Subspace topology

Definition

1. Subspace topology

   Let $(X,\tau)$ be a topological space and $Y\subseteq X$. Then

   $$\tau_Y=\{Y\cap U\mid U\in \tau\}$$

   is a topology on $Y$, called subspace topology.

2. Proof
   • Since $\emptyset=Y\cap\emptyset$, and $Y=Y\cap X$, we have (T1).
   • Let $V_{\alpha}\in\tau_{Y}$ for $\alpha\in I$. For each $\alpha\in I$, there is $U_{\alpha}\in\tau$ such that $V_{\alpha}=Y\cap U_{\alpha}$.

   • Then:

     $$Y\cap (\cup U_{\alpha})=\cup (Y\cap U_{\alpha})=\cup V_{\alpha},$$

     and we obtain (T2). The proof of (T3) is an exercise.

3. Open sets relative to $Y$

   If $Y$ is a subspace of $X$, and $U\subseteq Y$, it could be that $U$ is open in $Y$ but not open in $X$.

Base of the subspace topology

1. Lemma

   Let $\mathcal{B}$ be a base for the topology of $X$. Then

   $$\mathcal{B}_Y=\{Y\cap B\mid B\in \mathcal{B}\}$$

   is a base for the subspace topology on $Y$.

2. Proof

   Let $U$ be open in $X$ and $y\in Y\cap U$. Let $B\in \mathcal{B}$ such that $y\in B\subseteq U$. Then $Y\cap B\in \mathcal{B}_{Y}$, and $y\in Y\cap B\subseteq Y\cap U$. $\Box$

Example

1. Subspace of the real line
   • Consider $Y=[0,1]\subseteq X=\mathbb{R}$, where $X$ has the standard topology.
   • A base for the subspace topology on $Y$ is the collection of all sets of the form $(a,b)\cap Y$.
   • They all are of one of the following forms: $(a,b)$, $[0,b)$, $(a,1]$, $Y$, $\emptyset$.
   • We obtain that these elements generate the order topology on $[0,1]$.

Example

1. Subspace and order topology not always the same
   • Consider now $Y=[0,1)\cup\{2\}\subseteq X=\mathbb{R}$.
   • In the subspace topology, the set $\{2\}$ is open.
   • But in the order topology on $Y$, considering that $2$ is $\max Y$, any open set that contains $2$ has to contain other points.

Order topology in intervals

1. Order topology in intervals

   If $X$ is an ordered set that has the order topology, and $Y\subseteq X$ is an interval, then the order topology and the subspace topology on $Y$ are the same.

2. Proof

   Exercise

Exercises

1. Show that if we consider in $\mathbb{R}$ the usual topology, then the topology on $\mathbb{N}$ as a subspace of $\mathbb{R}$ is discrete.
2. Let $Y\subseteq X$ be open. Show that $A\subseteq Y$ is open in Y if and only if it is open in $X$.
3. Show that if $X$ is a topological space, and $A\subseteq Y\subseteq X$, then the topology of $A$ as a subspace of $Y$ is the same as a subspace of $X$ ($Y$ is a subspace of $X$).

Links

• Subspace topology - Wikipedia, the free encyclopedia