Subspace topology

Grade: A

This page is currently being refactored (along with many others)

Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.
The message provided is:

Needed urgently, ready to plough on with it now though!

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

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

---

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

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

Grade: C

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:

Really easy, hence low importance

This proof has been marked as an page requiring an easy proof

Claim 2: Equivalent formulation of the relatively open sets

Grade: C

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
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

---

See also

• Topological embedding

---

TODO: Link to more things

---

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

---

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...

---

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 $U\subseteq X$ 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 $X$

---

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 $X$