185: If Union of Disjoint Subsets Is Closed, Each Subset Is Not Necessarily Closed

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A description/proof of that if union of disjoint subsets is closed, each subset is not necessarily closed

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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 topological space.

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Target Context

• The reader will have a description and a proof of the proposition that if the union of some disjoint subsets is closed, each subset is not necessarily closed.

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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 topological space, $T$, and some disjoint subsets, $\{S_i|S_i\subseteq T\}$, such that the union, $S:={\cup }_i{S}_i$, is closed, each $S_i$ is not necessarily closed.

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2: Proof

A counterexample will suffice. Think of the $\mathbb{R}$ Euclidean topological space, and 2 subsets, $[-1,0],(0,1]$, which are disjoint and the union is $[-1,1]$, closed.

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References

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