<The previous article in this series | The table of contents of this series |
description/proof of that for infinite sequence on partially-ordered set and element of set, if there is any large index whose value is equal to or larger than element, limit superior is equal to or larger than element
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 infinite sequence on any partially-ordered set and any element of the set, if the limit superior exists and there is any large index whose value is equal to or larger than the element, the limit superior is equal to or larger than the element.
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:
\(J\): \(\subseteq \mathbb{N}\), such that \(\vert J \vert = \infty\)
\(S\): \(\in \{\text{ the partially-ordered sets }\}\), with any partial ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq S\)
\(s'\): \(\in S\)
//
Statements:
\(\exists lim sup s \land \forall j \in J (\exists j' \in J \text{ such that } j \lt j' (s' \le s (j)))\)
\(\implies\)
\(s' \le lim sup s\)
//
2: Note
If \(J\) is finite, it is just a matter of \(s' \le s (J_{\vert J \vert})\) if and only if \(s' \le lim sup s\), because \(lim sup s = s (J_{\vert J \vert})\).
3: Proof
Whole Strategy: Step 1: see that \(s' \le lim sup s\).
Step 1:
\(lim sup s = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
For each \(m \in \mathbb{N} \setminus \{0\}\), there is an \(n \in \mathbb{N} \setminus \{0\}\) such that \(m \lt n\) and \(s' \le s (J_n)\), by the supposition.
So, for each \(m \in \mathbb{N} \setminus \{0\}\), \(s' \le s (J_n) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
So, \(s' \in Lb (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
So, \(s' \le Max (Lb (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})) = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = lim sup s\).
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 infinite sequence on partially-ordered set and element of set, if there is any large index whose value is equal to or smaller than element, limit inferior is equal to or smaller than element
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 infinite sequence on any partially-ordered set and any element of the set, if the limit inferior exists and there is any large index whose value is equal to or smaller than the element, the limit inferior is equal to or smaller than the element.
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:
\(J\): \(\subseteq \mathbb{N}\), such that \(\vert J \vert = \infty\)
\(S\): \(\in \{\text{ the partially-ordered sets }\}\), with any partial ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq S\)
\(s'\): \(\in S\)
//
Statements:
\(\exists lim inf s \land \forall j \in J (\exists j' \in J \text{ such that } j \lt j' (s (j) \le s'))\)
\(\implies\)
\(lim inf s \le s'\)
//
2: Note
If \(J\) is finite, it is just a matter of \(s (J_{\vert J \vert}) \le s'\) if and only if \(lim inf s \le s'\), because \(lim inf s = s (J_{\vert J \vert})\).
3: Proof
Whole Strategy: Step 1: see that \(lim inf s \le s'\).
Step 1:
\(lim inf s = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
For each \(m \in \mathbb{N} \setminus \{0\}\), there is an \(n \in \mathbb{N} \setminus \{0\}\) such that \(m \lt n\) and \(s (J_n) \le s'\), by the supposition.
So, for each \(m \in \mathbb{N} \setminus \{0\}\), \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_n) \le s'\).
So, \(s' \in Ub (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
So, \(lim inf s = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = Min (Ub (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}))) \le s'\).
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 for sequence on linearly-ordered set, limit superior is equal to or larger than infimum of range of sequence
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 sequence on any linearly-ordered set, if the limit superior and the infimum of the range of the sequence exist, the limit superior is equal to or larger than the infimum of the range of the sequence.
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:
\(J\): \(\subseteq \mathbb{N}\), such that \(J \neq \emptyset\)
\(S\): \(\in \{\text{ the linearly-ordered sets }\}\), with any linear ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq S\)
//
Statements:
\(\exists lim sup s \land \exists Inf (Ran (s))\)
\(\implies\)
\(Inf (Ran (s)) \le lim sup s\)
//
2: Note
There is no so simple relation between the existence of \(lim sup s\) and the existence of \(Inf (Ran (s))\).
For example, let \(J = \mathbb{N}\) and \(S = \mathbb{Q}\) with the canonical linear ordering and \(s\) be a sequence that starts from \(1\) and increasingly approaches \(\sqrt{2}\), then, \(lim sup s\) does not exist but \(Inf (Ran (s))\) exists as \(1\).
For example, let \(J = \mathbb{N}\) and \(S = \mathbb{Q}\) with the canonical linear ordering and \(s\) be such that the index-even subsequence is constantly \(2\) and the index-odd subsequence starts from \(2\) and decreasingly approaches \(\sqrt{2}\), then, \(lim sup s\) exists as \(2\) but \(Inf (Ran (s))\) does not exist.
3: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite and suppose otherwise thereafter; Step 2: see that \(Inf (Ran (s)) \le lim sup s\).
Step 1:
Let us suppose that \(\vert J \vert \in \mathbb{N} \setminus \{0\}\).
\(lim sup s = s (J_{\vert J \vert})\).
\(Inf (Ran (s)) = Max (Lb (Ran (s))) \le s (J_{\vert J \vert})\).
So, \(Inf (Ran (s)) \le lim sup s\).
Let us suppose otherwise, hereafter.
Step 2:
\(lim sup s = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
Let \(s' \in S\) be any such that \(lim sup s \lt s'\).
If there is no such \(s'\), it is OK.
There is a \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) such that \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt s'\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller.
For any \(n \in \mathbb{N} \setminus \{0\}\) such that \(m \le n\), \(s (J_n) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
But \(Inf (Ran (s)) \le s (J_n)\).
So, \(Inf (Ran (s)) \le s (J_n) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt s'\), so, \(Inf (Ran (s)) \lt s'\).
So, \(Inf (Ran (s)) \le lim sup s\), by the proposition that for any linearly-ordered set and any \(2\) elements, if each element that is larger than the 2nd element is larger than the 1st element, the 1st element is equal to or smaller than the 2nd element.
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 for sequence on linearly-ordered set, limit inferior is equal to or smaller than supremum of range of sequence
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 sequence on any linearly-ordered set, if the limit inferior and the supremum of the range of the sequence exist, the limit inferior is equal to or smaller than the supremum of the range of the sequence.
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:
\(J\): \(\subseteq \mathbb{N}\), such that \(J \neq \emptyset\)
\(S\): \(\in \{\text{ the linearly-ordered sets }\}\), with any linear ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq S\)
//
Statements:
\(\exists lim inf s \land \exists Sup (Ran (s))\)
\(\implies\)
\(lim inf s \le Sup (Ran (s))\)
//
2: Note
There is no so simple relation between the existence of \(lim inf s\) and the existence of \(Sup (Ran (s))\).
For example, let \(J = \mathbb{N}\) and \(S = \mathbb{Q}\) with the canonical linear ordering and \(s\) be a sequence that starts from \(2\) and decreasingly approaches \(\sqrt{2}\), then, \(lim inf s\) does not exist but \(Sup (Ran (s))\) exists as \(2\).
For example, let \(J = \mathbb{N}\) and \(S = \mathbb{Q}\) with the canonical linear ordering and \(s\) be such that the index-even subsequence is constantly \(1\) and the index-odd subsequence starts from \(1\) and increasingly approaches \(\sqrt{2}\), then, \(lim inf s\) exists as \(1\) but \(Sup (Ran (s))\) does not exist.
3: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite and suppose otherwise thereafter; Step 2: see that \(lim inf s \le Sup (Ran (s))\).
Step 1:
Let us suppose that \(\vert J \vert \in \mathbb{N} \setminus \{0\}\).
\(lim inf s = s (J_{\vert J \vert})\).
\(s (J_{\vert J \vert}) \le Min (Ub (Ran (s))) = Sup (Ran (s))\).
So, \(lim inf s \le Sup (Ran (s))\).
Let us suppose otherwise, hereafter.
Step 2:
\(lim inf s = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
Let \(s' \in S\) be any such that \(s' \lt lim inf s\).
If there is no such \(s'\), it is OK.
There is an \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) such that \(s' \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger.
For any \(n \in \mathbb{N} \setminus \{0\}\) such that \(m \le n\), \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_n)\).
But \(s (J_n) \le Sup (Ran (s))\).
So, \(s' \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_n) \le Sup (Ran (s))\), so, \(s' \lt Sup (Ran (s))\).
So, \(lim inf s \le Sup (Ran (s))\), by the proposition that for any linearly-ordered set and any \(2\) elements, if each element that is smaller than the 1st element is smaller than the 2nd element, the 1st element is equal to or smaller than the 2nd element.
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 for map into measurable space and \(\sigma\)-algebra induced on domain, for measurable subset on domain, for point of codomain s.t. \(1\)-point subset is measurable, point preimage is contained in measurable subset or is disjoint from measurable subset
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 for any map into any measurable space and the \(\sigma\)-algebra induced on the domain, for each measurable subset on the domain, for each point of the codomain such that the \(1\)-point subset is measurable, the point preimage is contained in the measurable subset or is disjoint from the measurable subset.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f\): \(: S_1 \to M_2\)
\(\sigma (f)\): \(= \text{ the } \sigma \text{ -algebra induced on } S_1 \text{ of } f\)
//
Statements:
\(\forall a \in \sigma (f) (\forall m_2 \in M_2 \text{ such that } \{m_2\} \in A_2 (f^{-1} (m_2) \subseteq a \lor f^{-1} (m_2) \cap a = \emptyset))\)
//
2: Note
In fact, this is a special case of the proposition that for the \(\sigma\)-algebra induced on the domain of any maps into any measurable space, for each measurable subset, each intersection of point preimages is contained in the measurable subset or is disjoint from the measurable subset, in which \(\{m_2\} \in A_2\) is not required.
When \((M_2, A_2) = (\mathbb{R}^d, B (\mathbb{R}^d))\), the Euclidean measurable space, \(\{m_2\} \in A_2\) is guaranteed for each \(m_2 \in M_2\).
3: Proof
Whole Strategy: Step 1: see that \(f^{-1} (m_2) \cap a = f^{-1} (a_2)\) for an \(a_2 \in A_2\), and see that \(f^{-1} (m_2) \subseteq a \lor f^{-1} (m_2) \cap a = \emptyset\).
Step 1:
As \(\{m_2\} \in A_2\), \(f^{-1} (m_2) \in \sigma (f)\), by the proposition that for any map from any set into any measurable space, the smallest \(\sigma\)-algebra of the domain that makes the map measurable is the set of the preimages of the measurable subsets of the codomain, so, \(f^{-1} (m_2) \cap a \in \sigma (f)\).
So, \(f^{-1} (m_2) \cap a = f^{-1} (a_2)\) for an \(a_2 \in A_2\), by the definition of \(\sigma\)-algebra induced on domain of map into measurable space.
So, \(f^{-1} (m_2) \subseteq a\) or \(f^{-1} (m_2) \cap a = \emptyset\), by the proposition that for any map, any subset of the domain, and any point of the codomain, if the intersection of the preimage of the point and the subset is a preimage, the preimage of the point is contained in the subset or is disjoint from the subset.
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 for map, subset of domain, and point of codomain, if intersection of preimage of point and subset is preimage, preimage of point is contained in subset or is disjoint from subset
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 map, any subset of the domain, and any point of the codomain, if the intersection of the preimage of the point and the subset is a preimage, the preimage of the point is contained in the subset or is disjoint from the subset.
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:
\(S'_1\): \(\in \{\text{ the sets }\}\)
\(S'_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(: S'_1 \to S'_2\)
\(S_1\): \(\subseteq S'_1\)
\(s'_2\): \(\in S'_2\)
//
Statements:
\(\exists S_2 \subseteq S'_2 (f^{-1} (s'_2) \cap S_1 = f^{-1} (S_2))\)
\(\implies\)
\(f^{-1} (s'_2) \subseteq S_1 \lor f^{-1} (s'_2) \cap S_1 = \emptyset\)
//
2: Note
\(s'_2\) cannot be replaced by a subset, \({S'_2}^` \subseteq S'_2\), for this proposition: for each \(s'_2 \in {S'_2}^`\), this proposition holds, but for some \(s'_2, \widetilde{s'_2} \in {S'_2}^`\) such that \(s'_2 \neq \widetilde{s'_2}\), it may be that \(f^{-1} (s'_2) \subseteq S_1\) and \(f^{-1} (\widetilde{s'_2}) \cap S_1 = \emptyset\), for example, then, \(f^{-1} ({S'_2}^`) \subseteq S_1 \lor f^{-1} ({S'_2}^`) \cap S_1 = \emptyset\) does not hold.
For example, let \(f: \{0, 1\} \to \{0, 1\} = id\), \(S_1 = \{1\}\), and \(S_2 = \{0, 1\}\), then, \(f^{-1} (\{0, 1\}) = \{0, 1\}\) and \(f^{-1} (\{0, 1\}) \cap \{1\} = \{1\} = f^{-1} (\{1\})\), but not \(\{0, 1\} \subseteq \{1\}\) nor \(\{0, 1\} \cap \{1\} = \emptyset\) holds: for \(s'_2 = 0\), \(f^{-1} (0) = \{0\}\) and \(f^{-1} (0) \cap \{1\} = \emptyset = f^{-1} (\emptyset)\), and \(f^{-1} (0) \cap \{1\} = \emptyset\) holds; for \(s'_2 = 1\), \(f^{-1} (1) = \{1\}\) and \(f^{-1} (1) \cap \{1\} = \{1\} = f^{-1} (\{1\})\), and \(f^{-1} (1) \subseteq \{1\}\) holds.
3: Proof
Whole Strategy: Step 1: suppose that \(f^{-1} (s'_2) \cap S_1 \neq \emptyset\), and see that \(f^{-1} (s'_2) \subseteq S_1\).
Step 1:
Let us suppose that \(f^{-1} (s'_2) \cap S_1 \neq \emptyset\).
Let us suppose that \(s'_2 \notin S_2\).
\(f^{-1} (s'_2) \cap f^{-1} (S_2) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint.
Let \(s_1 \in f^{-1} (s'_2) \cap S_1\) be any.
\(s_1 \in f^{-1} (s'_2) \cap S_1 = f^{-1} (S_2)\), a contradiction against \(f^{-1} (s'_2) \cap f^{-1} (S_2) = \emptyset\).
So, \(s'_2 \in S_2\).
So, \(f^{-1} (s'_2) \subseteq f^{-1} (S_2) = f^{-1} (s'_2) \cap S_1 \subseteq S_1\).
So, \(f^{-1} (s'_2) \cap S_1 = \emptyset\) or \(f^{-1} (s'_2) \subseteq S_1\).
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 for \(\sigma\)-algebra induced on domain of maps into measurable space, for measurable subset, intersection of point preimages is contained in measurable subset or is disjoint from measurable subset
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 for the \(\sigma\)-algebra induced on the domain of any maps into any measurable space, for each measurable subset, each intersection of point preimages is contained in the measurable subset or is disjoint from the measurable subset.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(J\): \(\in \{\text{ the index sets }\}\), such that \(J \neq \emptyset\)
\(\{f_j: S_1 \to M_2 \vert j \in J\}\):
\(\sigma (\{f_j \vert j \in J\})\): \(= \text{ the } \sigma \text{ -algebra induced on } S_1 \text{ by } \{f_j \vert j \in J\}\)
//
Statements:
\(\forall a \in \sigma (\{f_j \vert j \in J\}) (\forall m \in \times_{j \in J} M_2 (\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a \lor \cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset))\)
//
2: Note
\(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a \lor \cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset\) is really a natural thing that \(\sigma (\{f_j \vert j \in J\})\) does not need to contain any measurable subset that divides \(\cap_{j \in J} {f_j}^{-1} (m_j)\), which is natural because each preimage under \(f_j\) contains the whole of \({f_j}^{-1} (m_j)\) or is disjoint from \({f_j}^{-1} (m_j)\), and an \(a\) is necessary because it is the result of some set operations on some preimages under \(f_j\) s, but it looks unlikely that the result divides \(\cap_{j \in J} {f_j}^{-1} (m_j)\) (Proof proves that it is really the case).
3: Proof
Whole Strategy: Step 1: define \(A := \{a \in \sigma (\{f_j \vert j \in J\}) \vert \forall m \in \times_{j \in J} M_2 (\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a \lor \cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset)\}\) and see that \(A\) is a \(\sigma\)-algebra that makes all the \(f_j\) s measurable.
Step 1:
Let us define \(A := \{a \in \sigma (\{f_j \vert j \in J\}) \vert \forall m \in \times_{j \in J} M_2 (\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a \lor \cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset)\}\).
In other words, \(A\) has eliminated from \(\sigma (\{f_j \vert j \in J\})\) the elements that divide a \(\cap_{j \in J} {f_j}^{-1} (m_j)\).
Let us see that \(A\) is a \(\sigma\)-algebra.
1) \(S_1 \in A\): while \(S_1 \in \sigma (\{f_j \vert j \in J\})\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq S_1\).
2) \(\forall a \in A (S_1 \setminus a \in A)\): as \(a \in \sigma (\{f_j \vert j \in J\})\), \(S_1 \setminus a \in \sigma (\{f_j \vert j \in J\})\), and when \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \cap (S_1 \setminus a) = \emptyset\), and when \(\cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq S_1 \setminus a\), so, \(S_1 \setminus a \in A\).
3) \(\forall s: \mathbb{N} \to A (\cup_{n \in \mathbb{N}} s (n) \in A)\): as \(s\) is into \(\sigma (\{f_j \vert j \in J\})\), \(\cup_{n \in \mathbb{N}} s (n) \in \sigma (\{f_j \vert j \in J\})\), and when \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq s (n)\) for an \(n \in \mathbb{N}\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq \cup_{n \in \mathbb{N}} s (n)\), and when \(\cap_{j \in J} {f_j}^{-1} (m_j) \cap s (n) = \emptyset\) for each \(n \in \mathbb{N}\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \cap \cup_{n \in \mathbb{N}} s (n) = \emptyset\).
So, \(A\) is a \(\sigma\)-algebra.
\(A\) makes each \(f_j\) measurable, because for each \(a_2 \in A_2\), \({f_j}^{-1} (a_2) \in \sigma (\{f_j \vert j \in J\})\), because \(\sigma (\{f_j \vert j \in J\})\) makes all the \(f_j\) s measurable, but when \(m_j \in a_2\), \({f_j}^{-1} (m_j) \subseteq {f_j}^{-1} (a_2)\), so, \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq {f_j}^{-1} (a_2)\), and when \(m_j \notin a_2\), \({f_j}^{-1} (m_j) \cap {f_j}^{-1} (a_2) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint, so, \(\cap_{j \in J} {f_j}^{-1} (m_j) \cap {f_j}^{-1} (a_2) = \emptyset\), so, \({f_j}^{-1} (a_2) \in A\) anyway.
So, \(A\) is a \(\sigma\)-algebra that makes all the \(f_j\) s measurable.
So, \(\sigma (\{f_j \vert j \in J\}) \subseteq A\).
But as \(A \subseteq \sigma (\{f_j \vert j \in J\})\), \(A = \sigma (\{f_j \vert j \in J\})\), which means that \(A\) did not really eliminate anything.
So, for each \(a \in \sigma (\{f_j \vert j \in J\})\), for each \(m \in \times_{j \in J} M_2\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a \lor \cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset\).
References
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description/proof of that for linearly-ordered set and \(2\) elements, if each element that is larger than 2nd element is larger than 1st element, 1st element is equal to or smaller than 2nd element
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 linearly-ordered set and any \(2\) elements, if each element that is larger than the 2nd element is larger than the 1st element, the 1st element is equal to or smaller than the 2nd element.
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:
\(S\): \(\in \{\text{ the linearly-ordered sets }\}\), with any linear ordering, \(\lt\)
\(s_1\): \(\in S\)
\(s_2\): \(\in S\)
//
Statements:
\(\forall s \in S \text{ such that } s_2 \lt s (s_1 \lt s)\)
\(\implies\)
\(s_1 \le s_2\)
//
2: Note
Compare with the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number, which requires only \(r_1 \le r_2 + \epsilon\), while this proposition requires that \(s_1 \lt s\): \(s_1 \le s\) is not enough.
For example, let \(S = \mathbb{Z}\) with the canonical ordering, \(s_1 = 1\), and \(s_2 = 0\), then, for each \(s_2 \lt s\), \(s_1 \le s\), but "\(s_1 \le s_2\)" does not hold.
3: Proof
Whole Strategy: Step 1: suppose that \(s_2 \lt s_1\), and find a contradiction.
Step 1:
Let us suppose that \(s_2 \lt s_1\).
\(s_1 \lt s_1\), by the supposition, a contradiction.
So, \(s_1 \le s_2\).
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 for linearly-ordered set and \(2\) elements, if each element that is smaller than 1st element is smaller than 2nd element, 1st element is equal to or smaller than 2nd element
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 linearly-ordered set and any \(2\) elements, if each element that is smaller than the 1st element is smaller than the 2nd element, the 1st element is equal to or smaller than the 2nd element.
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:
\(S\): \(\in \{\text{ the linearly-ordered sets }\}\), with any linear ordering, \(\lt\)
\(s_1\): \(\in S\)
\(s_2\): \(\in S\)
//
Statements:
\(\forall s \in S \text{ such that } s \lt s_1 (s \lt s_2)\)
\(\implies\)
\(s_1 \le s_2\)
//
2: Note
Compare with the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number, which requires only \(r_1 \le r_2 + \epsilon\), while this proposition requires that \(s \lt s_2\): \(s \le s_2\) is not enough.
For example, let \(S = \mathbb{Z}\) with the canonical ordering, \(s_1 = 1\), and \(s_2 = 0\), then, for each \(s \lt s_1\), \(s \le s_2\), but "\(s_1 \le s_2\)" does not hold.
3: Proof
Whole Strategy: Step 1: suppose that \(s_2 \lt s_1\), and find a contradiction.
Step 1:
Let us suppose that \(s_2 \lt s_1\).
\(s_2 \lt s_2\), by the supposition, a contradiction.
So, \(s_1 \le s_2\).
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 for sequence on \(1\)-dimensional Euclidean metric space with canonical ordering, if limit superior exists, there is subsequence that converges to limit superior
Topics
About:
metric space
The table of contents of this article
Starting Context
-
The reader knows a definition of limit superior of sequence on partially-ordered set.
-
The reader knows a definition of subsequence of sequence.
-
The reader knows a definition of convergence of sequence on metric space.
-
The reader admits the proposition that for any linearly-ordered set and any subset, any element of the set is the infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller.
-
The reader admits the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger.
-
The reader admits 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.
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 with the canonical ordering, if the limit superior exists, there is a subsequence that converges to the limit superior.
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:
\(J\): \(\subseteq \mathbb{N}\), such that \(J \neq \emptyset\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\) with the canonical ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\)
//
Statements:
\(\exists lim sup s\)
\(\implies\)
\(s^` \in \{\text{ the subsequences of } s\} (lim s^` = lim sup s)\)
//
2: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: choose \(l_1, l_2, ... \in \mathbb{N} \setminus \{0\}\) inductively such that \(l_1 \lt l_2 \lt ...\) and \(lim sup s - (1 / 2)^n \lt s (J_{l_n}) \lt lim sup s + (1 / 2)^n\) for each \(n \in \mathbb{N} \setminus \{0\}\); Step 3: take \(J^` = \mathbb{N} \setminus \{0\}\) and \(f: J^` \to J, n \mapsto J_{l_n}\).
Step 1:
Let us suppose that \(\vert J \vert = n \in \mathbb{N} \setminus \{0\}\).
\(lim sup s = s (J_n)\) inevitably exist, and \(lim s = s (J_n)\) exists, and \(lim s = lim sup s\).
So, let \(s^` := s = s \circ f\) with \(f: J^` \to J = id\), then, \(lim s^` = lim s = lim sup s\).
Let us suppose otherwise, hereafter.
Step 2:
Let us choose \(l_1, l_2, ... \in \mathbb{N} \setminus \{0\}\) inductively as this.
Let \(n = 1\).
As \(lim sup s = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), there is an \(m \in \mathbb{N} \setminus \{0\}\) such that \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt lim sup s + (1 / 2)^n\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller: \(\mathbb{R}\) is linearly-ordered.
\(lim sup s \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
There is an \(l_n \in \mathbb{N} \setminus \{0\}\) such that \(m \le l_n\) and \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) - (1 / 2)^n \lt s (J_{l_n})\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger.
\(s (J_{l_n}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
So, \(lim sup s - (1 / 2)^n \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) - (1 / 2)^n \lt s (J_{l_n}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt lim sup s + (1 / 2)^n\).
So, \(lim sup s - (1 / 2)^n \lt s (J_{l_n}) \lt lim sup s + (1 / 2)^n\).
Let us suppose that \(l_1, ..., l_{n' - 1}\) have been chosen such that \(l_1 \lt ... \lt l_{n' - 1}\) and \(lim sup s - (1 / 2)^n \lt s (J_{l_n}) \lt lim sup s + (1 / 2)^n\) for each \(n \in \{1, ..., n' - 1\}\).
There is an \(m \in \mathbb{N} \setminus \{0\}\) such that \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt lim sup s + (1 / 2)^{n'}\), as before, but \(m\) can be chosen such that \(l_{n' - 1} \lt m\), because if \(m \le l_{n' - 1}\), take any \(m' \in \mathbb{N} \setminus \{0\}\) such that \(l_{n' - 1} \lt m'\), then, \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m' \le n\}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt lim sup s + (1 / 2)^{n'}\), by 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: \(\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m' \le n\} \subseteq \{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}\).
\(lim sup s \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
There is an \(l_{n'} \in \mathbb{N} \setminus \{0\}\) such that \(m \le l_{n'}\) and \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) - (1 / 2)^{n'} \lt s (J_{l_{n'}})\), as before, but \(l_{n' - 1} \lt m \le l_{n'}\).
\(s (J_{l_{n'}}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
So, \(lim sup s - (1 / 2)^{n'} \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) - (1 / 2)^{n'} \lt s (J_{l_{n'}}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt lim sup s + (1 / 2)^{n'}\).
So, \(lim sup s - (1 / 2)^{n'} \lt s (J_{l_{n'}}) \lt lim sup s + (1 / 2)^{n'}\).
So, we have chosen \(l_1, l_2, ... \in \mathbb{N} \setminus \{0\}\) inductively such that \(l_1 \lt l_2 \lt ...\) and \(lim sup s - (1 / 2)^n \lt s (J_{l_n}) \lt lim sup s + (1 / 2)^n\) for each \(n \in \mathbb{N} \setminus \{0\}\).
Step 3:
Let us take \(J^` = \mathbb{N} \setminus \{0\}\).
Let us take \(f: J^` \to J, n \mapsto J_{l_n}\).
Then, \(s^` = s \circ f: J^` \to \mathbb{R}\) is a subsequence of \(s\), because \(\forall j^`_1, j^`_2 \in J^` \text{ such that } j^`_1 \lt j^`_2 (f (j^`_1) \lt f (j^`_2)) \land \forall j \in J (\exists j^` \in J^` (j \le f (j^`)))\): \(j = J_m\) and as \(l_1 \lt l_2 \lt ...\), \(m \le l_n\) for an \(n\), and \(j = J_m \le J_{l_n} = f (n)\).
\(lim s^` = lim sup s\), because for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N} \setminus \{0\}\) such that \((1 / 2)^N \lt \epsilon\), and for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \((1 / 2)^n \lt (1 / 2)^N \lt \epsilon\), and \(lim sup s - \epsilon \lt lim sup s - (1 / 2)^n \lt s^` (n) = s \circ f (n) = s (J_{l_n}) \lt lim sup s + (1 / 2)^n \lt lim sup s + \epsilon\), so, \(\vert s^` (n) - lim sup s \vert \lt \epsilon\).
References
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