Union of Connected Sets with Non-Empty Intersections is Connected/Corollary
Corollary to Union of Connected Sets with Non-Empty Intersections is Connected
Let $T = \struct {S, \tau}$ be a topological space.
Let $I$ be an indexing set.
Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$.
Let $B$ be a connected set of $T$ such that:
- $\forall C \in \AA: B \cap C \ne \O$
Then $\ds B \cup \bigcup \AA$ is connected.
Proof 1
Let $C \in \AA$.
From Union of Connected Sets with Non-Empty Intersections is Connected applied to $B$ and $C$, the union $B \cup C$ is connected.
Thus the set $\tilde \AA = \set {B \cup C: C \in \AA}$ satisfies the conditions of the theorem.
Hence the result.
$\blacksquare$
Proof 2
Let $\ds H = B \cup \bigcup \AA$
Aiming for a contradiction, suppose that $H$ is not connected.
That is, that $H$ is disconnected.
We have that:
- $\forall C \in \AA: B \cap C \ne \O$
Thus:
- $\exists x \in H: x \in B \cap C$
From the definition of disconnected, there exist separated sets $U, V$ whose union is $H$.
So:
- $\forall s \in S: s \in U \lor s \in V$
Aiming for a contradiction, suppose $s \in V$.
Then $H, V$ serve as separated sets whose union is a cover for $\family {A_\alpha}_{\alpha \mathop \in I}$.
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Note that the disconnection is valid as:
- $\exists x \in U \cap \family {A_\alpha}_{\alpha \mathop \in I}$
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As $s \in V$, it follows that the $\family {A_\alpha}_{\alpha \mathop \in I}$ are disconnected.
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From this contradiction it follows that $s \notin V$.
Hence $s \in U$.
This holds for arbitrary $s \in H$.
Hence:
- $H \subseteq U$
But as $U$ and $V$ are separated, $U \cap H$ is empty, as required per the definition of a disconnected set.
This contradicts our deduction that $H \subseteq U$.
Hence the sets $U, V$ are not separated sets in $H$.
Thus $H$ is connected in $T$.
$\blacksquare$