Subspace topology
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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]
 Claim 1: this is indeed a topology^{[1]}
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
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This proof has been marked as an page requiring an easy proof
Claim 2: Equivalent formulation of the relatively open sets
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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
 ↑ ^{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 [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]
References
 ↑ ^{1.0} ^{1.1} ^{1.2} Topology  Second Edition  Munkres