Revision as of 22:28, 25 September 2016

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 "the subspace topology on $S$"^[1] (AKA: relative topology on $S$^[1]) 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^[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]

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^[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

• 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)^{[Note 3]}

Proof of claims

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

Claim 2: Equivalent formulation of the relatively open sets

See next

TODO: Theorems and propositions involving subspaces

See also

• Topological embedding

TODO: Link to more things

Notes

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

1. ↑ ^{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:

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

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 $U\subseteq X$ 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 $U\subseteq Y$^[1]
• 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

References

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