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>. | ||
| + | ===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 {{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. | ||
| − | == | + | ==Example== |
| − | === | + | {{M|(0,1)}} is not closed, as take the point {{M|0}}. |
| − | + | ====Proof==== | |
| − | === | + | 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}} | ||
| + | |||
| + | ==See also== | ||
| + | * [[Relatively closed]] | ||
| + | * [[Open set]] | ||
| + | * [[Neighbourhood]] | ||
| + | ==Notes== | ||
| + | <references group="Note"/> | ||
| + | ==References== | ||
| + | <references/> | ||
| − | {{Definition|Topology}} | + | {{Definition|Topology|Metric Space}} |
Latest revision as of 15:36, 24 November 2015
Definition
A closed set in a topological space [math](X,\mathcal{J})[/math] is a set [math]A[/math] where [math]X-A[/math] 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 [ilmath]A[/ilmath] of the metric space [ilmath](X,d)[/ilmath] is closed if it contains all of its limit points[Note 1]
For convenience only: recall [ilmath]x[/ilmath] is a limit point if every neighbourhood of [ilmath]x[/ilmath] contains points of [ilmath]A[/ilmath] other than [ilmath]x[/ilmath] itself.
Example
[ilmath](0,1)[/ilmath] is not closed, as take the point [ilmath]0[/ilmath].
Proof
Let [ilmath]N[/ilmath] be any neighbourhood of [ilmath]x[/ilmath], 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 [ilmath]0[/ilmath] is certainly a limit point, but [ilmath]0\notin(0,1)[/ilmath]
TODO: This proof could be nonsense
See also
Notes
- ↑ Maurin proves this as an [ilmath]\iff[/ilmath] theorem. However he assumes the space is complete.
