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Definition

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

JS:={U∩S | U∈J} - the open sets of (S,JS) are precisely the intersection of open sets of (X,J) with S
- Claim 1: this is indeed a topology^[1]

Alternatively:

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

We get with this a map, called the canonical injection of the subspace topology, often denoted $i_S:S\rightarrow X$ or $\iota_S:S\rightarrow X$ given by $i_S:s\mapsto s$ . This is an example of an inclusion map, and it is continuous.

Note that if one proves $i_S$ 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 $i_S$ comes from it as a corollary

Terminology

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

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

Characteristic property

Diagram
$\xymatrix{ Y \ar[r]^f \ar[dr]_{i_S\circ f} & S \ar@{^{(}->}[d]^{i_S}\\ & X}$

Let

$(X,\mathcal{ J })$ be a topological space and let

$(S,\mathcal{J}_S)$ be any subspace of

$(X,\mathcal{ J })$ ^{[Note 2]}. The characteristic property of the subspace topology^[1] is that:

Given any topological space (Y,K) and any map f:Y→S we have:
- $(f:Y\rightarrow S$ is continuous $)\iff(i_S\circ f:Y\rightarrow X$ is continuous $)$

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

Proof of claims

Claim 1: $\mathcal{J}_S$ is a topology

Grade: C

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Claim 2: Equivalent formulation of the relatively open sets

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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

Notes

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

References

↑ ^{Jump up to: 1.0} ^1.1 ^1.2 ^1.3 ^1.4 ^1.5 ^1.6 Introduction to Topological Manifolds - John M. Lee

OLD PAGE

Definition

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

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

We may say any one of:

Let $Y$ be a subspace of $X$
Let $Y$ be a subspace of $(X,\mathcal{J})$

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

Proof of claims

[Expand]
Claim 1: The subspace topology is indeed a topology

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

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

$\emptyset\in\mathcal{K}$ and $Y\in\mathcal{K}$
For any $U,V\in\mathcal{K}$ we must have $U\cap V\in\mathcal{K}$
For an arbitrary family {Uα}α∈I of open sets (that is ∀α∈I[Uα∈K]) we have:
- $\bigcup_{\alpha\in I}A_\alpha\in\mathcal{K}$

Proof:

First we must show that ∅,Y∈K
Recall that ∅,X∈J and notice that:
- $\emptyset\cap Y=\emptyset$ , so by the definition of $\mathcal{K}$ we have $\emptyset\in\mathcal{K}$
- $X\cap Y=Y$ , so by the definition of $\mathcal{K}$ we have $Y\in\mathcal{K}$
Next we must show...

TODO: Easy work just takes time to write!

Terminology

A closed subspace (of $X$ ) is a subset of $X$ which is closed in $X$ and is imbued with the subspace topology
A open subspace (of $X$ ) is a subset of $X$ which is open in $X$ and is imbued with the subspace topology

TODO: Find reference

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

Immediate theorems

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

This may be easier to read symbolically:

if $U\in\mathcal{K}$ and $Y\in\mathcal{J}$ then $U\in\mathcal{J}$

Proof:

Since

$U$ is open in

$Y$ we know that:

$U=Y\cap V$ for some $V$ open in $X$ (for some $V\in\mathcal{J}$ )

Since

$Y$ and

$V$ are both open in

$X$ we know:

$Y\cap V$ is open in $X$

it follows that

$U$ is open in

$X$

References

↑ ^{Jump up to: 1.0} ^1.1 ^1.2 Topology - Second Edition - Munkres

[1] Jump up ↑ Recall $\mathcal{P}(X)$ denotes the power set of $X$ and $S\in\mathcal{P}(X)\iff S\subseteq X$ , so it's another way of saying let $S$ be a subset of $X$ , possibly empty, possibly equal to $X$ itself

[3] Jump up ↑ This means $S\in\mathcal{P}(X)$ , or $S\subseteq X$ of course

[4] Jump up ↑ This leads to two ways to prove the statement:
If we show $i_S:S\rightarrow X$ is continuous, then we can use the composition of continuous maps is continuous to show if $f$ continuous then so is $i_S\circ f$

We can show the property the "long way" and then show $i_S:S\rightarrow X$ is continuous as a corollary

[4] If we show $i_S:S\rightarrow X$ is continuous, then we can use the composition of continuous maps is continuous to show if $f$ continuous then so is $i_S\circ f$

[5] We can show the property the "long way" and then show $i_S:S\rightarrow X$ is continuous as a corollary

[ITTMJML-2] {Jump up to: 1.0} ^1.1 ^1.2 ^1.3 ^1.4 ^1.5 ^1.6 Introduction to Topological Manifolds - John M. Lee

[Topology-5] {Jump up to: 1.0} ^1.1 ^1.2 Topology - Second Edition - Munkres

[Note 1]

[1]

[Note 2]

[Note 3]

[1]

@@ Line 1: / Line 1: @@
+{{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==
 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>

Difference between revisions of "Subspace topology"

Latest revision as of 23:04, 25 September 2016

Contents

Definition

Terminology

Characteristic property

Proof of claims

Claim 1: $\mathcal{J}_S$ is a topology

Claim 2: Equivalent formulation of the relatively open sets

See next

See also

Notes

References

OLD PAGE

Definition

Proof of claims

Terminology

Immediate theorems

References

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Difference between revisions of "Subspace topology"

Latest revision as of 23:04, 25 September 2016

Contents

Definition

Terminology

Characteristic property

Proof of claims

Claim 1: JS\mathcal{J}_S is a topology

Claim 2: Equivalent formulation of the relatively open sets

See next

See also

Notes

References

OLD PAGE

Definition

Proof of claims

Terminology

Immediate theorems

References

Navigation menu

Search

Claim 1: $\mathcal{J}_S$ is a topology