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## Main Question or Discussion Point

Hi, I'm having trouble understanding this proof.

Then [tex]\cup_{i \in I} S_{i}[/tex] is connected.

expresses [tex]S_{i}[/tex] as a union of disjoint

(and the proof continues)

How can I show that [tex]A \cap S_{i}[/tex] (or [tex]B \cap S_{i}[/tex]) is indeed an open subset of [tex]S_{i}[/tex]?

**Theorem.**Let [tex]\{ S_{i} \} _{i \in I}[/tex] be a collection of connected subsets of a metric space [tex]E[/tex]. Suppose there exists [tex]i_{0} \in I[/tex] such that for each [tex]i \in I[/tex], [tex]S_{i} \cap S_{i_{0}} \neq \emptyset[/tex].Then [tex]\cup_{i \in I} S_{i}[/tex] is connected.

**Proof.**Suppose [tex]S = \cup_{i \in I} S_{i} = A \cup B[/tex], where [tex]A[/tex] and [tex]B[/tex] are disjoint open subsets of [tex]S[/tex]. For each [tex]i \in I[/tex],[tex]S_{i} = ( A \cap S_{i} ) \cup ( B \cap S_{i} )[/tex]

expresses [tex]S_{i}[/tex] as a union of disjoint

**subsets.**__open__(and the proof continues)

How can I show that [tex]A \cap S_{i}[/tex] (or [tex]B \cap S_{i}[/tex]) is indeed an open subset of [tex]S_{i}[/tex]?