G-Delta Sets Closed under Union
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Theorem
Let $T = \struct {S, \tau}$ 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 $\sequence {U_n}_{n \mathop \in \N}$ and $\sequence {U'_n}_{n \mathop \in \N}$ of open sets of $T$ such that:
- $G = \ds \bigcap_{n \mathop \in \N} U_n$
- $G' = \ds \bigcap_{n \mathop \in \N} U'_n$
Now compute:
| \(\ds G \cup G'\) | \(=\) | \(\ds \bigcap_{n \mathop \in \N} \paren {U_n \cup G'}\) | Intersection Distributes over Union: General Result | |||||||||||
| \(\ds \) | \(=\) | \(\ds \bigcap_{n \mathop \in \N} \bigcap_{m \mathop \in \N} \paren {U_n \cup U'_m}\) | 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.
$\blacksquare$