4

My notes from the lecture says "Let $X$ be a topological space and let $R$ be an equivalence relation. Then, $A \subseteq X$ is called saturated with respect to $R$ if it is a union of equivalence classes." This statement doesn't make sense to me. I have learnt about quotient spaces $X/R$ which is a set of all equivalence classes. So, since $A$ is a subset of $X$, how can it be a union of equivalence classes? I would understand the statement if it said $A$ is a subset of $X/R$?

Thanks.

Focus
  • 1,204
user120227
  • 995

1 Answers1

5

Since equivalence classes are subsets of $X$, a union of equivalence classes is a subset of $X$. A subset of $X/R$ would be a set of equivalence classes.

Every saturated set $A$ is of the form $q^{-1}(q(A))$ where $q$ is the quotient map $X\to X/R, x\mapsto [x]$. More generally, if $q:X\to Y$ is any function, one may call a set $A\subseteq X$ q-saturated when it is of the form $q^{-1}(B)$ for some $B\subseteq Y$. But this is equivalent to saturated sets for equivalence relation since the fibers $q^{-1}(y)$ can be thought of as equivalence classes. A preimage is then saturated with respect to that relation.

Stefan Hamcke
  • 27,733