description/proof of that Borel \(\sigma\)-algebra of \(1\)-dimensional Euclidean topological space is generated by set of upper-open-or-closed-bounded intervals, set of lower-open-or-closed-bounded intervals, or set of lower-open-or-closed-upper-open-or-closed-bounded intervals
Topics
About: measurable 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 Borel \(\sigma\)-algebra of topological space.
- The reader knows a definition of \(\sigma\)-algebra of set generated by set of subsets.
- The reader admits the proposition that any half or both closed interval is the intersection of the sequence of some open intervals.
- The reader admits the proposition that for any Euclidean topological space, the set of all the open balls with rational centers and rational radii is a basis.
- The reader admits some criteria for any collection of open sets to be a basis.
- The reader admits the proposition that any half or both open interval is the union of the sequence of some closed intervals.
Target Context
- The reader will have a description and a proof of the proposition that the Borel \(\sigma\)-algebra of the \(1\)-dimensional Euclidean topological space is generated by the set of the upper-open-bounded intervals, the set of the upper-closed-bounded intervals, the set of the lower-open-bounded intervals, the set of the lower-closed-bounded intervals, the set of the lower-open-upper-open-bounded intervals, the set of the lower-open-upper-closed-bounded intervals, the set of the lower-closed-upper-open-bounded intervals, or the set of the lower-closed-upper-closed-bounded 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:
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\) with the topology, \(O\)
\(S_1\): \(= \{(- \infty, r) \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\)
\(S_2\): \(= \{(- \infty, r] \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\)
\(S_3\): \(= \{(r, \infty) \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\)
\(S_4\): \(= \{[r, \infty) \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\)
\(S_5\): \(= \{(r_1, r_2) \subseteq \mathbb{R} \vert r_1, r_2 \in \mathbb{R} \text{ such that } r_1 \lt r_2\}\)
\(S_6\): \(= \{(r_1, r_2] \subseteq \mathbb{R} \vert r_1, r_2 \in \mathbb{R} \text{ such that } r_1 \lt r_2\}\)
\(S_7\): \(= \{[r_1, r_2) \subseteq \mathbb{R} \vert r_1, r_2 \in \mathbb{R} \text{ such that } r_1 \lt r_2\}\)
\(S_8\): \(= \{[r_1, r_2] \subseteq \mathbb{R} \vert r_1, r_2 \in \mathbb{R} \text{ such that } r_1 \le r_2\}\)
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Statements:
\(B (\mathbb{R}) = \sigma (S_1) = \sigma (S_2) = \sigma (S_3) = \sigma (S_4) = \sigma (S_5) = \sigma (S_6) = \sigma (S_7) = \sigma (S_8)\)
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2: Proof
Whole Strategy: Step 1: see that \(B (\mathbb{R}) = \sigma (O)\); Step 2: see that \(B (\mathbb{R}) = \sigma (S_1)\); Step 3: see that \(B (\mathbb{R}) = \sigma (S_2)\); Step 4: see that \(B (\mathbb{R}) = \sigma (S_3)\); Step 5: see that \(B (\mathbb{R}) = \sigma (S_4)\); Step 6: see that \(B (\mathbb{R}) = \sigma (S_5)\); Step 7: see that \(B (\mathbb{R}) = \sigma (S_6)\); Step 8: see that \(B (\mathbb{R}) = \sigma (S_7)\); Step 9: see that \(B (\mathbb{R}) = \sigma (S_8)\).
Step 1:
\(B (\mathbb{R}) = \sigma (O)\), by the definition of Borel \(\sigma\)-algebra.
In general, the \(\sigma\)-algebra generated by any set of subsets is the intersection of all the \(\sigma\)-algebras that contain the set of the subsets, by Note for the definition of \(\sigma\)-algebra of set generated by set of subsets.
So, if any set of some subsets, \(S\), is contained in any \(\sigma\)-algebra, \(A\), which means that \(S \subseteq A\), \(\sigma (S) \subseteq A\), which we use without explicitly mentioning it hereafter.
Step 2:
Let us see that \(B (\mathbb{R}) = \sigma (S_1)\).
For each \((- \infty, r) \in S_1\), \((- \infty, r) \in O\), so, \((- \infty, r) \in \sigma (O)\).
So, \(S_1 \subseteq \sigma (O)\).
So, \(\sigma (S_1) \subseteq \sigma (O)\).
For each \(r \in \mathbb{R}\), \((- \infty, r] \in \sigma (S_1)\), by the proposition that any half or both closed interval is the intersection of the sequence of some open intervals.
For each \(r \in \mathbb{R}\), \((r, \infty) = \mathbb{R} \setminus (- \infty, r] \in \sigma (S_1)\).
So, for each \(r_1, r_2 \in \mathbb{R}\) such that \(r_1 \lt r_2\), \((r_1, r_2) = (- \infty, r_2) \cap (r_1, \infty) \in \sigma (S_1)\).
Then, each open ball with rational center and rational radius on \(\mathbb{R}\) is in \(\sigma (S_1)\), because it is \((r_1, r_2)\).
Each \(U \in O\) is the union of some open balls with rational centers and rational radii, by the proposition that for any Euclidean topological space, the set of all the open balls with rational centers and rational radii is a basis and Description 1 of some criteria for any collection of open sets to be a basis.
As the basis is countable, \(U \in \sigma (S_1)\).
So, \(O \subseteq \sigma (S_1)\).
So, \(\sigma (O) \subseteq \sigma (S_1)\).
So, \(\sigma (O) = \sigma (S_1)\).
So, \(B (\mathbb{R}) = \sigma (S_1)\).
Step 3:
Let us see that \(B (\mathbb{R}) = \sigma (S_2)\).
For each \((- \infty, r] \in S_2\), \((- \infty, r] \in \sigma (S_1)\), by the proposition that any half or both closed interval is the intersection of the sequence of some open intervals.
So, \(S_2 \subseteq \sigma (S_1)\).
So, \(\sigma (S_2) \subseteq \sigma (S_1)\).
For each \((- \infty, r) \in S_1\), \((- \infty, r) \in \sigma (S_2)\), by the proposition that any half or both open interval is the union of the sequence of some closed intervals.
So, \(S_1 \subseteq \sigma (S_2)\).
So, \(\sigma (S_1) \subseteq \sigma (S_2)\).
So, \(B (\mathbb{R}) = \sigma (S_1) = \sigma (S_2)\).
Step 4:
Let us see that \(B (\mathbb{R}) = \sigma (S_3)\).
For each \((r, \infty) \in S_3\), \((r, \infty) \in O\), so, \((r, \infty) \in \sigma (O)\).
So, \(S_3 \subseteq \sigma (O)\).
So, \(\sigma (S_3) \subseteq \sigma (O)\).
For each \(r \in \mathbb{R}\), \([r, \infty) \in \sigma (S_3)\), by the proposition that any half or both closed interval is the intersection of the sequence of some open intervals.
For each \(r \in \mathbb{R}\), \((- \infty, r) = \mathbb{R} \setminus [r, \infty) \in \sigma (S_3)\).
So, for each \(r_1, r_2 \in \mathbb{R}\) such that \(r_1 \lt r_2\), \((r_1, r_2) = (- \infty, r_2) \cap (r_1, \infty) \in \sigma (S_3)\).
Then, each open ball with rational center and rational radius on \(\mathbb{R}\) is in \(\sigma (S_3)\), because it is \((r_1, r_2)\).
Each \(U \in O\) is the union of some open balls with rational centers and rational radii, by the proposition that for any Euclidean topological space, the set of all the open balls with rational centers and rational radii is a basis and Description 1 of some criteria for any collection of open sets to be a basis.
As the basis is countable, \(U \in \sigma (S_3)\).
So, \(O \subseteq \sigma (S_3)\).
So, \(\sigma (O) \subseteq \sigma (S_3)\).
So, \(\sigma (O) = \sigma (S_3)\).
So, \(B (\mathbb{R}) = \sigma (S_3)\).
Step 5:
Let us see that \(B (\mathbb{R}) = \sigma (S_4)\).
For each \([r, \infty) \in S_4\), \([r, \infty) \in \sigma (S_3)\), by the proposition that any half or both closed interval is the intersection of the sequence of some open intervals.
So, \(S_4 \subseteq \sigma (S_3)\).
So, \(\sigma (S_4) \subseteq \sigma (S_3)\).
For each \((r, \infty) \in S_3\), \((r, \infty) \in \sigma (S_4)\), by the proposition that any half or both open interval is the union of the sequence of some closed intervals.
So, \(S_3 \subseteq \sigma (S_4)\).
So, \(\sigma (S_3) \subseteq \sigma (S_4)\).
So, \(B (\mathbb{R}) = \sigma (S_3) = \sigma (S_4)\).
Step 6:
Let us see that \(B (\mathbb{R}) = \sigma (S_5)\).
For each \((r_1, r_2) \in S_5\), \((r_1, r_2) \in O\), so, \((r_1, r_2) \in \sigma (O)\).
So, \(S_5 \subseteq \sigma (O)\).
So, \(\sigma (S_5) \subseteq \sigma (O)\).
Each open ball with rational center and rational radius on \(\mathbb{R}\) is in \(\sigma (S_5)\), because it is \((r_1, r_2)\).
Each \(U \in O\) is the union of some open balls with rational centers and rational radii, by the proposition that for any Euclidean topological space, the set of all the open balls with rational centers and rational radii is a basis and Description 1 of some criteria for any collection of open sets to be a basis.
As the basis is countable, \(U \in \sigma (S_5)\).
So, \(O \subseteq \sigma (S_5)\).
So, \(\sigma (O) \subseteq \sigma (S_5)\).
So, \(\sigma (O) = \sigma (S_5)\).
So, \(B (\mathbb{R}) = \sigma (S_5)\).
Step 7:
Let us see that \(B (\mathbb{R}) = \sigma (S_6)\).
For each \((r_1, r_2] \in S_6\), \((r_1, r_2] \in \sigma (S_5)\), by the proposition that any half or both closed interval is the intersection of the sequence of some open intervals.
So, \(S_6 \subseteq \sigma (S_5)\).
So, \(\sigma (S_6) \subseteq \sigma (S_5)\).
For each \((r_1, r_2) \in S_5\), \((r_1, r_2) \in \sigma (S_6)\), by the proposition that any half or both open interval is the union of the sequence of some closed intervals.
So, \(S_5 \subseteq \sigma (S_6)\).
So, \(\sigma (S_5) \subseteq \sigma (S_6)\).
So, \(B (\mathbb{R}) = \sigma (S_5) = \sigma (S_6)\).
Step 8:
Let us see that \(B (\mathbb{R}) = \sigma (S_7)\).
For each \([r_1, r_2) \in S_7\), \([r_1, r_2) \in \sigma (S_5)\), by the proposition that any half or both closed interval is the intersection of the sequence of some open intervals.
So, \(S_7 \subseteq \sigma (S_5)\).
So, \(\sigma (S_7) \subseteq \sigma (S_5)\).
For each \((r_1, r_2) \in S_5\), \((r_1, r_2) \in \sigma (S_7)\), by the proposition that any half or both open interval is the union of the sequence of some closed intervals.
So, \(S_5 \subseteq \sigma (S_7)\).
So, \(\sigma (S_5) \subseteq \sigma (S_7)\).
So, \(B (\mathbb{R}) = \sigma (S_5) = \sigma (S_7)\).
Step 9:
Let us see that \(B (\mathbb{R}) = \sigma (S_8)\).
For each \([r_1, r_2] \in S_8\), \([r_1, r_2] \in \sigma (S_5)\), by the proposition that any half or both closed interval is the intersection of the sequence of some open intervals.
So, \(S_8 \subseteq \sigma (S_5)\).
So, \(\sigma (S_8) \subseteq \sigma (S_5)\).
For each \((r_1, r_2) \in S_5\), \((r_1, r_2) \in \sigma (S_8)\), by the proposition that any half or both open interval is the union of the sequence of some closed intervals.
So, \(S_5 \subseteq \sigma (S_8)\).
So, \(\sigma (S_5) \subseteq \sigma (S_8)\).
So, \(B (\mathbb{R}) = \sigma (S_5) = \sigma (S_8)\).
