Difference between revisions of "Subspace topology"

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==Definition==
 
==Definition==
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We define the subspace [[Topological space|topology]] as follows.
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Given a [[Topological space|topological space]] {{M|(X,\mathcal{J})}} and given a {{M|Y\subset X}} ({{M|Y}} is a subset of {{M|X}}) we define the ''subspace topology'' as follows:<ref name="Topology">Topology - Second Edition - Munkres</ref>
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* {{M|(Y,\mathcal{K})}} is a topological space where the [[Open set|open sets]], {{M|\mathcal{K} }}, are given by {{M|1=\mathcal{K}:=\{Y\cap V\vert\ V\in\mathcal{J}\} }}
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We may say any one of:
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# Let {{M|Y}} be a subspace of {{M|X}}
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# Let {{M|Y}} be a subspace of {{M|(X,\mathcal{J})}}
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and it is taken implicitly to mean {{M|Y}} is considered as a topological space with the ''subspace topology'' inherited from {{M|(X,\mathcal{J})}}
  
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Given a topological space <math>(X,\mathcal{J})</math> and any <math>Y\subset X</math> we can define a topology on <math>Y,\ (Y,\mathcal{J}_Y)</math> where <math>\mathcal{J}_Y=\{Y\cap U|U\in\mathcal{J}\}</math>
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==Proof of claims==
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{{Begin Theorem}}
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Claim 1: The subspace topology is indeed a topology
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{{Begin Proof}}
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Here {{M|(X,\mathcal{J})}} is a topological space and {{M|Y\subset X}} and {{M|\mathcal{K} }} is defined as above, we will prove that {{M|(Y,\mathcal{K})}} is a topology.
  
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We may say "<math>Y</math> is a subspace of <math>X</math> (or indeed <math>(X,\mathcal{J})</math>" to implicitly mean this topology.
 
  
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==Closed subspace==
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Recall that to be a topology {{M|(Y,\mathcal{K})}} must have the following properties:
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If {{m|Y}} is a "closed subspace" of {{m|(X,\mathcal{J})}} then it means that {{M|Y}} is [[Closed set|closed]] in {{M|X}} and should be considered with the subspace topology.
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# {{M|\emptyset\in\mathcal{K} }} and {{M|Y\in\mathcal{K} }}
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# For any {{M|U,V\in\mathcal{K} }} we must have {{M|U\cap V\in\mathcal{K} }}
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# For an [[Indexing set|arbitrary family]] {{M|\{U_\alpha\}_{\alpha\in I} }} of open sets (that is {{M|\forall\alpha\in I[U_\alpha\in\mathcal{K}]}}) we have:
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#* {{MM|\bigcup_{\alpha\in I}A_\alpha\in\mathcal{K} }}
  
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==Open subspace==
 
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{{Todo|same as closed, but with the word "open"}}
 
  
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==Open sets in open subspaces are open==
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'''Proof:'''
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{{Todo|easy}}
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# First we must show that {{M|\emptyset,Y\in\mathcal{K} }}
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#: Recall that {{M|\emptyset,X\in\mathcal{J} }} and notice that:
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#:* {{M|1=\emptyset\cap Y=\emptyset}}, so by the definition of {{M|\mathcal{K} }} we have {{M|\emptyset\in\mathcal{K} }}
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#:* {{M|1=X\cap Y=Y}}, so by the definition of {{M|\mathcal{K} }} we have {{M|Y\in\mathcal{K} }}
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# Next we must show...
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 +
{{Todo|Easy work just takes time to write!}}
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{{End Proof}}{{End Theorem}}
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==Terminology==
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* A '''closed subspace''' (of {{M|X}}) is a subset of {{M|X}} which is closed in {{M|X}} and is imbued with the subspace topology
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* A '''open subspace''' (of {{M|X}}) is a subset of {{M|X}} which is open in {{M|X}} and is imbued with the subspace topology
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{{Todo|Find reference}}
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* A set {{M|U\subseteq X}} is '''open relative to {{M|Y}}''' (or [[Relatively open|''relatively open'']] if it is obvious we are talking about a subspace {{M|Y}} of {{M|X}}) if {{M|U}} is open in {{M|Y}}
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** This implies that {{M|U\subseteq Y}}<ref name="Topology"/>
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* A set {{M|U\subseteq X}} is '''closed relative to {{M|Y}}''' (or [[Relatively closed|''relatively closed'']] if it is obvious we are talking about a subspace {{M|Y}} of {{M|X}}) if {{M|U}} is [[Closed set|closed]] in {{M|Y}}
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** This also implies that {{M|U\subseteq Y}}
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==Immediate theorems==
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{{Begin Theorem}}
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Theorem: Let {{M|Y}} be a subspace of {{M|X}}, if {{M|U}} is open in {{M|Y}} and {{M|Y}} is open in {{M|X}} then {{M|U}} is open in {{M|X}}<ref name="Topology"/>
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{{Begin Proof}}
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This may be easier to read symbolically:
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* if {{M|U\in\mathcal{K} }} and {{M|Y\in\mathcal{J} }} then {{M|U\in\mathcal{J} }}
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 +
 
 +
'''Proof:'''
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: Since {{M|U}} is open in {{M|Y}} we know that:
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:* {{M|1=U=Y\cap V}} for some {{M|V}} open in {{M|X}} (for some {{M|V\in\mathcal{J} }})
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: Since {{M|Y}} and {{M|V}} are both open in {{M|X}} we know:
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:* {{M|Y\cap V}} is open in {{M|X}}
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: it follows that {{M|U}} is open in {{M|X}}
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{{End Proof}}{{End Theorem}}
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==References==
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<references/>
  
 
{{Definition|Topology}}
 
{{Definition|Topology}}

Revision as of 03:12, 22 June 2015

Definition

Given a topological space [ilmath](X,\mathcal{J})[/ilmath] and given a [ilmath]Y\subset X[/ilmath] ([ilmath]Y[/ilmath] is a subset of [ilmath]X[/ilmath]) we define the subspace topology as follows:[1]

  • [ilmath](Y,\mathcal{K})[/ilmath] is a topological space where the open sets, [ilmath]\mathcal{K} [/ilmath], are given by [ilmath]\mathcal{K}:=\{Y\cap V\vert\ V\in\mathcal{J}\}[/ilmath]

We may say any one of:

  1. Let [ilmath]Y[/ilmath] be a subspace of [ilmath]X[/ilmath]
  2. Let [ilmath]Y[/ilmath] be a subspace of [ilmath](X,\mathcal{J})[/ilmath]

and it is taken implicitly to mean [ilmath]Y[/ilmath] is considered as a topological space with the subspace topology inherited from [ilmath](X,\mathcal{J})[/ilmath]

Proof of claims

Claim 1: The subspace topology is indeed a topology

Here [ilmath](X,\mathcal{J})[/ilmath] is a topological space and [ilmath]Y\subset X[/ilmath] and [ilmath]\mathcal{K} [/ilmath] is defined as above, we will prove that [ilmath](Y,\mathcal{K})[/ilmath] is a topology.


Recall that to be a topology [ilmath](Y,\mathcal{K})[/ilmath] must have the following properties:

  1. [ilmath]\emptyset\in\mathcal{K} [/ilmath] and [ilmath]Y\in\mathcal{K} [/ilmath]
  2. For any [ilmath]U,V\in\mathcal{K} [/ilmath] we must have [ilmath]U\cap V\in\mathcal{K} [/ilmath]
  3. For an arbitrary family [ilmath]\{U_\alpha\}_{\alpha\in I} [/ilmath] of open sets (that is [ilmath]\forall\alpha\in I[U_\alpha\in\mathcal{K}][/ilmath]) we have:
    • [math]\bigcup_{\alpha\in I}A_\alpha\in\mathcal{K} [/math]


Proof:

  1. First we must show that [ilmath]\emptyset,Y\in\mathcal{K} [/ilmath]
    Recall that [ilmath]\emptyset,X\in\mathcal{J} [/ilmath] and notice that:
    • [ilmath]\emptyset\cap Y=\emptyset[/ilmath], so by the definition of [ilmath]\mathcal{K} [/ilmath] we have [ilmath]\emptyset\in\mathcal{K} [/ilmath]
    • [ilmath]X\cap Y=Y[/ilmath], so by the definition of [ilmath]\mathcal{K} [/ilmath] we have [ilmath]Y\in\mathcal{K} [/ilmath]
  2. Next we must show...


TODO: Easy work just takes time to write!


Terminology

  • A closed subspace (of [ilmath]X[/ilmath]) is a subset of [ilmath]X[/ilmath] which is closed in [ilmath]X[/ilmath] and is imbued with the subspace topology
  • A open subspace (of [ilmath]X[/ilmath]) is a subset of [ilmath]X[/ilmath] which is open in [ilmath]X[/ilmath] and is imbued with the subspace topology

TODO: Find reference

  • A set [ilmath]U\subseteq X[/ilmath] is open relative to [ilmath]Y[/ilmath] (or relatively open if it is obvious we are talking about a subspace [ilmath]Y[/ilmath] of [ilmath]X[/ilmath]) if [ilmath]U[/ilmath] is open in [ilmath]Y[/ilmath]
    • This implies that [ilmath]U\subseteq Y[/ilmath][1]
  • A set [ilmath]U\subseteq X[/ilmath] is closed relative to [ilmath]Y[/ilmath] (or relatively closed if it is obvious we are talking about a subspace [ilmath]Y[/ilmath] of [ilmath]X[/ilmath]) if [ilmath]U[/ilmath] is closed in [ilmath]Y[/ilmath]
    • This also implies that [ilmath]U\subseteq Y[/ilmath]

Immediate theorems

Theorem: Let [ilmath]Y[/ilmath] be a subspace of [ilmath]X[/ilmath], if [ilmath]U[/ilmath] is open in [ilmath]Y[/ilmath] and [ilmath]Y[/ilmath] is open in [ilmath]X[/ilmath] then [ilmath]U[/ilmath] is open in [ilmath]X[/ilmath][1]

This may be easier to read symbolically:

  • if [ilmath]U\in\mathcal{K} [/ilmath] and [ilmath]Y\in\mathcal{J} [/ilmath] then [ilmath]U\in\mathcal{J} [/ilmath]


Proof:

Since [ilmath]U[/ilmath] is open in [ilmath]Y[/ilmath] we know that:
  • [ilmath]U=Y\cap V[/ilmath] for some [ilmath]V[/ilmath] open in [ilmath]X[/ilmath] (for some [ilmath]V\in\mathcal{J} [/ilmath])
Since [ilmath]Y[/ilmath] and [ilmath]V[/ilmath] are both open in [ilmath]X[/ilmath] we know:
  • [ilmath]Y\cap V[/ilmath] is open in [ilmath]X[/ilmath]
it follows that [ilmath]U[/ilmath] is open in [ilmath]X[/ilmath]


References

  1. ↑ 1.0 1.1 1.2 Topology - Second Edition - Munkres
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