Subspace topology

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Definition

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

• $\mathcal{J}_S:=\{U\cap S\ \vert\ U\in\mathcal{J}\}$ - the open sets of $(S,\mathcal{J}_S)$ are precisely the intersection of open sets of $(X,\mathcal{ J })$ with $S$
  • Claim 1: this is indeed a topology

Alternatively:

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

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 $U\in\mathcal{P}(S)$ 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 - indicating we mean open in the subspace, or
  2. $U$ is relatively closed - indicating we mean closed in the subspace

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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

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Characteristic property

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

Let $(X,\mathcal{ J })$ be a topological space and let $(S,\mathcal{J}_S)$ be any subspace of $(X,\mathcal{ J })$. The characteristic property of the subspace topology is that:

• Given any topological space $(Y,\mathcal{ K })$ and any map $f:Y\rightarrow S$ 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)

Proof of claims

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

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Claim 2: Equivalent formulation of the relatively open sets

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See next

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TODO: Theorems and propositions involving subspaces

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See also

• Topological embedding

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TODO: Link to more things

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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:

• $(Y,\mathcal{K})$ 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:

1. Let $Y$ be a subspace of $X$
2. 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

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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:

1. $\emptyset\in\mathcal{K}$ and $Y\in\mathcal{K}$
2. For any $U,V\in\mathcal{K}$ we must have $U\cap V\in\mathcal{K}$
3. For an arbitrary family $\{U_\alpha\}_{\alpha\in I}$ of open sets (that is $\forall\alpha\in I[U_\alpha\in\mathcal{K}]$) we have:
   • $$\bigcup_{\alpha\in I}A_\alpha\in\mathcal{K}$$

Proof:

1. First we must show that $\emptyset,Y\in\mathcal{K}$

   Recall that $\emptyset,X\in\mathcal{J}$ 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 $Y\in\mathcal{K}$

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 $X$) 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

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TODO: Find reference

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• A set $U\subseteq X$ 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 $U\subseteq Y$
• A set $U\subseteq X$ 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 $X$

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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 $X$