Subspace topology

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:

• [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

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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...

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TODO: Easy work just takes time to write!

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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

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TODO: Find reference

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• 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]
• 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]

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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]