description/proof of that for \(1\)-dimensional Euclidean topological space and equivalence relation that \(2\) elements are equivalent iff their difference is rational, quotient topology is trivial
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of quotient set.
- The reader knows a definition of quotient topology on set with respect to map.
- The reader knows a definition of trivial topological space.
- The reader admits a way for systematically choosing a rational number that is larger than any real number and is equal to or smaller than another any real number.
Target Context
- The reader will have a description and a proof of the proposition that for the \(1\)-dimensional Euclidean topological space and the equivalence relation that any \(2\) elements are equivalent if and only if their difference is rational, the quotient topology is trivial.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(\sim\): \(\in \{\text{ the equivalence relations on } \mathbb{R}\}\), such that for each \(r_1, r_2 \in \mathbb{R}\), \(r_1 \sim r_2 \iff r_1 - r_2 \in \mathbb{Q}\)
\(\mathbb{R} / \sim\): \(= \text{ the quotient set }\) with the quotient topology with respect to \(f\)
\(f\): \(: \mathbb{R} \to \mathbb{R} / \sim\), such that for each \(r \in \mathbb{R}\), \(r \in f (r)\)
\(\) //
Statements:
\(\mathbb{R} / \sim \in \{\text{ the trivial topological spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: for any nonempty open \(U \subseteq \mathbb{R} / \sim\), see that for each \(r' \in \mathbb{R}\), \(r' \in f^{-1} (U)\), by taking any \(r \in f^{-1} (U)\) and \(B_{r, \epsilon} \subseteq f^{-1} (U)\) and seeing that \(r' - q \in B_{r, \epsilon}\) for a \(q \in \mathbb{Q}\).
Step 1:
Let \(U \subseteq \mathbb{R} / \sim\) be any nonempty open subset.
Let \(r \in f^{-1} (U)\) be any, which exists, because \(U\) is nonempty and \(f\) is a surjection.
There is a \(B_{r, \epsilon} \subseteq \mathbb{R}\) such that \(B_{r, \epsilon} \subseteq f^{-1} (U)\), because \(f^{-1} (U) \subseteq \mathbb{R}\) is open.
Let \(r' \in \mathbb{R}\) be any.
\(r' - r - \epsilon \lt r' - r + \epsilon\).
So, there is a \(q \in \mathbb{Q}\) such that \(r' - r - \epsilon \lt q \lt r' - r + \epsilon\), by Note for the way for systematically choosing a rational number that is larger than any real number and is equal to or smaller than another any real number.
\(- r' + r - \epsilon \lt - q \lt - r' + r + \epsilon\), \(r - \epsilon \lt r' - q \lt r + \epsilon\), so, \(r' - q \in B_{r, \epsilon}\), so, \(r' - q \in f^{-1} (U)\).
\(f (r') = f (r' - q) \in U\), so, \(r' \in f^{-1} (U)\).
So, \(f^{-1} (U) = \mathbb{R}\).
So, \(U = \mathbb{R} / \sim\), because if \([r] \notin U\), \(r \notin f^{-1} (U)\), a contradiction.
That means that the open subsets of \(\mathbb{R} / \sim\) are \(\emptyset\) and \(\mathbb{R} / \sim\).
So, \(\mathbb{R} / \sim\) is trivial.
