386: Connected Topological Subspaces of 1-Dimensional Euclidean Topological Space Are Intervals

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A description/proof of that connected topological subspaces of 1-dimensional Euclidean topological space are intervals

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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: Note
• 2: Description
• 3: Proof

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Starting Context

• The reader knows a definition of connected topological space.
• The reader knows a definition of Euclidean topological space.
• The reader knows a definition of ${\mathbb{R}}^n$ interval.
• The reader knows a definition of subspace topology.
• The reader admits the proposition that any ${\mathbb{R}}^n$ interval is a connected topological subspace.

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Target Context

• The reader will have a description and a proof of the proposition that the set of all the connected topological subspaces of the $\mathbb{R}$ Euclidean topological space is the set of all the intervals.

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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: Note

'interval' here includes any singleton, $\{r\}$, as $[r,r]$. There can be another definition that any interval has to have at least 2 points, which we have not adopted, because excluding $[r,r]$ seems more unnatural and more cumbersome than convenient for our purposes.

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2: Description

The set of all the connected topological subspaces of the $\mathbb{R}$ Euclidean topological space is the set of all the intervals.

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3: Proof

Let us prove that any connected topological subspace is an interval, by proving that any non-interval is not any connected topological subspace. Let $S\subseteq \mathbb{R}$ be any subset that is not any interval. For some points, $r_1,r_2\in S$, such that $r_1<r_2$, there is a point, $r_3\in \mathbb{R}$, such that $r_1<r_3<r_2$ and $r_3\notin S$. Let us define $S_1:=S\cap (-\mathrm{\infty },r_3)$ and $S_2:=S\cap (r_3,\mathrm{\infty })$. $S_i$ is nonempty, and is open on $S$ by the definition of subspace topology. $S=S_1\cup S_2$ and $S_1\cap S_2=\mathrm{\varnothing }$. So, $S$ is not a connected topological subspace.

Any interval is a connected topological subspace, by the proposition that any ${\mathbb{R}}^n$ interval is a connected topolgical subspace.

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References

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