2023-10-15

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

Topics


About: topological space

The table of contents of this article


Starting Context



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.

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.


Main Body


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.


2: Description


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


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 \(r_1 \lt r_2\), there is a point, \(r_3 \in \mathbb{R}\), such that \(r_1 \lt r_3 \lt r_2\) and \(r_3 \notin S\). Let us define \(S_1:= S \cap (-\infty, r_3)\) and \(S_2:= S \cap (r_3, \infty)\). \(S_i\) is nonempty, and is open on \(S\) by the definition of subspace topology. \(S = S_1 \cup S_2\) and \(S_1 \cap S_2 = \emptyset\). 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.


References


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