# 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 Dec 10 '13 at 4:26
• En la Universidad Nacional de Ingeniería en Perú. – Julio Cáceres Dec 10 '13 at 4:29
• interesante, esta clase de topologia es pregrado o postgrado en tu universidad? – user113609 Dec 10 '13 at 4:30
• Pregrado, el problema es del J. Munkres-Topología (2ed) – Julio Cáceres Dec 10 '13 at 4:32
• – Julio Cáceres Dec 10 '13 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.
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 $A$.
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 $X$.