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!}} | ||
+ | __TOC__ | ||
+ | ==Definition== | ||
+ | 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: | ||
+ | * {{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}} | ||
+ | ** '''Claim 1: ''' this is indeed a [[topology]]<ref name="ITTMJML"/> | ||
+ | Alternatively: | ||
+ | * '''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"/> | ||
+ | We get with this a map, called the ''[[canonical injection of the subspace topology]]'', often denoted {{M|i_S:S\rightarrow X}} or {{M|\iota_S:S\rightarrow X}} given by {{M|i_S:s\mapsto s}}. This is an example of an [[inclusion map]], and it is [[continuous]]. | ||
+ | Note that if one proves {{M|i_S}} is continuous then the [[characteristic property of the subspace topology|characteristic property]] boils down to little more than [[the composition of continuous maps is continuous]], if one proves the characteristic property first, then continuity of {{m|i_S}} comes from it as a [[corollary]] | ||
+ | |||
+ | ==Terminology== | ||
+ | * 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: | ||
+ | *# {{M|U}} is ''[[relatively open]]''<ref name="ITTMJML"/> - indicating we mean open in the subspace, or | ||
+ | *# {{M|U}} is ''[[relatively closed]]''<ref name="ITTMJML"/> - indicating we mean closed in the subspace | ||
+ | {{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''}} | ||
+ | ==[[Characteristic property of the subspace topology|Characteristic property]]== | ||
+ | {{:Characteristic property of the subspace topology/Statement}} | ||
+ | ==Proof of claims== | ||
+ | ===Claim 1: {{M|\mathcal{J}_S}} is a [[topology]]=== | ||
+ | {{Requires proof|grade=C|msg=Really easy, hence low importance|easy=true}} | ||
+ | ===Claim 2: Equivalent formulation of the relatively open sets=== | ||
+ | {{Requires proof|grade=C|msg=Really easy, hence low importance|easy=true}} | ||
+ | ==See next== | ||
+ | {{Todo|Theorems and propositions involving subspaces}} | ||
+ | ==See also== | ||
+ | * [[Topological embedding]] | ||
+ | {{Todo|Link to more things}} | ||
+ | ==Notes== | ||
+ | <references group="Note"/> | ||
+ | ==References== | ||
+ | <references/> | ||
+ | {{Topology navbox|plain}} | ||
+ | {{Definition|Topology}} | ||
+ | |||
+ | =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> |
Latest revision as of 23:04, 25 September 2016
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Contents
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]
We get with this a map, called the canonical injection of the subspace topology, often denoted [ilmath]i_S:S\rightarrow X[/ilmath] or [ilmath]\iota_S:S\rightarrow X[/ilmath] given by [ilmath]i_S:s\mapsto s[/ilmath]. This is an example of an inclusion map, and it is continuous.
Note that if one proves [ilmath]i_S[/ilmath] is continuous then the characteristic property boils down to little more than the composition of continuous maps is continuous, if one proves the characteristic property first, then continuity of [ilmath]i_S[/ilmath] comes from it as a corollary
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:
- [ilmath]U[/ilmath] is relatively open[1] - indicating we mean open in the subspace, or
- [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
- 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
The message provided is:
This proof has been marked as an page requiring an easy proof
Claim 2: Equivalent formulation of the relatively open sets
The message provided is:
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
- ↑ 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
- ↑ This means [ilmath]S\in\mathcal{P}(X)[/ilmath], or [ilmath]S\subseteq X[/ilmath] of course
- ↑ This leads to two ways to prove the statement:
- 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]
- We can show the property the "long way" and then show [ilmath]i_S:S\rightarrow X[/ilmath] is continuous as a corollary
References
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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:
- Let [ilmath]Y[/ilmath] be a subspace of [ilmath]X[/ilmath]
- 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:
- [ilmath]\emptyset\in\mathcal{K} [/ilmath] and [ilmath]Y\in\mathcal{K} [/ilmath]
- For any [ilmath]U,V\in\mathcal{K} [/ilmath] we must have [ilmath]U\cap V\in\mathcal{K} [/ilmath]
- 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:
- 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]
- Recall that [ilmath]\emptyset,X\in\mathcal{J} [/ilmath] and notice that:
- 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]