Quasicomponent is Intersection of Clopen Sets

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $p \in S$.


Then the quasicomponent containing $p$ equals the intersection of all sets which are both open and closed which contain $p$.


Proof

Let $C$ be the quasicomponent of $p$.

Let $Q$ be the set of clopen sets containing $p$ that are not equal to $S$.


First let $x \in C$.

Let $U \in Q$.

By definition of $Q$:

$p \in U$

Let $V = \complement_S \left({U}\right)$ be the complement of $U$ relative to $S$.

By Complement of Clopen Space is Clopen $V$ is also a clopen set of $T$.

By Clopen Set and Complement form Separation $U$ and $V$ separate $S$.

Thus $U \mid V$ is a separation of $S$ which contains a single open set $U$ containing

By the definition of quasicomponent, $U \mid V$ is a separation of $S$ which contains a single open set $U$ containing both $x$ and $p$.

Thus:

$x \in U$

As this holds for all such $U$:

$x \in \displaystyle \bigcap Q$

As this holds for all such $x$:

$C \subseteq Q$

$\Box$


Let $x \in \displaystyle \bigcap Q$.

Let $U \mid V$ be a separation of $S$.

Let $p \in U$.

Then $U \in Q$.

Thus $x \in U$.

Thus $x$ lies on the same side as $p$ of any separation of $X$.

So $x \in C$.

$\blacksquare$


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