Please have a look at this page.

What does "relatively closed" mean? Let $S=\{x\in U \mid u(x)=M \}$ I have the same problem but I don't quite understand how the set S is both open and relatively closed in $U$.

  • $\begingroup$ you have two sets $U$ here. $\endgroup$ – Henno Brandsma Mar 15 '14 at 12:06

We have a subset $U$ here and a subset $C = \{x \in U: u(x) = M \}$. This set is by definition relatively closed iff it is closed in the subspace topology that $U$ inherits from the whole space (which is here $\mathbb{C}$ I believe, in the original context). The latter just means that there is a closed subset $C'$ of the whole space such that $C = C' \cap U$.

To apply this to your linked case, we take $C' = u^{-1}[\{M\}] = \{x \in \mathbb{C}: u(x) = M \}$, which is closed because $u$ is continuous (even differentiable) and $\{M\}$ is closed. And clearly we have $C = U \cap C'$ by the definitions. Alternatively, if $u$ is only defined on $U$, then your set $C$ is just the inverse image of $\{M\}$ and thus by definition of continuity closed in the domain $U$, and so relatively closed.

One only says relatively closed to avoid confusion with closed sets of the whole space (here $\mathbb{C}$); the set $C$ need not be closed in $\mathbb{C}$, generally.

  • $\begingroup$ finally got, why it is mentioned at all! thanks $\endgroup$ – cesare borgia Jun 28 '17 at 14:46

"Relatively closed" means closed in the subspace topology. If you have a subset $V=\{x\in U | u(x)=M \}$ then "relatively closed" means that $V$ is closed in $U$, that is $V=A\cap U$ for some closed (in the given space!) subset $A$.


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