TITLE: About compactness in topological quotient space
QUESTION [0 upvotes]: Let $(X,T)$ compact space with the equivalence relation such that:
$C=\{(x,y)\in X \times X: x \  \mbox{is related with}\  y\}$ is closed in $X\times X$. First I don't know if I understand this definition well, two elements are related iff $(x,y)$ is closed?.
Now the questions are:
Show that if $A$ is a closed subspace of $X$ then the set $ D=\{ x:\ \exists a \in A:\ (x,a)\in C\}$ is closed. I think that this set is closed because if x is related with $a$, then the set of points of the form $(x,a)$ is closed, then is this set is closed for properties of closure and A is closed then the set of points $x$ is a union of every $x$ has to be closed.
Second one: Show that the quotient space $(X/ \sim, T(\sim))$ is compact. I need a hint because I don't know how to start to proceed this is compact space.

REPLY [2 votes]: For the first part, consider the set $(X\times A)\cap C=\{(x,a)\in X\times A:(x,a)\in C\}$ which is closed in $X\times X$ being the intersection of two closed sets. 
Now, $D$ is clearly the image of this set under the projection map $\pi_1:X\times X\to X$.
By this result, since $X$ is compact, the projection map is closed. Thus $D$ being the image of a closed set is closed.
For the second part, consider the map $X\to X/\sim$ given by sending any $x\in X$ to its corresponding equivalence class $[x]\in X/\sim$. This map is clearly surjective and is continuous as $X/\sim$ has the quotient topology. Since the image of a compact set under a continuous map is compact we have that $X/\sim$ is compact.
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