Quasicomponents and components in compact Hausdorff space

Question

Let $X$ be a compact Hausdorff space, $x,y\in X$ and $\mathcal{A}$ a colection of closed subspaces of $X$ such that for every $A\in \mathcal{A}$ then $x$ and $y$ are in the same quasicomponent of $A$. If $\mathcal{B}$ is a subcolection of $\mathcal{A}$, then $x$ and $y$ are in the same quasicomponent of $$D=\bigcap_{B\in\mathcal{B}} B $$

($x$ and $y$ are in the same quasicomponent of $A$ if and only if there's no two disjoint open sets $U,V$ such that $A=U\cup V$, $x\in U$ and $y\in V$)

This is the first part of a question that wants me to show that if $X$ is a compact Hausdorff space then $x,y$ are in the same quasicomponent if and only if they are in the same component. The second part is to show that $\mathcal{A}$ has a minimal element which I have done, and the last part is to show that the minimal element is connected which I haven't done.

What I've tried so far is assuming theres a separation $U',V'$ for $x$ and $y$ in $D$. Then tried to extend that separation to any $B$ because $U'=B\cap U$ where $U$ is open in $B$. I fail to prove that $U$ is also closed in $B$.

interesante, esta clase de topologia es pregrado o postgrado en tu universidad? — , Dec 10 '13 at 04:30
mira aqui un solucionario de munkres si te interesa : http://dbfin.com/2011/02/2000-munkres-topology-solutions-chapter-2/ — , Dec 10 '13 at 04:33

score 11 · Accepted Answer · answered Dec 17 '13 at 16:17

to show that if $X$ is a compact Hausdorff space then $x,y$ are in the same quasicomponent if and only if they are in the same component.

This is from "General topology" by Riszard Engelking.

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It seems the following.

to show that the minimal element is connected

Let $A$ be a minimal element of the family $\mathcal{A}$. Suppose that the set $A$ is not connected. Then $A$ is a union of two its disjoint clopen non-empty subsets . Since $x$ and $y$ are in the same quasicomponent of $A$, the set $\{x,y\}$ is contained in one of these two sets. Denote this set of $B$. Since the set $B$ is a clopen subset of $A$, each quasicomponent of the set $B$ is a quasicomponent of the set $A$. Hence $B\in\mathcal A$. But $B$ is a proper subset of the set $A$, which contradicts to the minimality of the set $A$.

The proof from Engelking is also typed out in this answer. – Post No Bulls Dec 20 '13 at 03:51 — Post No Bulls, Dec 20 '13 at 03:51

score -1 · Answer 2 · edited Jan 09 '15 at 00:29

At first prove that family $A$ from closed subset of compact and hausdoorf topological space $X$ which for $x,y$ be given in $X$ are in a quasicomponet of this closed subsets then $x,y$ are in a quasicoponet of intersection of elements of $A$. secound if $A$ is chain with icloudtion then intersection of $A$ is conected at last is simple that see "componets of $X$ are quasicomponets of $X$.

Quasicomponents and components in compact Hausdorff space

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