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 $X$.

$T / \RR$

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:

$\VV = \set {q_\RR^{-1} \sqbrk U : U \in \UU}$

is a set of open sets of $T$.

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:

$\set {U_i}_{1 \le i \le n} \subseteq \UU$

As $\UU$ was arbitrary, $T / \RR$ is compact by definition.

$\blacksquare$