G-Delta Sets Closed under Union

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $G, G'$ be $G_\delta$ sets of $T$.


Then their union $G \cup G'$ is also a $G_\delta$ set of $T$.


Proof

By definition of $G_\delta$ set, there exist sequences $\left({U_n}\right)_{n \in \N}$ and $\left({U'_n}\right)_{n \in \N}$ of open sets of $T$ such that:

$G = \displaystyle \bigcap_{n \mathop \in \N} U_n$
$G' = \displaystyle \bigcap_{n \mathop \in \N} U'_n$

Now compute:

\(\displaystyle G \cup G'\) \(=\) \(\displaystyle \bigcap_{n \mathop \in \N} \left({U_n \cup G'}\right)\) Intersection Distributes over Union: General Result
\(\displaystyle \) \(=\) \(\displaystyle \bigcap_{n \mathop \in \N} \bigcap_{m \mathop \in \N} \left({U_n \cup U'_m}\right)\) Intersection Distributes over Union: General Result

By axiom $(1)$ of a topology, $U_n \cup U'_m$ is open, for all $n, m \in \N$.

By Cartesian Product of Countable Sets is Countable, $\N \times \N$ is countable.


Thus $G \cup G'$ is seen to be a $G_\delta$ set.

$\blacksquare$


Also see