385: Union of Dichotomically Nondisjoint Set of Real Intervals Is Real Interval

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A description/proof of that union of dichotomically nondisjoint set of real intervals is real interval

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Topics

About: set

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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 dichotomically disjoint set of sets.
• The reader knows a definition of $\mathbb{R}$ interval.

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Target Context

• The reader will have a description and a proof of the proposition that the union of any possibly uncountable dichotomically nondisjoint set of $\mathbb{R}$ intervals is a $\mathbb{R}$ interval.

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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 dichotomically nondisjoint set of $\mathbb{R}$ intervals, $\{I_{\alpha }|\alpha \in A\}$ where $A$ is any possibly uncountable indices set, the union, ${\cup }_{\alpha \in A}{I}_{\alpha }$, is an $\mathbb{R}$ interval.

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2: Proof

For any points, $r_1,r_2\in {\cup }_{\alpha \in A}{I}_{\alpha }$, such that $r_1<r_2$, for any $r_3\in \mathbb{R}$ such that $r_1<r_3<r_2$, $r_3\in {\cup }_{\alpha \in A}{I}_{\alpha }$? Let us suppose that $r_3\notin {\cup }_{\alpha \in A}{I}_{\alpha }$. Then, each $I_{\alpha }$ would be entirely smaller than $r_3$ or entirely larger than $r_3$, because if $I_{\alpha }$ contained both a point smaller than $r_3$ and a point larger than $r_3$, $r_3$ would be contained in $I_{\alpha }$, and so, in the union. So, there would be the dichotomy such that all the intervals that are smaller than $r_3$ are in a part and all the intervals that are larger than $r_3$ are in the other part. The dichotomy would be disjoint, a contradiction.

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References

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