2
$\begingroup$

Let $\pi : X \to Y$ be any map, and $U$ be a subset of $X$.

The problem is:

"$\forall x\in U$, $ \pi (x) = \pi(x') $, then $x' \in U$" then $U$ is saturated.

(U is saturated $\iff$ $\exists V \subset Y $ s.t. $U = \pi ^{-1}(V)$ )

My understanding is:

To show $U$ is saturated, it is enough to show $U = \pi ^{-1}(\pi(U))$.

$U \subset \pi ^{-1}(\pi(U))$ is satisfied for any map $\pi$.

We will show $U \supset \pi ^{-1}(\pi(U))$.

Suppose $ x\in \pi ^{-1}(\pi(U))$.

$\iff \pi(x) \in \pi(U)$

By definition of $\pi(U)$, there is some $a \in U$ s.t. $\pi(x) = \pi(a)$,

so hypothesis says $x\in U$.

My proof is fine? Thanks.

$\endgroup$
2
  • $\begingroup$ Yes, your proof is fine. $\endgroup$ Jun 21, 2015 at 20:01
  • $\begingroup$ My pleasure. $\,$ $\endgroup$ Jun 22, 2015 at 21:29

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.