Difference between revisions of "Closed set"
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| − | + | ==Definition== | |
| − | + | A closed set in a [[Topological space|topological space]] <math>(X,\mathcal{J})</math> is a set <math>A</math> where <math>X-A</math> is open<ref>Introduction to topology - Third Edition - Mendelson</ref><ref name="KMAPI">Krzyzstof Maurin - Analysis - Part I: Elements</ref>. | |
| − | == | + | |
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| − | A closed set | + | |
===Metric space=== | ===Metric space=== | ||
| − | A subset {{M|A}} of the [[Metric space|metric space]] {{M|(X,d)}} is closed if it contains all of its [[Limit point|limit points]] | + | * '''Note: ''' as every [[metric space]] is also a [[topological space]] it is still true that a set is closed if its complement is open. |
| + | A subset {{M|A}} of the [[Metric space|metric space]] {{M|(X,d)}} is closed if it contains all of its [[Limit point|limit points]]<ref group="Note">Maurin proves this as an {{M|\iff}} theorem. However he assumes the space is complete.</ref> | ||
For convenience only: recall {{M|x}} is a limit point if every [[Open set#Neighbourhood|neighbourhood]] of {{M|x}} contains points of {{M|A}} other than {{M|x}} itself. | For convenience only: recall {{M|x}} is a limit point if every [[Open set#Neighbourhood|neighbourhood]] of {{M|x}} contains points of {{M|A}} other than {{M|x}} itself. | ||
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{{M|(0,1)}} is not closed, as take the point {{M|0}}. | {{M|(0,1)}} is not closed, as take the point {{M|0}}. | ||
====Proof==== | ====Proof==== | ||
| − | Let {{M|N}} be any [[Open set#Neighbourhood|neighbourhood]] of {{M|x}}, then <math>\exists \delta>0:B_\delta(x)\subset N</math> | + | Let {{M|N}} be any [[Open set#Neighbourhood|neighbourhood]] of {{M|x}}, then <math>\exists \delta>0:B_\delta(x)\subset N</math>, then: |
| − | + | * Take <math>y=\text{Max}\left(\frac{1}{2}\delta,\frac{1}{2}\right)</math>, then <math>y\in(0,1)</math> and <math>y\in N</math> thus {{M|0}} is certainly a limit point, but {{M|0\notin(0,1)}} | |
| − | + | {{Todo|This proof could be nonsense}} | |
| − | Take <math>y=\text{Max}\left(\frac{1}{2}\delta,\frac{1}{2}\right)</math>, then <math>y\in(0,1)</math> and <math>y\in N</math> thus {{M|0}} is certainly a limit point, but {{M|0\notin(0,1)}} | + | |
| − | + | ||
==See also== | ==See also== | ||
* [[Relatively closed]] | * [[Relatively closed]] | ||
* [[Open set]] | * [[Open set]] | ||
| − | + | * [[Neighbourhood]] | |
| + | ==Notes== | ||
| + | <references group="Note"/> | ||
==References== | ==References== | ||
<references/> | <references/> | ||
| − | {{Definition|Topology}} | + | {{Definition|Topology|Metric space}} |
Revision as of 15:12, 24 November 2015
Definition
A closed set in a topological space (X,J) is a set A where X−A is open[1][2].
Metric space
- Note: as every metric space is also a topological space it is still true that a set is closed if its complement is open.
A subset A of the metric space (X,d) is closed if it contains all of its limit points[Note 1]
For convenience only: recall x is a limit point if every neighbourhood of x contains points of A other than x itself.
Example
(0,1) is not closed, as take the point 0.
Proof
Let N be any neighbourhood of x, then ∃δ>0:Bδ(x)⊂N, then:
- Take y=Max(12δ,12), then y∈(0,1) and y∈N thus 0 is certainly a limit point, but 0∉(0,1)
TODO: This proof could be nonsense
See also
Notes
- Jump up ↑ Maurin proves this as an ⟺ theorem. However he assumes the space is complete.
