Quotient Space of Compact Space is Compact
Theorem
Let $T = \struct {X, \tau}$ be a compact topological space.
Let $\RR \subseteq X \times X$ be an equivalence relation on $X$.
Then, the quotient space:
- $T / \RR$
is compact.
Proof
Let $\UU$ be an open cover of $T / \RR$.
By definition of quotient topology, for every $U \in \UU$:
- $q_\RR^{-1} \sqbrk U \in \tau$
where $q_\RR$ is the quotient mapping induced by $\RR$.
Therefore:
- $\VV = \set {q_\RR^{-1} \sqbrk U : U \in \UU}$
Let $x \in X$ be arbitrary.
By definition of cover, there is some $U \in \UU$ such that:
- $\eqclass x \RR \in U$
Therefore:
- $x \in q_\RR^{-1} \sqbrk U$
As $x \in X$ was arbitrary, it follows that $\VV$ is a cover of $X$.
As $\VV$ is a cover consisting of open sets, it is by definition an open cover of $T$.
Thus, by definition of compact space, there is some finite subcover $\set {V_i}_{1 \le i \le n} \subseteq \VV$.
But by definition of $\VV$, for every $1 \le i \le n$, there is some open set $U_i \in \UU$ such that:
- $V_i = q_\RR^{-1} \sqbrk U_i$
It remains to show that:
- $\set {U_i}_{1 \le i \le n}$
is a cover of $X / \RR$, the quotient set of $X$ induced by $\RR$.
Let $y \in X / \RR$ be arbitrary.
By definition of quotient set, there is some $x \in X$ for which:
- $y = \eqclass x \RR$
By definition of cover there is some $1 \le i \le n$ such that:
- $x \in V_i$
By definition of preimage:
- $\eqclass x \RR = \map {q_\RR} x \in U_i$
As $y \in X / \RR$ was arbitrary, it follows that:
- $\set {U_i}_{1 \le i \le n} \subseteq \UU$
is a finite subcover of $\UU$.
As $\UU$ was arbitrary, $T / \RR$ is compact by definition.
$\blacksquare$