Topological Subspaces Examples 2

Topological Subspaces Examples 2

Recall from the Topological Subspaces page that if $(X, \tau)$ is a topological space and $A \subseteq X$ then the subspace topology on $A$ is defined to be:

\begin{align} \quad \tau_A = \{ A \cap U : U \in \tau \} \end{align}

We verified that $\tau_A$ is indeed a topology for any subset $A$ of $X$.

We will now look at some examples of subspace topologies.

Example 1

Let $X$ be a set with the discrete topology $\tau$, and let $A \subseteq X$. Prove that the subspace $(A, \tau_A)$ also has the discrete topology (on $A$).

Let $A \subseteq X$. To show that $A$ has the discrete topology on $A$, all we need to show is that any subset of $A$ is open in $X$.

Let $U \subseteq A$ be open in $X$. Then $U \subseteq X$. So $U$ is open in $X$. Moreover, $A \cap U = U$ is open in $A$. Hence $\tau_A$ is the discrete topology on $A$.

Example 2

Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Prove that $\tau_A \subseteq \tau$ if and only if $A \in \tau$.

$\Rightarrow$ Suppose that $\tau_A \subseteq \tau$. Since $A$ is open in $A$, this means that $A \in \tau_A$, so $A \in \tau$.

$\Leftarrow$ Suppose that $A \in \tau$. Let $U \subseteq \tau_A$. Then $U$ is open in $A$. So there exists an open set $V \in X$ such that $U = A \cap V$. But $A$ is open in $X$ since $A \in \tau$, so $A \cap V = U$ is open in $X$. Therefore $U \in \tau$. This shows that:

\begin{align} \quad \tau_A \subseteq \tau \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License