description/proof of that for \(1\)-dimensional Euclidean topological space and open subset, open subset is disjoint union of countable open intervals
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of countable set.
- The reader admits the proposition that the union of any possibly uncountable dichotomically nondisjoint set of \(\mathbb{R}\) intervals is a \(\mathbb{R}\) interval.
- The reader admits the proposition that any finite composition of injections is an injection.
- The reader admits the proposition that for any infinite set, if there is any injection from the set into the natural numbers set, there is a bijection from the natural numbers set onto the set.
Target Context
- The reader will have a description and a proof of the proposition that for the \(1\)-dimensional Euclidean topological space and any open subset, the open subset is the union of some disjoint countable open 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(U\): \(\in \{\text{ the open subsets of } T\}\)
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Statements:
\(\exists J \in \{\text{ the countable index sets }\}, \exists \{I_j \in \{\text{ the open intervals }\} \vert j \in J\} (\forall j, j' \in J \text{ such that } j \neq j' (I_j \cap I_{j'} = \emptyset) \land U = \cup_{j \in J} I_j)\)
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2: Proof
Whole Strategy: Step 1: for each \(u \in U\), take the maximum open interval around \(u\) contained in \(U\), \(I_u\), and take \(\{I_u \vert u \in U\}\); Step 2: see that \(\{I_u \vert u \in U\}\) is disjoint, take a rational number in each \(I_u\), take an injection, \(f: \{I_u \vert u \in U\} \to \mathbb{N}\), and a bijection, \(f': \mathbb{N} \to \{I_u \vert u \in U\}\), take \(J = \mathbb{N}\) and \(f' (J) = \{I_j \vert j \in J\}\), and see that \(U = \cup_{j \in J} I_j\).
Step 1:
For each \(u \in U\), take the maximum open interval around \(u\) contained in \(U\), \(I_u \subseteq \mathbb{R}\), which is determined uniquely, because that is the union of the set of the open intervals around \(u\) contained in \(U\): the set is nonempty, because as \(U\) is open, there is at least \(1\) \((u - \epsilon, u + \epsilon) \subseteq U\), and the union is indeed an interval, by the proposition that the union of any possibly uncountable dichotomically nondisjoint set of \(\mathbb{R}\) intervals is a \(\mathbb{R}\) interval: the set is dichotomically nondisjoint, because \(u\) is contained in each element, and the union is an open interval, because it is open as the union of some open subsets, and the union contains \(u\) and is contained in \(U\), and the union is the maximum, because any open interval around \(u\) contained in \(U\) is an element of the set and is contained in the union.
Let us take \(\{I_u \vert u \in U\}\).
Note that for some \(u \neq u'\), \(I_u = I_{u'}\) is possible, and they are a single element, so, \(\{I_u \vert u \in U\}\) does not necessarily really have the points of \(U\) number elements.
Step 2:
Let us see that \(\{I_u \vert u \in U\}\) is disjoint.
Let \(I_u, I_{u'} \in \{I_u \vert u \in U\}\) be any such that \(I_u \neq I_{u'}\).
Let us suppose that \(I_u \cap I_{u'} \neq \emptyset\).
There would be an \(r \in I_u \cap I_{u'}\).
There would be \(I_r\) and \(I_u, I_{u'} \subseteq I_r\), because \(I_u\) and \(I_{u'}\) were some open intervals around \(r\) contained in \(U\) and \(I_r\) was the union of all the such open intervals, which would imply that \(u, u' \in I_r\), so, \(I_r \subseteq I_u, I_{u'}\), because \(I_r\) was an open interval around \(u\) contained in \(U\) while \(I_u\) was the union of all the such open intervals, and \(I_r\) was an open interval around \(u'\) contained in \(U\) while \(I_{u'}\) was the union of all the such open intervals.
So, \(I_r \subseteq I_u, I_{u'} \subseteq I_r\), so, \(I_u = I_{u'} = I_r\), a contradiction against \(I_u \neq I_{u'}\).
So, \(I_u \cap I_{u'} = \emptyset\).
So, \(\{I_u \vert u \in U\}\) is disjoint.
There is the map, \(g: \{I_u \vert u \in U\} \to Pow (\mathbb{Q})\), that maps \(I_u\) to the set of the points of \(\mathbb{Q}\) contained in \(I_u\).
Each \(g (I_u)\) is nonempty, because \(I_u\) contains at least \(1\) rational number.
So, by the axiom of choice, there is a map, \(g': \{I_u \vert u \in U\} \to \mathbb{Q}\), where \(g' (I_j) \in g (I_u)\), which is an injection, because \(\{I_u \vert u \in U\}\) is disjoint.
As \(\mathbb{Q}\) is countable, there is a bijection, \(h: \mathbb{N} \to \mathbb{Q}\).
Let \(f: \{I_u \vert u \in U\} \to \mathbb{N} = h^{-1} \circ g'\), which is an injection, by the proposition that any finite composition of injections is an injection.
So, there is a bijection, \(f': \mathbb{N} \to \{I_u \vert u \in U\}\), by the proposition that for any infinite set, if there is any injection from the set into the natural numbers set, there is a bijection from the natural numbers set onto the set.
Let \(J := \mathbb{N}\).
Let \(f' (J) = \{I_j = f' (j) \vert j \in J\}\).
\(\{I_j \vert j \in J\} = \{I_u \vert u \in U\}\), because \(f'\) is a bijection onto \(\{I_u \vert u \in U\}\).
\(\{I_j \vert j \in J\}\) is disjoint, because \(\{I_u \vert u \in U\}\) is so.
Let us see that \(U = \cup_{j \in J} I_j\).
For each \(u \in U\), \(u \in I_u\), but there is a \(j \in J\) such that \(I_j = I_u\), because \(\{I_j \vert j \in J\} = \{I_u \vert u \in U\}\), so, \(u \in I_j\).
So, \(u \in \cup_{j \in J} I_j\).
So, \(U \subseteq \cup_{j \in J} I_j\).
For each \(r \in \cup_{j \in J} I_j\), \(r \in I_j\) for a \(j \in J\), but \(I_j = I_u\) for a \(u \in U\), and \(r \in I_j = I_u \subseteq U\).
So, \(r \in U\).
So, \(\cup_{j \in J} I_j \subseteq U\).
So, \(U = \cup_{j \in J} I_j\).
