Hi, I'm having trouble understanding this proof.(adsbygoogle = window.adsbygoogle || []).push({});

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 disjointopensubsets.

(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]?

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# Proof about connected sets

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