Open and Closed Sets in Topological Quotients

Recall from the Topological Quotients page that if $X$ is a topological space, $\sim$ is an equivalence relation on $X$, $X \: / \sim$ denotes the set of equivalence classes on $X$, and $q : X \to X \: / \sim$ is the canonical quotient map defined for all $x \in X$ by $q(x) = [x]$ (each element $x \in X$ is sent to its equivalence class $[x]$) then the quotient topology on $X \: / \sim$ is define to the final topology induced by $q$ and hence:

(1)

\begin{align} \quad \tau = \{ U \subseteq X \: / \sim : q^{-1}(U) \: \mathrm{is \: open \: in \:} X \} \end{align}

The following theorems will tell us when a set $U \subseteq X \: / \sim$ is open/closed in $X \: / \sim$.

Theorem 1: Let $(X, \tau)$ be a topological space and $\sim$ be an equivalence relation on $X$. Furthermore, let $X \: / \sim$ have the quotient topology with respect to $\sim$. Then $U \subseteq X \: / \sim$ is open in $X \: / \sim$ if and only if $\displaystyle{\bigcup_{[x] \in U} [x]}$ is open in $X$.

• Proof: By definition, the quotient topology $\tau^*$ on $X \: / \sim$ is the final topology on the quotient map $q : X \to X \: / \sim$ and can be described as the following set:

(2)

\begin{align} \quad \tau^* = \{ U \subseteq X \: / \sim : q^{-1}(U) \in \tau \} \end{align}

• $\Rightarrow$ Suppose that $U \subseteq X \: / \sim$ is open in $X \: / \sim$. Then $q^{-1}(U)$ is open in $X$. But:

(3)

\begin{align} \quad q^{-1}(U) = \bigcup_{[x] \in U} [x] \end{align}

• So $\bigcup_{[x] \in U} [x]$ is open in $X$.

• $\Leftarrow$ Let $U \subseteq X \: / \sim$ and suppose that $\bigcup_{[x] \in U} [x]$ is open in $X$. Then $q \left ( \bigcup_{[x] \in U} [x] \right ) = U$. So $q^{-1}(U)$ is open and since the quotient topology is the finest topology making $q : X \to X \: / \sim$ continuous we have that $U$ is open in $X \: / \sim$. $\blacksquare$

Theorem 2: Let $(X, \tau)$ be a topological space and $\sim$ be an equivalence relation on $X$. Furthermore, let $X \: / \sim$ have the quotient topology with respect to $\sim$. Then $C \subseteq X \: / \sim$ is closed in $X \: / \sim$ if and only if $\displaystyle{\bigcup_{[x] \in C} [x]}$ is closed in $X$.

• Proof: $\Rightarrow$ Suppose that $C$ is closed in $X \: / \sim$. Then $(X \: / \sim) \setminus C$ is open in $X \: / \sim$, so by Theorem 1 we have that:

(4)

\begin{align} \quad (X \: / \sim) \setminus C = \bigcup_{[x] \in (X \: / \sim) \setminus C} [x] \end{align}

• Taking the complement on both sides gives us:

(5)

\begin{align} \quad C = \bigcup_{[x] \in C} [x] \end{align}

• $\Leftarrow$ Suppose that $\displaystyle{\bigcup_{[x] \in C} [x]}$ is closed in $X$. Then $\displaystyle{X \setminus \bigcup_{[x] \in C} [x] = \bigcup_{[x] \in X \setminus C} [x]}$ is open in $X$. By Theorem 1 this means that $X \setminus C$ is open in $X \: / \sim$. So $C$ is closed in $X \: / \sim$. $\blacksquare$