<The previous article in this series | The table of contents of this series |
description/proof of that for sequence on \(1\)-dimensional Euclidean metric space, if sequence converges, sequence of arithmetic means of leading elements converges with convergence
Topics
About:
metric 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 for any sequence on the \(1\)-dimensional Euclidean metric space, if the sequence converges, the sequence of the arithmetic means of the leading elements converges with the convergence.
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:
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s\): \(: \mathbb{N} \to \mathbb{R}\)
\(s'\): \(: \mathbb{N} \to \mathbb{R}, n \mapsto \sum_{j \in \{0, ..., n\}} s (j) / (n + 1)\)
\(r\): \(\in \mathbb{R}\)
//
Statements:
\(lim s = r\)
\(\implies\)
\(lim s' = r\)
//
2: Note
In fact, for any \(m \in \mathbb{N}\), \(s'': \mathbb{N} \to \mathbb{R}, n \mapsto \sum_{j \in \{0, ..., n\}} s (j) / (n + m)\) converges to \(r\), because \(s'' (n) = \sum_{j \in \{0, ..., n\}} s (j) / (n + m) = \sum_{j \in \{0, ..., n\}} s (j) / (n + 1) (n + 1) / (n + m) = s' (n) (n + 1) / (n + m)\), and \((n + 1) / (n + m)\) converges to \(1\), so, \(s''\) converges to \(r 1 = r\).
The reverse of this proposition does not necessarily hold: as a counterexample, let \(s (j) = 1\) for each even \(j\) and \(s (j) = -1\) for each odd \(j\), then, \(s' (j) = 1 / (j + 1)\) for each even \(j\) and \(s' (j) = 0\) for each odd \(j\), so, \(s' (j) \lt 1 / (2 (j + 1))\) for each \(j\), so, \(s'\) converges to \(0\), but \(s\) does not converge.
3: Proof
Whole Strategy: Step 1: take an \(N_1\) such that for each \(N_1 \lt j\), \(\vert s (j) - r \vert \lt \epsilon / 2\) and an \(N_2\) such that \(N_1 \lt N_2\) and for each \(N_2 \lt l\), \(1 / (l + 1) \sum_{j \in \{0, ..., N_1\}} \vert s (j) - r \vert \lt \epsilon / 2\), and see that for each \(N_2 \lt l\), \(\vert s' (l) - r \vert \lt \epsilon\).
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N_1 \in \mathbb{N}\) such that for each \(j \in \mathbb{N}\) such that \(N_1 \lt j\), \(\vert s (j) - r \vert \lt \epsilon / 2\), because \(s\) converges to \(r\).
There is an \(N_2 \in \mathbb{N}\) such that \(N_1 \lt N_2\) and for each \(l \in \mathbb{N}\) such that \(N_2 \lt l\), \(1 / (l + 1) \sum_{j \in \{0, ..., N_1\}} \vert s (j) - r \vert \lt \epsilon / 2\), because as \(\sum_{j \in \{0, ..., N_1\}} \vert s (j) - r \vert\) is fixed, making \(l\) larger makes it any smaller.
For each \(N_2 \lt l\), \(\vert s' (l) - r \vert = \vert \sum_{j \in \{0, ..., l\}} s (j) / (l + 1) - r \vert = \vert \sum_{j \in \{0, ..., l\}} (s (j) / (l + 1) - r / (l + 1)) \vert = 1 / (l + 1) \vert \sum_{j \in \{0, ..., l\}} (s (j) - r) \vert \le 1 / (l + 1) (\sum_{j \in \{0, ..., N_1\}} \vert s (j) - r \vert + \sum_{j \in \{N_1 + 1, ..., l\}} \vert s (j) - r \vert) = 1 / (l + 1) \sum_{j \in \{0, ..., N_1\}} \vert s (j) - r \vert + 1 / (l + 1) \sum_{j \in \{N_1 + 1, ..., l\}} \vert s (j) - r \vert \lt \epsilon / 2 + 1 / (l + 1) \sum_{j \in \{N_1 + 1, ..., l\}} \epsilon / 2 = \epsilon / 2 + 1 / (l + 1) (l - N_1) \epsilon / 2 \lt \epsilon / 2 + 1 / (l + 1) (l + 1) \epsilon / 2 = \epsilon\).
So, \(lim s' = r\).
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for non-negative measurable map into \(1\)-dimensional Euclidean measurable space, composition of floor Map after map is measurable
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any non-negative measurable map into the \(1\)-dimensional Euclidean measurable space, the composition of the floor Map after the map as into the \(1\)-dimensional Euclidean measurable space is measurable.
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:
\(M_1\): \(\in \{\text{ the measurable spaces }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\([0, \infty)\): \(\subseteq \mathbb{R}\), with the subspace \(\sigma\)-algebra
\(f\): \(: M_1 \to \mathbb{R}\), \(\in \{\text{ the measurable maps }\}\), such that \(0 \le f\)
\(f'\): \(: M_1 \to [0, \infty)\), \(= \text{ the codomain restriction of } f\)
\(fl\): \(: [0, \infty) \to \mathbb{N}\)
\(fl'\): \(: [0, \infty) \to \mathbb{R}\), \(= \text{ the codomain extension of } fl\)
\(fl' \circ f'\): \(: M_1 \to \mathbb{R}\)
//
Statements:
\(fl' \circ f' \in \{\text{ the measurable maps }\}\)
//
2: Note
\(fl' \circ f'\) is \(fl \circ f\) loosely speaking, but we need \(f'\) and \(fl'\) formally, because \(fl \circ f\) does not make sense strictly speaking, because any composition is allowed only when the codomain of a map is contained in the domain of the succeeding map and \(fl \circ f'\) is \(: M \to \mathbb{N}\).
3: Proof
Whole Strategy: Step 1: see that \(f'\) is measurable; Step 2: see that \(fl'\) is measurable; Step 3: conclude the proposition.
Step 1:
\(f'\) is measurable, by the proposition that for any measurable map between any measurable spaces, the restriction on any domain subspace and any codomain subspace is measurable.
Step 2:
\(fl'\) is measurable, by the proposition that the floor map with the codomain extended to the \(1\)-dimensional Euclidean measurable space is measurable.
Step 3:
\(fl' \circ f'\) is measurable, by the proposition that for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that floor map with codomain extended to \(1\)-dimensional Euclidean measurable space is measurable
Topics
About:
measurable 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 floor map with the codomain extended to the \(1\)-dimensional Euclidean measurable space is measurable.
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 Borel \(\sigma\)-algebra
\([0, \infty)\): \(\subseteq \mathbb{R}\), with the subspace \(\sigma\)-algebra
\(fl\): \(: [0, \infty) \to \mathbb{N}, r \mapsto Max (\{n \in \mathbb{N} \vert n \le r\})\)
\(fl'\): \(: [0, \infty) \to \mathbb{R}\), \(= \text{ the codomain extension of } fl\)
//
Statements:
\(fl' \in \{\text{ the measurable maps }\}\)
//
2: Note
As an immediate corollary, \(fl\) is measurable with \(\mathbb{N}\) as the subspace \(\sigma\)-algebra, by the proposition that for any measurable map between any measurable spaces, the restriction on any domain subspace and any codomain subspace is measurable.
3: Proof
Whole Strategy: Step 1: see that \(B (\mathbb{R})\) is generated by \(\{(- \infty, r] \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\); Step 2: see that \({fl'}^{-1} ((- \infty, r])\) is measurable; Step 3: conclude the proposition.
Step 1:
\(B (\mathbb{R})\) is generated by \(\{(- \infty, r] \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\), by 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.
Step 2:
Let us see that \({fl'}^{-1} ((- \infty, r]) \subseteq [0, \infty)\) is measurable.
When \(r \lt 0\), \({fl'}^{-1} ((- \infty, r]) = \emptyset\) is measurable.
Otherwise, \({fl'}^{-1} ((- \infty, r]) = [0, fl (r + 1))\), because for each \(r' \in {fl'}^{-1} ((- \infty, r])\), \(fl' (r') \in (- \infty, r]\), so, \(fl (r') = fl' (r') \le r\), so, \(r' \lt fl (r + 1)\), by the property 5) mentioned in Note for the definition of floor map, so, \(r' \in [0, fl (r + 1))\); for each \(r' \in [0, fl (r + 1))\), \(r' \lt fl (r + 1)\), so, \(fl' (r') = fl (r') \le r\), by the property 5) mentioned in Note for the definition of floor map, so, \(fl' (r') \in (- \infty, r]\), so, \(r' \in {fl'}^{-1} ((- \infty, r])\), and \([0, fl (r + 1)) = (- \infty, fl (r + 1)) \cap [0, \infty)\) is measurable.
Step 3:
So, \(fl'\) is measurable, by the proposition that for any map between any measurable spaces, if the preimage of each element of any generator of the codomain \(\sigma\)-algebra is measurable, the map is measurable.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of floor map
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of floor map.
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:
\(*fl\): \(: [0, \infty) \to \mathbb{N}, r \mapsto Max (\{n \in \mathbb{N} \vert n \le r\})\)
//
Conditions:
//
2: Note
Let us see that \(fl\) satisfies some properties.
1): For each \(r \in [0, \infty)\), \(fl (r) \le r\), and if and only if \(r \in \mathbb{N}\), \(fl (r) = r\), because \(r = n + r'\) where \(0 \le r' \lt 1\), and \(fl (r) = fl (n + r') = n\), and \(fl (r) = n \le n + r' = r\); if \(r \in \mathbb{N}\), \(r' = 0\), and \(fl (r) = n = r\), and if \(fl (r) = r\), \(n = r\), so, \(r \in \mathbb{N}\).
2): For each \(r \in [0, \infty)\), \(fl \circ fl (r) = fl (r)\), because \(fl (r) \in \mathbb{N}\) and 1) applies.
3): For each \(r, r' \in [0, \infty)\) such that \(r \le r'\), \(fl (r) \le fl (r')\) but not necessarily \(r \lt r'\) implies \(fl (r) \lt fl (r')\), because if \(r \le r'\), \(\{n \in \mathbb{N} \vert n \le r\} \subseteq \{n \in \mathbb{N} \vert n \le r'\}\) and the proposition that for any partially-ordered set, any subset, and any subset of the subset, if the infimum of the subset and the infimum of the subset of the subset exist, the infimum of the subset is equal to or smaller than the infimum of the subset of the subset, and if the supremum of the subset and the supremum of the subset of the subset exist, the supremum of the subset is equal to or larger than the supremum of the subset of the subset applies: the maximums are the supremums; if \(r \lt r'\), if \(r = 1.1\) and \(r' = 1.2\), \(fl (r) = 1 = fl (r')\), so, "\(fl (r) \lt fl (r')\)" does not necessarily hold.
4): For each \(r \in [0, \infty)\), \(r \lt fl (r + 1)\), because \(r = n + r'\) where \(0 \le r' \lt 1\), and \(fl (r + 1) = fl (n + 1 + r') = n + 1\) and \(n + r' \lt n + 1\).
5): For each \(r, r' \in [0, \infty)\), if and only if \(fl (r) \le r'\), \(r \lt fl (r' + 1)\), because if \(fl (r) \le r'\), if \(fl (r' + 1) \le r\), \(fl (r' + 1) = fl \circ fl (r' + 1) \le fl (r)\), by 2) and 3), \(\le r'\), a contradiction against 4); if \(r \lt fl (r' + 1)\), \(fl (r) \le r \lt fl (r' + 1)\), by 1), but \(r' = n + r''\) where \(0 \le r'' \lt 1\), and \(fl (r' + 1) = fl (n + 1 + r'') = n + 1\), so, \(fl (r) \lt n + 1\), but as \(fl (r) \in \mathbb{N}\), \(fl (r) \le n \lt n + 1\), and \(fl (r) \le n \le n + r'' = r'\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
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
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, \(O\)
\(S_1\): \(= \{(- \infty, r) \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\)
\(S_2\): \(= \{(- \infty, r] \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\)
\(S_3\): \(= \{(r, \infty) \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\)
\(S_4\): \(= \{[r, \infty) \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\)
\(S_5\): \(= \{(r_1, r_2) \subseteq \mathbb{R} \vert r_1, r_2 \in \mathbb{R} \text{ such that } r_1 \lt r_2\}\)
\(S_6\): \(= \{(r_1, r_2] \subseteq \mathbb{R} \vert r_1, r_2 \in \mathbb{R} \text{ such that } r_1 \lt r_2\}\)
\(S_7\): \(= \{[r_1, r_2) \subseteq \mathbb{R} \vert r_1, r_2 \in \mathbb{R} \text{ such that } r_1 \lt r_2\}\)
\(S_8\): \(= \{[r_1, r_2] \subseteq \mathbb{R} \vert r_1, r_2 \in \mathbb{R} \text{ such that } r_1 \le r_2\}\)
//
Statements:
\(B (\mathbb{R}) = \sigma (S_1) = \sigma (S_2) = \sigma (S_3) = \sigma (S_4) = \sigma (S_5) = \sigma (S_6) = \sigma (S_7) = \sigma (S_8)\)
//
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, \(B (\mathbb{R}) = \sigma (S_5) = \sigma (S_8)\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that half or both open interval is union of sequence of closed intervals
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any half or both open interval is the union of the sequence of some closed 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:
\(I\): \(\in \{\text{ the half or both open intervals }\}\), \(= (- \infty, r_2), [r_1, r_2), (r_1, \infty), (r_1, r_2], (- \infty, \infty), \text{ or } (r_1, r_2)\)
//
Statements:
\((- \infty, r_2) = \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}] = \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\)
\(\land\)
\([r_1, r_2) = \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\)
\(\land\)
\((r_1, \infty) = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})] = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\)
\(\land\)
\((r_1, r_2] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\)
\(\land\)
\((- \infty, \infty) = \cup_{n \in \mathbb{N}} [- n, n]\)
\(\land\)
\((r_1, r_2) = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\)
//
2: Note
While \(\cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) or \(\cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\) is not really "union of the sequence of some closed intervals", they are included here, because they are sometimes used.
3: Proof
Whole Strategy: Step 1: see that \((- \infty, r_2) = \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}] = \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\); Step 2: see that \([r_1, r_2) = \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\); Step 3; see that \((r_1, \infty) = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})] = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\); Step 4: see that \((r_1, r_2] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\); Step 5: \((- \infty, \infty) = \cup_{n \in \mathbb{N}} [- n, n]\); Step 6: \((r_1, r_2) = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
Step 1:
\([Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\) is valid for each \(n \in \mathbb{N}\), because \(Min (\{- n, r_2 - 2^{- n}\}) \le r_2 - 2^{- n}\).
Let us see that \((- \infty, r_2) = \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}] = \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\).
Let \(r \in (- \infty, r_2)\) be any.
There is an \(N_1 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_1 \lt n\), \(Min (\{- n, r_2 - 2^{- n}\}) \le - n \le r\).
\(r \lt r_2\), and there is an \(N_2 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_2 \lt n\), \(r \le r_2 - 2^{- n}\).
So, for each \(n \in \mathbb{N}\) such that \(N_1, N_2 \lt n\), \(Min (\{- n, r_2 - 2^{- n}\}) \le r \le r_2 - 2^{- n}\), so, \(r \in [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\).
So, \(r \in \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\).
So, \((- \infty, r_2) \subseteq \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\).
Let \(r \in \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\) be any.
\(r \in [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\) for an \(n \in \mathbb{N}\).
So, \(r \in [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}] \subseteq (- \infty, r_2)\).
So, \(\cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}] \subseteq (- \infty, r_2)\).
So, \((- \infty, r_2) = \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\).
Let \(r \in (- \infty, r_2)\) be any.
\(- \infty \lt r\).
As \(r \lt r_2\), there is an \(n \in \mathbb{N}\) such that \(r \le r_2 - 2^{- n}\).
So, \(r \in (- \infty, r_2 - 2^{- n}]\).
So, \(r \in \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\).
So, \((- \infty, r_2) \subseteq \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\).
Let \(r \in \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\) be any.
\(r \in (- \infty, r_2 - 2^{- n}]\) for an \(n \in \mathbb{N}\).
\(r \in (- \infty, r_2 - 2^{- n}] \subseteq (- \infty, r_2)\).
So, \(\cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}] \subseteq (- \infty, r_2)\).
So, \((- \infty, r_2) = \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\).
Step 2:
\([r_1, Max (\{r_1, r_2 - 2^{- n}\})]\) is valid for each \(n \in \mathbb{N}\), because \(r_1 \le Max (\{r_1, r_2 - 2^{- n}\})\).
Let us see that \([r_1, r_2) = \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\).
Let \(r \in [r_1, r_2)\) be any.
\(r_1 \le r\).
\(r \lt r_2\), and there is an \(n \in \mathbb{N}\) such that \(r \le r_2 - 2^{- n}\), then, \(r \le r_2 - 2^{- n} \le Max (\{r_1, r_2 - 2^{- n}\})\).
So, \(r \in [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\).
So, \(r \in \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\).
So, \([r_1, r_2) \subseteq \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\).
Let \(r \in \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\) be any.
\(r \in [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\) for an \(n \in \mathbb{N}\).
So, \(r \in [r_1, Max (\{r_1, r_2 - 2^{- n}\})] \subseteq [r_1, r_2)\), because \(r_1 \lt r_2\) and \(r_2 - 2^{- n} \lt r_2\).
So, \(\cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})] \subseteq [r_1, r_2)\).
So, \([r_1, r_2) = \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\).
Step 3:
\([r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\) is valid for each \(n \in \mathbb{N}\), because \(r_1 + 2^{- n} \le Max (\{n, r_1 + 2^{- n}\})\).
Let us see that \((r_1, \infty) = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})] = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\).
Let \(r \in (r_1, \infty)\) be any.
\(r_1 \lt r\), and there is an \(N_1 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_1 \lt n\), \(r_1 + 2^{- n} \le r\).
There is an \(N_2 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_2 \lt n\), \(r \le n \le Max (\{n, r_1 + 2^{- n}\})\).
So, for each \(n \in \mathbb{N}\) such that \(N_1, N_2 \lt n\), \(r_1 + 2^{- n} \le r \le Max (\{n, r_1 + 2^{- n}\})\), so, \(r \in [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\).
So, \(r \in \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\).
So, \((r_1, \infty) \subseteq \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\).
Let \(r \in \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\) be any.
\(r \in [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\) for an \(n \in \mathbb{N}\).
So, \(r \in [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})] \subseteq (r_1, \infty)\).
So, \(\cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})] \subseteq (r_1, \infty)\).
So, \((r_1, \infty) = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\).
Let \(r \in (r_1, \infty)\) be any.
\(r \lt \infty\).
As \(r_1 \lt r\), there is an \(n \in \mathbb{N}\) such that \(r_1 + 2^{- n} \le r\).
So, \(r \in [r_1 + 2^{- n}, \infty)\).
So, \(r \in \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\).
So, \((r_1, \infty) \subseteq \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\).
Let \(r \in \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\) be any.
\(r \in [r_1 + 2^{- n}, \infty)\) for an \(n \in \mathbb{N}\).
\(r \in [r_1 + 2^{- n}, \infty) \subseteq (r_1, \infty)\).
So, \(\cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty) \subseteq (r_1, \infty)\).
So, \((r_1, \infty) = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\).
Step 4:
Let us see that \((r_1, r_2] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\).
Let \(r \in (r_1, r_2]\) be any.
\(r_1 \lt r\), and there is an \(n \in \mathbb{N}\) such that \(r_1 + 2^{- n} \le r\), then, \(Min (\{r_1 + 2^{- n}, r_2\}) \le r_1 + 2^{- n} \le r\).
\(r \le r_2\).
So, \(r \in [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\).
So, \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\).
So, \((r_1, r_2] \subseteq \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\).
Let \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\) be any.
\(r \in [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\) for an \(n \in \mathbb{N}\).
So, \(r \in [Min (\{r_1 + 2^{- n}, r_2\}), r_2] \subseteq (r_1, r_2]\), because \(r_1 \lt r_1 + 2^{- n}\) and \(r_1 \lt r_2\).
So, \(\cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2] \subseteq (r_1, r_2]\).
So, \((r_1, r_2] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\).
Step 5:
Let us see that \((- \infty, \infty) = \cup_{n \in \mathbb{N}} [- n, n]\).
Let \(r \in (- \infty, \infty)\) be any.
There is an \(n \in \mathbb{N}\) such that \(r \in [- n, n]\).
So, \(r \in \cup_{n \in \mathbb{N}} [- n, n]\).
So, \((- \infty, \infty) \subseteq \cup_{n \in \mathbb{N}} [- n, n]\).
Let \(r \in \cup_{n \in \mathbb{N}} [- n, n]\) be any.
\(r \in [- n, n]\) for an \(n \in \mathbb{N}\).
So, \(r \in [- n, n] \subseteq (- \infty, \infty)\).
So, \(\cup_{n \in \mathbb{N}} [- n, n] \subseteq (- \infty, \infty)\).
So, \((- \infty, \infty) = \cup_{n \in \mathbb{N}} [- n, n]\).
Step 6:
\([Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) is valid for each \(n \in \mathbb{N}\), because \(Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}) \le (r_1 + r_2) / 2 \le Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})\).
\((r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2]\) is valid for each \(n \in \mathbb{N}\), because \(r_1 \lt (r_1 + r_2) / 2 \le Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2)\).
\(\cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\) is valid for each \(n \in \mathbb{N}\), because \(Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}) \le (r_1 + r_2) / 2 \lt r_2\).
Let us see that \((r_1, r_2) = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
Let \(r \in (r_1, r_2)\) be any.
\(r_1 \lt r\), and there is an \(N_1 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_1 \lt n\), \(r_1 + 2^{- n} \le r\), then, \(Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}) \le r_1 + 2^{- n} \le r\).
\(r \lt r_2\), and there is an \(N_2 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_2 \lt n\), \(r \le r_2 - 2^{- n}\), then, \(r \le r_2 - 2^{- n} \le Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})\).
For each \(n \in \mathbb{N}\) such that \(N_1, N_2 \lt n\), \(Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}) \le r \le Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})\), so, \(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
So, \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
So, \((r_1, r_2) \subseteq \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
Let \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) be any.
\(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) for an \(N \in \mathbb{N}\).
\(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] \subseteq (r_1, r_2)\), because \(r_1 \lt r_1 + 2^{- n}\) and \(r_1 \lt (r_1 + r_2) / 2\) and \(r_2 - 2^{- n} \lt r_2\) and \((r_1 + r_2) / 2 \lt r_2\).
So, \(\cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] \subseteq (r_1, r_2)\).
So, \((r_1, r_2) = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
Let \(r \in (r_1, r_2)\) be any.
\(r_1 \lt r\).
\(r \lt r_2\), and there is an \(n \in \mathbb{N}\) such that \(r \le r_2 - 2^{- n}\), then, \(r \le r_2 - 2^{- n} \le Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})\).
So, \(r \in (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
So, \(r \in \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
So, \((r_1, r_2) \subseteq \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
Let \(r \in \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) be any.
\(r \in (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) for an \(n \in \mathbb{N}\).
\(r \in (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] \subseteq (r_1, r_2)\), because \(r_2 - 2^{- n} \lt r_2\) and \((r_1 + r_2) / 2 \lt r_2\).
So, \(\cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] \subseteq (r_1, r_2)\).
So, \((r_1, r_2) = \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
Let \(r \in (r_1, r_2)\) be any.
\(r_1 \lt r\), and there is an \(n \in \mathbb{N}\) such that \(Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}) \le r_1 + 2^{- n} \le r\).
\(r \lt r_2\).
So, \(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
So, \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
So, \((r_1, r_2) \subseteq \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
Let \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\) be any.
\(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\) for an \(n \in \mathbb{N}\).
\(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2) \subseteq (r_1, r_2)\), because \(r_1 \lt r_1 + 2^{- n}\) and \(r_1 \lt (r_1 + r_2) / 2\).
So, \(\cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2) \subseteq (r_1, r_2)\).
So, \((r_1, r_2) = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
References
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description/proof of that half or both closed interval is intersection of sequence of open intervals
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any half or both closed interval is the intersection of the sequence of some 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:
\(I\): \(\in \{\text{ the half or both closed intervals }\}\), \(= (- \infty, r_2], (r_1, r_2], [r_1, \infty), [r_1, r_2), \text{ or } [r_1, r_2]\)
//
Statements:
\((- \infty, r_2] = \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\)
\(\land\)
\((r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\)
\(\land\)
\([r_1, \infty) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\)
\(\land\)
\([r_1, r_2) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\)
\(\land\)
\([r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n}) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2] = \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\)
//
2: Note
While \(\cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\) or \(\cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\) is not really "intersection of the sequence of some open intervals", they are included here, because they are sometimes used.
3: Proof
Whole Strategy: Step 1: see that \((- \infty, r_2] = \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\); Step 2: see that \((r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\); Step 3: see that \([r_1, \infty) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\); Step 4: see that \([r_1, r_2) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\); Step 5: see that \([r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n}) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2] = \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\).
Step 1:
Let us see that \((- \infty, r_2] = \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\).
Let \(r \in (- \infty, r_2]\) be any.
For each \(n \in \mathbb{N}\), \(r \in (- \infty, r_2 + 2^{- n})\).
So, \(r \in \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\).
So, \((- \infty, r_2] \subseteq \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\).
Let \(r \in \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\) be any.
\(- \infty \lt r\).
Let us suppose that \(r_2 \lt r\).
There would be an \(n \in \mathbb{N}\) such that \(r_2 + 2^{- n} \le r\), then, \(r \notin (- \infty, r_2 + 2^{- n})\), then, \(r \notin \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\), a contradiction.
So, \(r \le r_2\).
So, \(r \in (- \infty, r_2]\).
So, \(\cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n}) \subseteq (- \infty, r_2]\).
So, \((- \infty, r_2] = \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\).
Step 2:
Let us see that \((r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\).
Let \(r \in (r_1, r_2]\) be any.
For each \(n \in \mathbb{N}\), \(r \in (r_1, r_2 + 2^{- n})\).
So, \(r \in \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\).
So, \((r_1, r_2] \subseteq \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\).
Let \(r \in \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\) be any.
\(r_1 \lt r\).
Let us suppose that \(r_2 \lt r\).
There would be an \(n \in \mathbb{N}\) such that \(r_2 + 2^{- n} \le r\), then, \(r \notin (r_1, r_2 + 2^{- n})\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\), a contradiction.
So, \(r \le r_2\).
So, \(r \in (r_1, r_2]\).
So, \(\cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n}) \subseteq (r_1, r_2]\).
So, \((r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\).
Step 3:
Let us see that \([r_1, \infty) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\).
Let \(r \in [r_1, \infty)\) be any.
For each \(n \in \mathbb{N}\), \(r \in (r_1 - 2^{- n}, \infty)\).
So, \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\).
So, \([r_1, \infty) \subseteq \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\).
Let \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\) be any.
\(r \lt \infty\).
Let us suppose that \(r \lt r_1\).
There would be an \(n \in \mathbb{N}\) such that \(r \le r_1 - 2^{- n}\), then, \(r \notin (r_1 - 2^{- n}, \infty)\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\), a contradiction.
So, \(r_1 \le r\).
So, \(r \in [r_1, \infty)\).
So, \(\cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty) \subseteq [r_1, \infty)\).
So, \([r_1, \infty) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\).
Step 4:
Let us see that \([r_1, r_2) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\).
Let \(r \in [r_1, r_2)\) be any.
For each \(n \in \mathbb{N}\), \(r \in (r_1 - 2^{- n}, r_2)\).
So, \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\).
So, \([r_1, r_2) \subseteq \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\).
Let \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\) be any.
\(r \lt r_2\).
Let us suppose that \(r \lt r_1\).
There would be an \(n \in \mathbb{N}\) such that \(r \le r_1 - 2^{- n}\), then, \(r \notin (r_1 - 2^{- n}, r_2)\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\), a contradiction.
So, \(r_1 \le r\).
So, \(r \in [r_1, r_2)\).
So, \(\cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2) \subseteq [r_1, r_2)\).
So, \([r_1, r_2) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\).
Step 5:
Let us see that \([r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n}) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2] = \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\).
Let \(r \in [r_1, r_2]\) be any.
For each \(n \in \mathbb{N}\), \(r \in (r_1 - 2^{- n}, r_2 + 2^{- n})\).
So, \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\).
So, \([r_1, r_2] \subseteq \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\).
Let \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\) be any.
Let us suppose that \(r \lt r_1\).
There would be an \(n \in \mathbb{N}\) such that \(r \le r_1 - 2^{- n}\), then, \(r \notin (r_1 - 2^{- n}, r_2 + 2^{- n})\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\), a contradiction.
So, \(r_1 \le r\).
Let us suppose that \(r_2 \lt r\).
There would be an \(n \in \mathbb{N}\) such that \(r_2 + 2^{- n} \le r\), then, \(r \notin (r_1 - 2^{- n}, r_2 + 2^{- n})\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\), a contradiction.
So, \(r \le r_2\).
So, \(r \in [r_1, r_2]\).
So, \(\cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n}) \subseteq [r_1, r_2]\).
So, \([r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\).
Let \(r \in [r_1, r_2]\) be any.
For each \(n \in \mathbb{N}\), \(r \in (r_1 - 2^{- n}, r_2]\).
So, \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\).
So, \([r_1, r_2] \subseteq \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\).
Let \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\) be any.
Let us suppose that \(r \lt r_1\).
There would be an \(n \in \mathbb{N}\) such that \(r \le r_1 - 2^{- n}\), then, \(r \notin (r_1 - 2^{- n}, r_2]\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\), a contradiction.
So, \(r_1 \le r\).
\(r \le r_2\).
So, \(r \in [r_1, r_2]\).
So, \(\cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2] \subseteq [r_1, r_2]\).
So, \([r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\).
Let \(r \in [r_1, r_2]\) be any.
For each \(n \in \mathbb{N}\), \(r \in [r_1, r_2 + 2^{- n})\).
So, \(r \in \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\).
So, \([r_1, r_2] \subseteq \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\).
Let \(r \in \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\) be any.
\(r_1 \le r\).
Let us suppose that \(r_2 \lt r\).
There would be an \(n \in \mathbb{N}\) such that \(r_2 + 2^{- n} \le r\), then, \(r \notin [r_1, r_2 + 2^{- n})\), then, \(r \notin \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\), a contradiction.
So, \(r \le r_2\).
So, \(r \in [r_1, r_2]\).
So, \(\cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n}) \subseteq [r_1, r_2]\).
So, \([r_1, r_2] = \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\).
References
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