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Let $ (X, \tau) $ be a Hausdorff topological space and $ K \subset X $ compact. If $ A, B \subset K $ closed non-empty disjoint such that no component of $ K $ intersects $ A $ and $ B, $ show that there exist $ K_A, K_B \subset K $ mutually separated such that $ K = K_A \cup K_B $, $ A \subset K_A $ and $ B \subset K_B $

Proof: Let $ A, B \subset K $ closed non-empty disjoint of a Hausdorff space then they are compact and disjoint of a Hausdorff space, therefore there exist $ K_A $ and $ K_B $ open and disjoint such that $ A \subset K_A $ y $ B \subset K_B $ and let $ x \in K $ such that $ C_{K} (x) $ the component connected with $ C_{K} (x) \cap A = \emptyset = C_{K} (x) \cap B $ then $ C_{K} (x) \cap K_A = \emptyset = C_{K} (x) \cap K_B $, so $K_A, \ K_B$ they are mutually separate ...

How could the demonstration continue?

Thanks!

Us12
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1 Answers1

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SKETCH: Work in the compact Hausdorff space $K$. For each $x\in K$ let $C(x)$ be the component of $K$ containing $x$. $C(x)$ is also the quasicomponent of $K$ containing $x$, so it is the intersection of the clopen nbhds of $x$ in $K$.

  • Use that to show that each $x\in A$ has a clopen nbhd $H(x)$ disjoint from $B$..

$A$ is compact, and $A\subseteq\bigcup_{x\in A}H(x)$, so there is a finite $F\subseteq A$ such that $A\subseteq\bigcup_{x\in F}H(x)$; let $K_A=\bigcup_{x\in F}H(x)$ and $K_B=K\setminus K_A$, and verify that these sets have the desired properties.

Brian M. Scott
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  • let $ C_ {K} (x) $ be the connected component of $ x $ in $ K $, since $ K $ is compact and Hausdorff by Theorem (\ref{quasi}) we have that $ QC_ {K} (x ) = C_K (X) $ – Us12 Nov 05 '20 at 02:48
  • , that is, so since $ A $ is compact, since it is closed of a Hausdorff space, for each $ x \in A $, we have that $ H_i (x) $ is an open closed, such that $ x \in H_i (x) $ for all $ i \in I $ with $ H_i (x) \cap B = \emptyset $ therefore $ {H_i (x) }_{i \in I } $ is an open coverage of $ A $, thus – Us12 Nov 05 '20 at 02:49
  • $ A \subset \cup {i \in I } H_i (x) $ and therefore, there is a finite coverage $ {1, \cdots, n } \subset I $ such that $ A \subset \cup{i = 1}^{n} H_i (x) $, we will take $ K_A = \cup_{i = 1}^{n} H_i (x) $ and let $ K_B = K \setminus \cup_{i = 1}^{n} H_i (x) $ such that $ K_A \cap \overline {K_B}= \cup_ {i = 1}^{n} H_i (x) \cap \overline {K \setminus \cup_ {i = 1}^{n} H_i (x)}=\cup_ {i = 1}^{n} H_i (x) \cap K \setminus (\cup_{i = 1}^{n} H_i (x))^{\circ}=\cup_ {i = 1}^{n} H_i (x) \cap K \setminus \cup_ {i = 1}^{n} H_i (x) = \emptyset$, and so too – Us12 Nov 05 '20 at 02:49
  • $ \overline {K_A} \cap K_B =\emptyset$ and $ K = K_A \cup K_B $, $A \subset K_A$ y $B \subset K_B$ mutually separated – Us12 Nov 05 '20 at 02:50