239: Disjoint Union of Closed Sets Is Closed in Disjoint Union Topology

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A description/proof of that disjoint union of closed sets is closed in disjoint union topology

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of disjoint union topology.
• The reader knows a definition of closed set.
• The reader admits the proposition that the disjoint union of any complements is the disjoint union of the whole sets minus the disjoint union of the subsets.

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Target Context

• The reader will have a description and a proof of the proposition that the disjoint union of any closed sets (at most 1 from each constituent topological space) is closed in the disjoint union topology.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any disjoint union topological space, $T:=\coprod _{\alpha }{T}_{\alpha }$, and any closed sets, $\{C_{\beta }\subseteq T_{\beta }\}$ where $\{\beta \}\subseteq \{\alpha \}$, $\coprod _{\beta }{C}_{\beta }$ is closed on $T$.

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2: Proof

$\coprod _{\beta }{C}_{\beta }=\coprod _{\beta }{T}_{\beta }\setminus U_{\beta }$ where $U_{\beta }\subseteq T_{\beta }$ is open on $T_{\beta }$. $\coprod _{\beta }(T_{\beta }\setminus U_{\beta })=\coprod _{\beta }{T}_{\beta }\setminus \coprod _{\beta }{U}_{\beta }$, by the proposition that the disjoint union of any complements is the disjoint union of the whole sets minus the disjoint union of the subsets. $\coprod _{\beta }{C}_{\beta }=\coprod _{\alpha }{T}_{\alpha }\setminus \coprod _{\gamma \in \{\alpha \}\setminus \{\beta \}}{T}_{\gamma }\setminus \coprod _{\beta }{U}_{\beta }=\coprod _{\alpha }{T}_{\alpha }\setminus (\coprod _{\gamma \in \{\alpha \}\setminus \{\beta \}}{T}_{\gamma }\cup \coprod _{\beta }{U}_{\beta })$. $\coprod _{\gamma \in \{\alpha \}\setminus \{\beta \}}{T}_{\gamma }\cup \coprod _{\beta }{U}_{\beta }$ is open.

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References

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