I'm trying to prove this lemma from exercise of "General Toplology" book(James Munkres):

Let X be a compact Hausdorff space. Then $x, y$ belong to the same quasicomponent of $X$ iff they belong to the same component of $X$.

The first step in the exercise is:

Let $\mathscr{A}$ be the collection of all closed subspaces $A$ of $X$ such that $x$ and $y$ lie in the same quasicomponent of $A$. Let $\mathscr{B}$ be a subcollection of $\mathscr{A}$ that is simply ordered by proper inclusion. Show that the intersection of the elements of $\mathscr{B}$ belongs to $\mathscr{A}$.

At this chapter, actually, I have no knowledge about what quasicomponent is. I just have definition from exercise that:

$x, y$ belong to the same quasicomponent of $A$ if there is no separation $A = C \cup D$ of $A$ into two disjoint sets open in $A$ such that $x \in C$ and $y \in D$.

So I will really appreciate if anyone can help me solve this problem using only this definition.

My trying so far is that: Let $M = \bigcap_{B \in \mathscr{B}}B$. Sure we have that $x, y \in M$. Suppose $M = C \cup D$, here $C, D$ open in $M$. Because $M$ is closed and $C, D$ are both open and closed in $M$, we must have $C, D$ is closed in $X$. Because $X$ is compact Hausdorff, there exist open disjoint subsets $U, V$ of $X$ such that $U$ contains $C$, $V$ contains $D$.

Up to here, my goal is to prove that for all $B \in \mathscr{B}$, we must have that $B - (C \cup D)$ is not empty. But I can't prove it. Please give me some hint to move on. I really appreciate.


You have done almost all the work. Using your notation etc. Since $C\cup D=M=\cap \{B\,|\, B\in \mathscr{B}\}$ it follows that $\cap \{B - C\cup D\,|\, B\in \mathscr{B}\}=M - (C\cup D) = \emptyset$ and hence $\cap \{B - U\cup V\,|\, B \in \mathscr{B}\}=\emptyset$, but since $\mathscr{B}$ is simply ordered and each $B-(U\cup V)$ is closed it follows by compactness see e.g. A question about intersection of closed set is non empty that there exists $B$ in $\mathscr{B}$ such that $B \subseteq U\cup V$. Therefore since $U'=B\cap U$ and $V'=B\cap V$ are disjoint open subsets of $B$ such that $B = U'\cup V'$ it follows that $x$ and $y$ are both in exactly one of $U'$ or $V'$ and consequently both in exactly one of $C$ or $D$.

  • $\begingroup$ absolutely right. Thanks so much for your clarification. I forgot to use the finite intersection property of $X$ ^^ $\endgroup$ – le duc quang Nov 28 '15 at 13:25

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