Subspace topology

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

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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

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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 $X$