<The previous article in this series | The table of contents of this series |
description/proof of that for partially-ordered set and nonempty subset, if infimum and supremum of subset exist, infimum is equal to or smaller than supremum
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 partially-ordered set and any nonempty subset, if the infimum and the supremum of the subset exist, the infimum is equal to or smaller than the supremum.
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:
\(S'\): \(\in \{\text{ the partially-ordered sets }\}\)
\(S\): \(\subseteq S'\), such that \(S \neq \emptyset\)
//
Statements:
\(\exists Inf (S) \land \exists Sup (S)\)
\(\implies\)
\(Inf (S) \le Sup (S)\)
//
2: Note
When \(S = \emptyset\) (although we do not particularly expect to have to deal with that case), if \(Inf (S)\) and \(Sup (S)\) exist, \(Sup (S) \le Inf (S)\), because \(Lb (S) = Ub (S) = S'\) (refer to Note for the definition of set of lower bounds of subset of partially-ordered set and Note for the definition of set of upper bounds of subset of partially-ordered set), and \(Sup (S) = Min (S') \le Max (S') = Inf (S)\).
3: Proof
Whole Strategy: Step 1: take any \(s \in S\), and see that \(Inf (S) \le s \le Sup (S)\).
Step 1:
Let \(s \in S\) be any, which exists, because \(S \neq \emptyset\).
\(Inf (S) \le s\), because \(Inf (S) \in Lb (S)\).
\(s \le Sup (S)\), because \(Sup (S) \in Ub (S)\).
So, \(Inf (S) \le Sup (S)\).
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 non-decreasing and non-increasing sequences on real numbers set s.t. 1st sequence is equal to or smaller than 2nd sequence, each element of 1st sequence is equal to or smaller than any element of 2nd sequence, and supremum of 1st sequence is equal to or smaller than infimum of 2nd 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 non-decreasing sequence and any non-increasing sequence on the real numbers set with any same domain such that the 1st sequence is equal to or smaller than the 2nd sequence, each element of the 1st sequence is equal to or smaller than any element of the 2nd sequence, and the supremum of the 1st sequence is equal to or smaller than the infimum of 2nd 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 \(J \neq \emptyset\)
\(\mathbb{R}\): with the canonical ordering, \(\lt\)
\(s\): \(\in \{\text{ the non-decreasing sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\)
\(s'\): \(\in \{\text{ the non-increasing sequences }\}\), such that \(Dom (s') = J\) and \(Ran (s') \subseteq \mathbb{R}\)
//
Statements:
\(\forall j \in J (s (j) \le s' (j))\)
\(\implies\)
(
\(\forall j, j' \in J (s (j) \le s' (j'))\)
\(\land\)
\(Sup (Ran (s)) \le Inf (Ran (s'))\)
)
//
2: Proof
Whole Strategy: Step 1: see that \(s (j) \le s' (j')\); Step 2: see that for each \(\epsilon\), \(Sup (Ran (s)) - \epsilon / 2 \lt s (j)\) and \(s' (j') \lt Inf (Ran (s')) + \epsilon / 2\).
Step 1:
Let \(j, j' \in J\) be any.
\(j \le j'\) or \(j' \lt j\).
When \(j \le j'\), \(s (j) \le s (j')\), because \(s\) is non-decreasing, \(\le s' (j')\), so, \(s (j) \le s' (j')\).
When \(j' \lt j\), \(s' (j) \le s' (j')\), because \(s'\) is non-increasing, so, \(s (j) \le s' (j) \le s' (j')\).
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is a \(j \in J\) such that \(Sup (Ran (s)) - \epsilon / 2 \lt s (j)\), 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.
There is a \(j' \in J\) such that \(s' (j) \lt Sup (Ran (s')) + \epsilon / 2\), 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.
\(Sup (Ran (s)) - \epsilon / 2 \lt s (j) \le s' (j) \lt Sup (Ran (s')) + \epsilon / 2\), by Step 1.
So, \(Sup (Ran (s)) \lt Sup (Ran (s')) + \epsilon\), so, \(Sup (Ran (s)) \le Sup (Ran (s')) + \epsilon\).
So, \(Sup (Ran (s)) \le Sup (Ran (s'))\), by 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.
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 \(2\) sequences on partially-ordered ring with same domain, limit superior of sum of sequences is not necessarily sum of limits superior of sequences, and limit inferior of sum of sequences is not necessarily sum of limits inferior of sequences
Topics
About:
ring
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 some \(2\) sequences on a partially-ordered ring with a same domain, the limit superior of the sum of the sequences is not necessarily the sum of the limits superior of the sequences, and the limit infimum of the sum of the sequences is not necessarily the sum of the limits inferior of the sequences.
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 \(J \neq \emptyset\)
\(R\): \(\in \{\text{ the partially-ordered rings }\}\) with any partial ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq R\)
\(s'\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s') = J\) and \(Ran (s') \subseteq R\)
\(s + s'\): \(: J \to R, j \mapsto s (j) + s' (j)\)
//
Statements:
Not necessarily "\(lim sup (s + s') = (lim sup s) + (lim sup s')\)"
\(\land\)
Not necessarily "\(lim inf (s + s') = (lim inf s) + (lim inf s')\)"
//
2: Note
Compare with the proposition that for any sequence on any partially-ordered ring, if the limit superior of the sequence exists, the limit superior of (the sequence plus any element) exists and equals (the limit superior of the sequence) plus the element, and if the limit inferior of the sequence exists, the limit inferior of (the sequence plus any element) exists and equals (the limit inferior of the sequence) plus the element.
3: Proof
Whole Strategy: Step 1: see an example that \(lim sup (s + s') \neq (lim sup s) + (lim sup s')\); Step 2: see an example that \(lim inf (s + s') \neq (lim inf s) + (lim inf s')\).
Step 1:
Let us see an example that \(lim sup (s + s') \neq (lim sup s) + (lim sup s')\).
Let \(J = \mathbb{N}\), \(R = \mathbb{R}\), and \(s: j \mapsto 1 \text{ when } j \text{ is even }; \mapsto - 1 \text{ when } j \text{ is odd }\) and \(s': j \mapsto - 1 \text{ when } j \text{ is even }; \mapsto 1 \text{ when } j \text{ is odd }\).
\(lim sup s = 1\) and \(lim sup s' = 1\).
But as \(s + s' = 0\), \(lim sup (s + s') = 0\).
So, \(lim sup (s + s') = 0 \neq 2 = (lim sup s) + (lim sup s')\).
Step 2:
Let us see an example that \(lim inf (s + s') \neq (lim inf s) + (lim inf s')\).
Let \(J = \mathbb{N}\), \(R = \mathbb{R}\), and \(s: j \mapsto 1 \text{ when } j \text{ is even }; \mapsto - 1 \text{ when } j \text{ is odd }\) and \(s': j \mapsto - 1 \text{ when } j \text{ is even }; \mapsto 1 \text{ when } j \text{ is odd }\).
\(lim inf s = - 1\) and \(lim inf s' = - 1\).
But as \(s + s' = 0\), \(lim inf (s + s') = 0\).
So, \(lim inf (s + s') = 0 \neq - 2 = (lim inf s) + (lim inf s')\).
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 partially-ordered ring, if limit superior exists, limit superior of (sequence plus element) exists and equals (limit superior of sequence) plus element, and if limit inferior exists, limit inferior of (sequence plus element) exists and equals (limit inferior of sequence) plus element
Topics
About:
ring
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 partially-ordered ring, if the limit superior of the sequence exists, the limit superior of (the sequence plus any element) exists and equals (the limit superior of the sequence) plus the element, and if the limit inferior of the sequence exists, the limit inferior of (the sequence plus any element) exists and equals (the limit inferior of the sequence) plus 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 \(J \neq \emptyset\)
\(R\): \(\in \{\text{ the partially-ordered rings }\}\) with any partial ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq S\)
\(r\): \(\in R\)
//
Statements:
(
\(\exists lim sup s\)
\(\implies\)
\(\exists lim sup (s + r) \land lim sup (s + r) = (lim sup s) + r\)
)
\(\land\)
(
\(\exists lim inf s\)
\(\implies\)
\(\exists lim inf (s + r) \land lim inf (s + r) = (lim inf s) + r\)
)
//
2: Note
For some \(2\) sequences, \(s, s': J \to R\), "\(lim sup (s + s') = (lim sup s) + (lim sup s')\)" or "\(lim inf (s + s') = (lim inf s) + (lim inf s')\)" does not hold in general, by the proposition that for some \(2\) sequences on a partially-ordered ring with a same domain, the limit superior of the sum of the sequences is not necessarily the sum of the limits superior of the sequences, and the limit inferior of the sum of the sequences is not necessarily the sum of the limits inferior of the sequences. So, it is crucial for this proposition that \(r\) is a constant element instead of a sequence.
3: Proof
Whole Strategy: apply the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element; Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: suppose that \(lim sup s\) exists; Step 3: see that \(lim sup (s + r) = (lim sup s) + r\); Step 4: suppose that \(lim inf s\) exists; Step 5: see that \(lim inf (s + r) = (lim inf s) + r\).
Step 1:
Let us suppose that \(\vert J \vert = n \in \mathbb{N} \setminus \{0\}\).
\(lim sup s\) inevitably exists and equals \(s (J_n)\).
\(lim sup (s + r)\) inevitably exists and equals \(s (J_n) + r\).
So, \(lim sup (s + r) = s (J_n) + r = (lim sup s) + r\).
\(lim inf s\) inevitably exists and equals \(s (J_n)\).
\(lim inf (s + r)\) inevitably exists and equals \(s (J_n) + r\).
So, \(lim inf (s + r) = s (J_n) + r = (lim inf s) + r\).
Let us suppose otherwise, hereafter.
Step 2:
Let us suppose that \(lim sup s\) exists.
Step 3:
\(lim sup s = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), and those supremums and the infimum exist.
\(Sup (\{s (J_n) + r \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) exists and \(= Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r\), by the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element.
\(Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r \vert m \in \mathbb{N} \setminus \{0\}\})\) exists and \(= Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) + r\), by the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element.
So, \(lim sup (s + r) = Inf (\{Sup (\{s (J_n) + r \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r \vert m \in \mathbb{N} \setminus \{0\}\}) = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) + r = (lim inf s) + r\), which exists.
Step 4:
Let us suppose that \(lim inf s\) exists.
Step 5:
\(lim inf s = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), and those infimums and the supremum exist.
\(Inf (\{s (J_n) + r \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) exists and \(= Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r\), by the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element.
\(Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r \vert m \in \mathbb{N} \setminus \{0\}\})\) exists and \(= Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) + r\), by the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element.
So, \(lim inf (s + r) = Sup (\{Inf (\{s (J_n) + r \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r \vert m \in \mathbb{N} \setminus \{0\}\}) = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) + r = (lim inf s) + r\), which exists.
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 real numbers set with canonical ordering, if limit superior of minus sequence exists, it is minus limit inferior of sequence, and if limit inferior of minus sequence exists, it is minus limit superior 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 the real numbers set with the canonical ordering, if the limit superior of minus the sequence exists, it is minus the limit inferior of the sequence, and if the limit inferior of minus the sequence exists, it is minus the limit superior 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:
\(\mathbb{R}\): \(= \text{ the real numbers set }\) with the canonical ordering
\(J\): \(\subseteq \mathbb{N}\), such that \(J \neq \emptyset\)
\(s\): \(: J \to \mathbb{R}\)
\(- s\): \(: J \to \mathbb{R}, j \mapsto - s(j)\)
//
Statements:
(
\(\exists lim sup - s\)
\(\implies\)
\(\exists lim inf s \land lim sup - s = - lim inf s\)
)
\(\land\)
(
\(\exists lim inf - s\)
\(\implies\)
\(\exists lim sup s \land lim inf - s = - lim sup s\)
)
//
2: Note
"\(lim sup - s = - lim sup s\)" or "\(lim inf - s = - lim inf s\)" does not hold in general.
For example, for \(s = (1, -1, 1, -1, ...)\), \(lim sup - s = 1\) while \(lim sup s = 1\) and \(lim inf - s = - 1\) while \(lim inf s = - 1\).
3: Proof
Whole Strategy: apply the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset; Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: suppose that \(lim sup - s\) exists; Step 3: see that \(lim sup - s = - lim inf s\); Step 4: suppose that \(lim inf - s\) exists; Step 5: see that \(lim inf - s = - lim sup s\).
Step 1:
Let us suppose that \(\vert J \vert = n \in \mathbb{N} \setminus \{0\}\).
\(lim sup - s\) inevitably exists as \(= - s (J_n)\).
\(- lim inf s\) inevitably exists as \(= - s (J_n)\).
So, \(lim sup - s = - lim inf s\).
\(lim inf - s\) inevitably exists as \(= - s (J_n)\).
\(- lim sup s\) inevitably exists as \(= - s (J_n)\).
So, \(lim inf - s = - lim sup s\).
Let us suppose otherwise hereafter.
Step 2:
Let us suppose that \(lim sup - s\) exists.
Step 3:
\(lim sup - s = Inf (\{Sup (\{- s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
\(= Inf (\{- Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), by the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset, \(= - Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), by the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset, \(= - lim inf s\).
Step 4:
Let us suppose that \(lim inf - s\) exists.
Step 5:
\(lim inf - s = Sup (\{Inf (\{- s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = Sup (\{- Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), by the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset, \(= - Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), by the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset, \(= - lim sup s\).
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 real numbers set with canonical ordering and subset, if supremum of minus subset exists, it is minus infimum of subset, and if infimum of minus subset exists, it is minus supremum of 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 the real numbers set with the canonical ordering and any subset, if the supremum of minus the subset exists, it is minus the infimum of the subset, and if the infimum of minus the subset exists, it is minus the supremum of 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:
\(\mathbb{R}\): \(= \text{ the real numbers set }\) with the canonical ordering
\(S\): \(\subseteq \mathbb{R}\)
\(- S\): \(= \{- s \in \mathbb{R} \vert s \in S\}\)
//
Statements:
(
\(\exists Sup (- S)\)
\(\implies\)
\(\exists Inf (S) \land Sup (- S) = - Inf (S)\)
)
\(\land\)
(
\(\exists Inf (- S)\)
\(\implies\)
\(\exists Sup (S) \land Inf (- S) = - Sup (S)\)
)
//
2: Proof
Whole Strategy: Step 1: suppose that \(Sup (- S)\) exists; Step 2: see that \(- Sup (- S)\) satisfies the conditions to be the infimum of \(S\); Step 3: suppose that \(Inf (- S)\) exists; Step 4: see that \(- Inf (- S)\) satisfies the conditions to be the supremum of \(S\).
Step 1:
Let us suppose that \(Sup (- S)\) exists.
Step 2:
For each \(- s \in - S\), \(- s \le Sup (- S)\), and for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is a \(- s \in - S\) such that \(Sup (- S) - \epsilon \lt - s\), 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.
So, for each \(s \in S\), \(- s \le Sup (- S)\), so, \(- Sup (- S) \le s\), and for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is a \(- s \in - S\) such that \(Sup (- S) - \epsilon \lt - s\), so, there is an \(s \in S\) such that \(s \lt - Sup (- S) + \epsilon\), which implies that \(- Sup (- S) = Inf (S)\), 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.
So, \(Sup (- S) = - Inf (S)\).
Step 3:
Let us suppose that \(Inf (- S)\) exists.
Step 4:
For each \(- s \in - S\), \(Inf (- S) \le - s\), and for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is a \(- s \in - S\) such that \(- s \lt Inf (- S) + \epsilon\), 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.
So, for each \(s \in S\), \(Inf (- S) \le - s\), so, \(s \le - Inf (- S)\), and for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is a \(- s \in - S\) such that \(- s \lt Inf (- S) + \epsilon\), so, there is an \(s \in S\) such that \(- Inf (- S) - \epsilon \lt s\), which implies that \(- Inf (- S) = Sup (S)\), 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.
So, \(Inf (- S) = - Sup (S)\).
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 metric space and subset, distance from subset map is uniformly continuous
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 metric space and any subset, the distance from the subset map is uniformly continuous.
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:
\(M\): \(\in \{\text{ the metric spaces }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(S\): \(\subseteq M\)
\(f\): \(: M \to \mathbb{R}, m \mapsto dist (S, m)\)
//
Statements:
\(f \in \{\text{ the uniformly continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: take any \(\epsilon \in \mathbb{R}\) and see that for \(\delta = \epsilon\), \(f (B_{m, \delta}) \subseteq B_{f (m), \epsilon}\).
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let us take \(\delta = \epsilon\).
Let \(m \in M\) be any.
Let us see that \(f (B_{m, \delta}) \subseteq B_{f (m), \epsilon}\).
Let \(m' \in B_{m, \delta}\) be any.
\(dist (S, m') \le dist (S, m) + dist (m, m')\), by the proposition that for any metric space and any subset, the distance between the subset and any point satisfies the triangle inequality with respect to any other point.
That equals \(f (m') \le f (m) + dist (m, m')\).
So, \(f (m') - f (m) \le dist (m, m') \lt \delta\).
By the symmetry, \(f (m) - f (m') \lt \delta\).
So, \(\vert f (m') - f (m) \vert \lt \delta = \epsilon\).
That means that \(f (m') \in B_{f (m), \epsilon}\).
So, \(f (B_{m, \delta}) \subseteq B_{f (m), \epsilon}\).
As \(\delta\) is determined independent of \(m\), \(f\) is uniformly continuous.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of for finite number of uniformly continuous maps from same metric space into \(1\)-dimensional Euclidean metric space, maximum or minimum map is uniformly continuous
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 finite number of uniformly continuous maps from any same metric space into the \(1\)-dimensional Euclidean metric space, the maximum or minimum map is uniformly continuous.
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:
\(M\): \(\in \{\text{ the metric spaces }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(\{f_1, ..., f_n\}\): \(f_j: M \to \mathbb{R} \in \{\text{ the uniformly continuous maps }\}\)
\(Max (\{f_1, ..., f_n\})\): \(: M \to \mathbb{R}, m \mapsto Max (\{f_1 (m), ..., f_n (m)\})\)
\(Min (\{f_1, ..., f_n\})\): \(: M \to \mathbb{R}, m \mapsto Min (\{f_1 (m), ..., f_n (m)\})\)
//
Statements:
\(Max (\{f_1, ..., f_n\}) \in \{\text{ the uniformly continuous maps }\}\)
\(\land\)
\(Min (\{f_1, ..., f_n\}) \in \{\text{ the uniformly continuous maps }\}\)
//
2: Proof
Whole Strategy: prove it inductively; Step 1: deal with the case that \(n = 1\); Step 2: when \(n = 2\), see that \(Max (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)\) and \(Min (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)\), and conclude the proposition for when \(n = 2\); Step 3: suppose that it holds when \(n = n' - 1\), and see that it holds when \(n = n'\).
Step 1:
It holds when \(n = 1\), because \(Max (\{f_1\}) = f_1\) and \(Min (\{f_1\}) = f_1\).
Step 2:
Let us suppose that \(n = 2\).
\(Max (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)\), because for each \(m \in M\), \(f_1 (m) \le f_2 (m)\) or \(f_2 (m) \lt f_1 (m)\), and when \(f_1 (m) \le f_2 (m)\), \(Max (\{f_1, f_2\}) (m) = f_2 (m)\) while \(1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert) (m) = 1 / 2 (f_1 (m) + f_2 (m) + \vert f_1 (m) - f_2 (m) \vert) = 1 / 2 (f_1 (m) + f_2 (m) + f_2 (m) - f_1 (m)) = 1 / 2 (f_2 (m) + f_2 (m)) = 1 / 2 2 f_2 (m) = f_2 (m)\); and when \(f_2 (m) \lt f_1 (m)\), \(Max (\{f_1, f_2\}) (m) = f_1 (m)\) while \(1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert) (m) = 1 / 2 (f_1 (m) + f_2 (m) + \vert f_1 (m) - f_2 (m) \vert) = 1 / 2 (f_1 (m) + f_2 (m) + f_1 (m) - f_2 (m)) = 1 / 2 (f_1 (m) + f_1 (m)) = 1 / 2 2 f_1 (m) = f_1 (m)\); so, \(Max (\{f_1, f_2\}) (m) = 1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert) (m)\) anyway.
\(Min (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)\), because for each \(m \in M\), \(f_1 (m) \le f_2 (m)\) or \(f_2 (m) \lt f_1 (m)\), and when \(f_1 (m) \le f_2 (m)\), \(Min (\{f_1, f_2\}) (m) = f_1 (m)\) while \(1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert) (m) = 1 / 2 (f_1 (m) + f_2 (m) - \vert f_1 (m) - f_2 (m) \vert) = 1 / 2 (f_1 (m) + f_2 (m) - f_2 (m) + f_1 (m)) = 1 / 2 (f_1 (m) + f_1 (m)) = 1 / 2 2 f_1 (m) = f_1 (m)\); and when \(f_2 (m) \lt f_1 (m)\), \(Min (\{f_1, f_2\}) (m) = f_2 (m)\) while \(1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert) (m) = 1 / 2 (f_1 (m) + f_2 (m) - \vert f_1 (m) - f_2 (m) \vert) = 1 / 2 (f_1 (m) + f_2 (m) - f_1 (m) + f_2 (m)) = 1 / 2 (f_2 (m) + f_2 (m)) = 1 / 2 2 f_2 (m) = f_2 (m)\); so, \(Min (\{f_1, f_2\}) (m) = 1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert) (m)\) anyway.
\(1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)\) is uniformly continuous, by the proposition that for any uniformly continuous map from any metric space into any Euclidean metric space, its norm map is uniformly continuous and the proposition that any linear combination of uniformly continuous maps into any Euclidean metric space is uniformly continuous.
So, \(Max (\{f_1, f_2\})\) is uniformly continuous.
\(1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)\) is uniformly continuous, by the proposition that for any uniformly continuous map from any metric space into any Euclidean metric space, its norm map is uniformly continuous and the proposition that any linear combination of uniformly continuous maps into any Euclidean metric space is uniformly continuous.
So, \(Min (\{f_1, f_2\})\) is uniformly continuous.
Step 3:
Let us suppose that it holds when \(n = n' - 1\) where \(2 \le n'\).
Let us see that it holds when \(n = n'\).
\(Max (\{f_1, ..., f_{n'}\}) = Max (\{Max (\{f_1, ..., f_{n' - 1}\}), f_{n'}\})\), because for each \(m \in M\), \(Max (\{f_1, ..., f_{n'}\}) (m) = f_j (m)\) for a \(j \in \{1, ..., n'\}\), and when \(j \in \{1, ..., n' - 1\}\), \(Max (\{Max (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m) = Max (\{Max (\{f_1, ..., f_{n' - 1}\}) (m), f_{n'} (m)\}) = Max (\{f_j (m), f_{n'} (m)\}) = f_j (m)\); and when \(j = n'\), \(Max (\{Max (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m) = Max (\{Max (\{f_1, ..., f_{n' - 1}\}) (m), f_{n'} (m)\}) = Max (\{f_l (m), f_{n'} (m)\})\) for a \(l \in \{1, ..., n' - 1\}\), \(= f_{n'} (m) = f_j (m)\); so, anyway, \(Max (\{f_1, ..., f_{n'}\}) (m) = Max (\{Max (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m)\).
\(Max (\{f_1, ..., f_{n' - 1}\})\) is uniformly continuous, by the induction hypothesis, and \(Max (\{Max (\{f_1, ..., f_{n' - 1}\}), f_{n'}\})\) is uniformly continuous, by Step 2.
So, \(Max (\{f_1, ..., f_{n'}\})\) is uniformly continuous.
By the induction principle, \(Max (\{f_1, ..., f_n\})\) is uniformly continuous for each \(n \in \mathbb{N} \setminus \{0\}\).
\(Min (\{f_1, ..., f_{n'}\}) = Min (\{Min (\{f_1, ..., f_{n' - 1}\}), f_{n'}\})\), because for each \(m \in M\), \(Min (\{f_1, ..., f_{n'}\}) (m) = f_j (m)\) for a \(j \in \{1, ..., n'\}\), and when \(j \in \{1, ..., n' - 1\}\), \(Min (\{Min (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m) = Min (\{Min (\{f_1, ..., f_{n' - 1}\}) (m), f_{n'} (m)\}) = Min (\{f_j (m), f_{n'} (m)\}) = f_j (m)\); and when \(j = n'\), \(Min (\{Min (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m) = Min (\{Min (\{f_1, ..., f_{n' - 1}\}) (m), f_{n'} (m)\}) = Min (\{f_l (m), f_{n'} (m)\})\) for a \(l \in \{1, ..., n' - 1\}\), \(= f_{n'} (m) = f_j (m)\); so, anyway, \(Min (\{f_1, ..., f_{n'}\}) (m) = Min (\{Min (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m)\).
\(Min (\{f_1, ..., f_{n' - 1}\})\) is uniformly continuous, by the induction hypothesis, and \(Min (\{Min (\{f_1, ..., f_{n' - 1}\}), f_{n'}\})\) is uniformly continuous, by Step 2.
So, \(Min (\{f_1, ..., f_{n'}\})\) is uniformly continuous.
By the induction principle, \(Min (\{f_1, ..., f_n\})\) is uniformly continuous for each \(n \in \mathbb{N} \setminus \{0\}\).
References
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