Show that $\partial_S(U) \subset (\partial_S(S') \cup \partial_{S'}(U))$ where $S'$ is subspace of the metric space $(S,d)$. $U$ is nonempty subset of $S'$.
To approach this problem, should I divide it into several cases (I think there are four cases)?
Or Should I start from
$\partial_S(U) = cl_S(U) \cap cl_S(S\backslash U)$?
I'm kinda stuck on both ways. For the second one, I have:
$\partial_S(U) = cl_S(U) \cap cl_S(S\backslash U)$
$\subset cl_S(S') \cap cl_S(S\backslash U)$.
Or is there a counterexample to the statement??