization
Closed (in this context) is translated:
In French: | fermé | (literally: closed) |