Difference between revisions of "Subspace topology"

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{{Refactor notice|grade=A|msg=Needed urgently, ready to plough on with it now though!}}
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==Definition==
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Let {{Top.|X|J}} be a [[topological space]] and let {{M|S\in\mathcal{P}(X)}}<ref group="Note">Recall {{M|\mathcal{P}(X)}} denotes the [[power set]] of {{M|X}} and {{M|S\in\mathcal{P}(X)\iff S\subseteq X}}, so it's another way of saying let {{M|S}} be a subset of {{M|X}}, possibly empty, possibly equal to {{M|X}} itself</ref> be given. We can construct a new topological space, {{M|(S,\mathcal{J}_S)}} where the [[topology]] {{M|\mathcal{J}_S}} is known as {{nowrap|"the ''subspace topology''}} on {{M|S}}"{{rITTMJML}} ({{AKA}}: ''relative topology'' on {{M|S}}<ref name="ITTMJML"/>) and is defined as follows:
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* {{M|1=\mathcal{J}_S:=\{U\cap S\ \vert\ U\in\mathcal{J}\} }} - the open sets of {{M|(S,\mathcal{J}_S)}} are precisely the intersection of open sets of {{Top.|X|J}} with {{M|S}}
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** '''Claim 1: ''' this is indeed a [[topology]]<ref name="ITTMJML"/>
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Alternatively:
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* '''Claim 2: ''' {{M|1=\forall U\in\mathcal{P}(S)\big[U\in\mathcal{J}_S\iff\exists V\in\mathcal{J}[U=S\cap V]\big]}}<ref name="ITTMJML"/>
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==Terminology==
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* Let {{M|U\in\mathcal{P}(S)}} be given. For clarity rather than saying {{M|U}} is [[open set|open]], or {{M|U}} is [[closed set|closed]] (which is surprisingly ambiguous when using subspaces) we instead say:
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*# {{M|U}} is ''[[relatively open]]''<ref name="ITTMJML"/> - indicating we mean open in the subspace, or
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*# {{M|U}} is ''[[relatively closed]]''<ref name="ITTMJML"/> - indicating we mean closed in the subspace
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{{Todo|Closed and open subspace terminology, For example if {{M|S\in\mathcal{P}(X)}} is ''closed'' with respect to the topology {{M|\mathcal{J} }} on {{M|X}}, then we call {{M|S}} imbued with the subspace topology a ''closed subspace''}}
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==[[Characteristic property of the subspace topology|Characteristic property]]==
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{{:Characteristic property of the subspace topology/Statement}}
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==Proof of claims==
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===Claim 1: {{M|\mathcal{J}_S}} is a [[topology]]===
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{{Requires proof|grade=C|msg=Really easy, hence low importance|easy=true}}
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===Claim 2: Equivalent formulation of the relatively open sets===
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{{Requires proof|grade=C|msg=Really easy, hence low importance|easy=true}}
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==See next==
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{{Todo|Theorems and propositions involving subspaces}}
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==See also==
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* [[Topological embedding]]
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{{Todo|Link to more things}}
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==Notes==
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<references group="Note"/>
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==References==
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<references/>
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{{Topology navbox|plain}}
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{{Definition|Topology}}
  
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=OLD PAGE=
 
==Definition==
 
==Definition==
 
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>
 
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>

Revision as of 22:28, 25 September 2016

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Definition

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]S\in\mathcal{P}(X)[/ilmath][Note 1] be given. We can construct a new topological space, [ilmath](S,\mathcal{J}_S)[/ilmath] where the topology [ilmath]\mathcal{J}_S[/ilmath] is known as "the subspace topology on [ilmath]S[/ilmath]"[1] (AKA: relative topology on [ilmath]S[/ilmath][1]) and is defined as follows:

  • [ilmath]\mathcal{J}_S:=\{U\cap S\ \vert\ U\in\mathcal{J}\}[/ilmath] - the open sets of [ilmath](S,\mathcal{J}_S)[/ilmath] are precisely the intersection of open sets of [ilmath](X,\mathcal{ J })[/ilmath] with [ilmath]S[/ilmath]

Alternatively:

  • Claim 2: [ilmath]\forall U\in\mathcal{P}(S)\big[U\in\mathcal{J}_S\iff\exists V\in\mathcal{J}[U=S\cap V]\big][/ilmath][1]

Terminology

  • Let [ilmath]U\in\mathcal{P}(S)[/ilmath] be given. For clarity rather than saying [ilmath]U[/ilmath] is open, or [ilmath]U[/ilmath] is closed (which is surprisingly ambiguous when using subspaces) we instead say:
    1. [ilmath]U[/ilmath] is relatively open[1] - indicating we mean open in the subspace, or
    2. [ilmath]U[/ilmath] is relatively closed[1] - indicating we mean closed in the subspace

TODO: Closed and open subspace terminology, For example if [ilmath]S\in\mathcal{P}(X)[/ilmath] is closed with respect to the topology [ilmath]\mathcal{J} [/ilmath] on [ilmath]X[/ilmath], then we call [ilmath]S[/ilmath] imbued with the subspace topology a closed subspace

Characteristic property

[ilmath]\xymatrix{ Y \ar[r]^f \ar[dr]_{i_S\circ f} & S \ar@{^{(}->}[d]^{i_S}\\ & X}[/ilmath]
Diagram
Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath](S,\mathcal{J}_S)[/ilmath] be any subspace of [ilmath](X,\mathcal{ J })[/ilmath][Note 2]. The characteristic property of the subspace topology[1] is that:
  • Given any topological space [ilmath](Y,\mathcal{ K })[/ilmath] and any map [ilmath]f:Y\rightarrow S[/ilmath] we have:
    • [ilmath](f:Y\rightarrow S [/ilmath] is continuous[ilmath])\iff(i_S\circ f:Y\rightarrow X [/ilmath] is continuous[ilmath])[/ilmath]

Where [ilmath]i_S:S\rightarrow X[/ilmath] given by [ilmath]i_S:s\mapsto s[/ilmath] is the canonical injection of the subspace topology (which is itself continuous)[Note 3]

Proof of claims

Claim 1: [ilmath]\mathcal{J}_S[/ilmath] is a topology

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Claim 2: Equivalent formulation of the relatively open sets

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This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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The message provided is:
Really easy, hence low importance

This proof has been marked as an page requiring an easy proof

See next

TODO: Theorems and propositions involving subspaces

See also

TODO: Link to more things

Notes

  1. ↑ Recall [ilmath]\mathcal{P}(X)[/ilmath] denotes the power set of [ilmath]X[/ilmath] and [ilmath]S\in\mathcal{P}(X)\iff S\subseteq X[/ilmath], so it's another way of saying let [ilmath]S[/ilmath] be a subset of [ilmath]X[/ilmath], possibly empty, possibly equal to [ilmath]X[/ilmath] itself
  2. ↑ This means [ilmath]S\in\mathcal{P}(X)[/ilmath], or [ilmath]S\subseteq X[/ilmath] of course
  3. ↑ This leads to two ways to prove the statement:
    1. If we show [ilmath]i_S:S\rightarrow X[/ilmath] is continuous, then we can use the composition of continuous maps is continuous to show if [ilmath]f[/ilmath] continuous then so is [ilmath]i_S\circ f[/ilmath]
    2. We can show the property the "long way" and then show [ilmath]i_S:S\rightarrow X[/ilmath] is continuous as a corollary

References

  1. ↑ 1.0 1.1 1.2 1.3 1.4 1.5 1.6 Introduction to Topological Manifolds - John M. Lee

v • d • e
 
Topology
Overview of the structures studied in topology
Primitives
Global properties of topological spaces
Local properties of topological spaces
Constructs
Subtypes of topological spaces
inner product spaces [ilmath]\subset[/ilmath] normed spaces [ilmath]\subset[/ilmath] metric spaces [ilmath]\subset[/ilmath] topological spaces
Maps between topological spaces

OLD PAGE

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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