description/proof of that for real numbers field with canonical linear ordering and countable set of non-negative subsets, if sum of each set of representatives of subsets is equal to or larger than number, sum of infimums of subsets is equal to or larger than number
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of real numbers field.
- The reader knows a definition of infimum of subset of partially-ordered set.
- The reader admits the proposition that for the real numbers field with the canonical linear ordering and any finite set of subsets, if the sum of each set of representatives of the subsets is equal to or larger than any number, the sum of the infimums of the subsets is equal to or larger than the number.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that for the real numbers field with the canonical linear ordering and any countable set of non-negative subsets, if the sum of each set of representatives of the subsets is equal to or larger than any number, the sum of the infimums of the subsets is equal to or larger than the number.
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:
\(\mathbb{R}\): \(= \text{ the real numbers field with the canonical linear ordering }\)
\(J\): \(\in \{\text{ the countable index sets }\}\)
\(\{S_j \subseteq [0, \infty) \vert j \in J\}\):
\(r\): \(\in \mathbb{R}\)
//
Statements:
\(\forall s \in \times_{j \in J} S_j (r \le \sum_{j \in J} s^j)\)
\(\implies\)
\(r \le \sum_{j \in J} Inf (S_j)\)
//
\(Inf (S_j)\) s inevitably exist.
As \(s^j\) s and \(Inf (S_j)\) s are non-negative, the order of \(\sum_{j \in J}\) does not matter, as is well known.
\(\sum_{j \in J} s^j\) or \(\sum_{j \in J} Inf (S_j)\) may not converge, in which case, the inequality is regarded to hold.
2: Proof
Whole Strategy: Step 1: see that each \(Inf (S_j)\) exists; Step 2: deal with the case that \(J\) is finite, and suppose otherwise hereafter; Step 3: deal with the case that each \(\sum_{j \in J} s^j\) does not converge; Step 4: suppose that \(\sum_{j \in J} Inf (S_j) \lt r\) and find a contradiction.
Step 1:
Let us see that each \(Inf (S_l)\) exists.
For each \(p \in S_l\), \(0 \le p\).
So, \(S_l\) is lower bounded, and \(Inf (S_l)\) exists, because any lower bounded subset of \(\mathbb{R}\) has the infimum, as is well known.
Step 2:
When \(J\) is finite, the proposition that for the real numbers field with the canonical linear ordering and any finite set of subsets, if the sum of each set of representatives of the subsets is equal to or larger than any number, the sum of the infimums of the subsets is equal to or larger than the number applies, which guarantees this proposition.
Hereafter, let us suppose that \(J\) is infinite.
Let us suppose that \(J = \{1, 2, ...\}\) just for convenience of our expressions, without loss of generality.
Step 3:
Let us suppose that each \(\sum_{j \in J} s^j\) does not converge.
Let us see that then, \(\sum_{j \in J} Inf (S_j)\) does not converge.
Let \(a \in \mathbb{R}\) be any such that \(0 \lt a \lt 1\).
For each \(j \in J\), there is a \(p_j \in S_j\) such that \(p_j \lt Inf (S_j) + a^j\), 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.
Let \(s \in \times_{j \in J} S_j\) be the one such that for each \(j \in J\), \(s^j = p_j\).
\(\sum_{j \in J} s^j \le \sum_{j \in J} (Inf (S_j) + a^j) = \sum_{j \in J} Inf (S_j) + \sum_{j \in J} a^j = \sum_{j \in J} Inf (S_j) + a / (1 - a)\).
So, \(\sum_{j \in J} s^j - a / (1 - a) \le \sum_{j \in J} Inf (S_j)\).
As the left hand side does not converge, the right hand side does not converge.
So, \(r \le \sum_{j \in J} Inf (S_j)\) holds.
Step 4:
Let us suppose that there is an \(s \in \times_{j \in J} S_j\) such that \(\sum_{j \in J} s^j\) converges.
\(\sum_{j \in J} Inf (S_j)\) converges, because \(Inf (S_j) \le s^j\).
Let us suppose that \(\sum_{j \in J} Inf (S_j) \lt r\).
Let \(a \in \mathbb{R}\) be any such that \(0 \lt a \lt 1\) and \(a / (1 - a) \lt r - \sum_{j \in J} Inf (S_j)\), which would be obviously possible.
For each \(j \in J\), there would be a \(p_j \in S_j\) such that \(p_j \lt Inf (S_j) + a^j\), 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.
Let \(s \in \times_{j \in J} S_j\) be the one such that for each \(j \in J\), \(s^j = p_j\).
\(\sum_{j \in J} s^j \lt \sum_{j \in J} (Inf (S_j) + a^j) = \sum_{j \in J} Inf (S_j) + \sum_{j \in J} a^j = \sum_{j \in J} Inf (S_j) + a / (1 - a) \lt \sum_{j \in J} Inf (S_j) + r - \sum_{j \in J} Inf (S_j) = r\), a contradiction.
So, \(r \le \sum_{j \in J} Inf (S_j)\).
