Open and Closed Sets in Topological Subspaces

# Open and Closed Sets in Topological Subspaces

Recall from the Topological Subspaces page that if $(X, \tau)$ is a topological space and $A \subseteq X$ then we can define a topology on $\tau_A$ on $A$ called the subspace topology on $A$ (from $X$) explicitly defined as:

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

Together, $(A, \tau_A)$ form what is called a topological subspace.

Note that by definition, a set $V \subseteq A$ is open in $A$ with the subspace topology if and only if there exists an open set $U$ of $X$ such that:

(2)\begin{align} \quad V = A \cap U \end{align}

So, what can be said about the closed sets in $A$? The following theorem gives us exactly what we'd expect.

Theorem 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. A set $C \subseteq A$ is closed in $A$ if and only if there exists a closed set $D$ of $X$ such that $C = A \cap D$. |

**Proof:**$\Rightarrow$ Let $C \subseteq A$ be closed in $A$. Then $A \setminus C$ is open in $A$. So, there exists an open set $U$ in $X$ such that:

\begin{align} \quad A \setminus C = A \cap U \end{align}

- Taking the complement of both sides with respect to $A$ shows us that:

\begin{align} \quad A \setminus (A \setminus C) = A \setminus (A \cap U) \\ \quad C = A \cap U^c \end{align}

- So let $D = U^c$. Then $D$ is closed in $X$ and $C = A \cap D$.

- $\Leftarrow$ Suppose that there exists a closed set $D$ of $X$ such that $C = A \cap D$. Then:

\begin{align} \quad A \setminus C = A \setminus (A \cap D) \\ \quad A \setminus C = A \cap D^c \\ \end{align}

- $D^c$ is an open set in $X$, so $A \cap D^c$ is open in $A$. In other words, $A \setminus C$ is open in $A$, so $C$ is closed in $A$. $\blacksquare$