# Quasicomponents and components in compact Hausdorff space

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 $B$.

• Hola Julio. Una pregunta, donde estudias?
– user113609
Commented Dec 10, 2013 at 4:26
• En la Universidad Nacional de Ingeniería en Perú. Commented Dec 10, 2013 at 4:29
– user113609
Commented Dec 10, 2013 at 4:30
• Pregrado, el problema es del J. Munkres-Topología (2ed) Commented Dec 10, 2013 at 4:32
• Commented Dec 10, 2013 at 4:41

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.

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. Commented Dec 20, 2013 at 3:51

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 $X$.