i have this theorem with it's proof but i don't understand the last part 
They use this proposition:
My question is Why $\varphi^c\cap U_i$ is closed and pairwise disjoint ? where $\varphi^c=\lbrace x, \varphi(x)\leq c\rbrace$
Please thank you.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ There's a lot of notation in there. What are the $(C)_{c'}$ condition, $K^c_\phi,H_k, M_k,\beta_k$? $\endgroup$– Kevin CarlsonAug 17, 2014 at 19:33
-
$\begingroup$ All this have no relation, i juste want to know why $U_i$ is open and $\varphi^c$ is closed implies that $\varphi^c\cap U_i$ is closed ? $\endgroup$– VrouvrouAug 17, 2014 at 19:47
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
18
The theorem is applied (see the middle in $6.6$ ) to $X=\varphi^c\cap C$ and $X_i=\varphi^c\cap U_i$ which is a closed subset in $X$.($X_i=X\cap(\cup_{j\neq i}U_j)^c$ where the c's power is the complementary ).
-
$\begingroup$ I don't understand why $\varphi^c\cap U_i =X\cap(\cup_{i\neq j} U_j)^c$ ? thank you $\endgroup$– VrouvrouAug 17, 2014 at 21:33
-
$\begingroup$ If $x\in \varphi^c\cap U_i$, then $x\in X$ and $x\notin U_j$ for $j\neq i$ (because $(U_i)_i$ are pairwise disjoit ).If $x\in X\cap(\cup_{i\neq j}U_j)$ then $x\in X$ and not in any $U_j$ for $j\neq i$, so $x\in U_i$ and that $x\in \varphi^c\cap U_i$. $\endgroup$– HamouAug 17, 2014 at 21:39
-
$\begingroup$ I don't understand if $x\in X\cap(\cap_{i\neq j} U_j)^c$ then $x\in X$ but why $x\in \varphi^c$ ? $\endgroup$– VrouvrouAug 17, 2014 at 21:44
-
$\begingroup$ $X=\varphi^c\cap C\subset \varphi^c$. and $x\in X\cap(\cup_{j\neq i}U_j)$ there is union not the intersection. $\endgroup$– HamouAug 17, 2014 at 21:50
-
$\begingroup$ $\cup_{i\neq j} U_j)^c $ is closed ok, but $\varphi^c$ why is closed ? $\endgroup$– VrouvrouAug 17, 2014 at 21:58