description/proof of that for set and subset, for finite cover of subset, there is subcover whose each element is indispensable, and for infinite cover of subset, there is not necessarily subcover whose each element is indispensable
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set.
Target Context
- The reader will have a description and a proof of the proposition that for any set and its any subset, for any finite cover of the subset, there is a subcover whose each element is indispensable, and for an infinite cover of the subset, there is not necessarily any subcover whose each element is indispensable.
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:
\(S'\): \(\in \{\text{ the sets }\}\)
\(S\): \(\subseteq S'\)
\(J\): \(\in \{\text{ the index sets }\}\)
\(\{S_j \subseteq S' \vert j \in J\}\): such that \(S \subseteq \cup_{j \in J} S_j\)
//
Statements:
(
\(J \in \{\text{ the finite sets }\}\)
\(\implies\)
\(\exists J^` \subseteq J (S \subseteq \cup_{j \in J^`} S_j \land \forall l \in J^` (\lnot S \subseteq \cup_{j \in J^` \setminus \{l\}} S_j))\)
)
(
\(J \in \{\text{ the infinite sets }\}\)
\(\implies\)
not necessarily "\(\exists J^` \subseteq J (S \subseteq \cup_{j \in J^`} S_j \land \forall l \in J^` (\lnot S \subseteq \cup_{j \in J^` \setminus \{l\}} S_j))\)"
)
//
2: Note
Making \(J\) countable infinite does not change the conclusions at all.
Making \(S'\) any topological space or something or making \(\{S_j \subseteq S' \vert j \in J\}\) any open cover or something of \(S\) does not change the conclusions at all.
Any cover might seem to be able to be reduced to a subcover whose each element is indispensable by disposing some unnecessary elements, but that is not really the case when the cover is infinite, this proposition says.
3: Proof
Whole Strategy: Step 1: suppose that \(J\) is finite; Step 2: for \(J = \{J_1, ..., J_n\}\), choose \(J^`\) by disposing or not \(J_j\) in the order; Step 3: suppose that \(J\) is infinite; Step 4: see an example that there is no such \(J^`\).
Step 1:
Let us suppose that \(J\) is finite.
Step 2:
Let \(J = \{J_1, ..., J_n\}\).
Let us modify \(J^`\) iteratively by deciding whether \(J_j\) is disposed in the order of \(J_1, J_2, ..., J_n\).
1st, let \(J^` = J\).
For \(J_1\), if \(S \subseteq \cup_{j \in J^` \setminus \{J_1\}} S_j\) holds, let \(J^` = J^` \setminus \{J_1\}\); otherwise, let \(J^` = J^`\).
Anyway, \(S \subseteq \cup_{j \in J^`} S_j\) holds.
Then, for \(J_2\), if \(S \subseteq \cup_{j \in J^` \setminus \{J_2\}} S_j\) holds, let \(J^` = J^` \setminus \{J_2\}\); otherwise, let \(J^` = J^`\).
Anyway, \(S \subseteq \cup_{j \in J^`} S_j\) holds.
And so on, through \(J_n\).
Now, \(S \subseteq \cup_{j \in J^`} S_j\) holds.
For each \(l \in J^`\), \(S \subseteq \cup_{j \in J^` \setminus \{l\}} S_j\) does not hold, because otherwise, in the process of deciding whether \(l\) should be disposed, \(S \subseteq \cup_{j \in J^` \setminus \{l\}} S_j\) would have held, because the right hand side was equal to or larger than the final \(\cup_{j \in J^` \setminus \{l\}} S_j\), then, \(l\) would have been disposed, a contradiction against that \(l\) was in the final \(J^`\).
So, \(J^`\) is a one that satisfies the conditions for the proposition.
Step 3:
Let us suppose that \(J\) is infinite.
Step 4:
Let us see an example that there is no such \(J^`\).
Let \(S' = \mathbb{R}\), \(S = (-1, 1)\), \(J = \mathbb{N}\), and \(S_j = (-1 + (1 / 2)^{j + 1}, 1 - (1 / 2)^{j + 1})\).
\(S \subseteq \cup_{j \in J} S_j\), because for each \(s \in S\), there is a \(j \in J\) such that \(s \in S_j = (-1 + (1 / 2)^{j + 1}, 1 - (1 / 2)^{j + 1})\), because \(-1 + (1 / 2)^{j + 1}\) limits to \(-1\) and \(1 - (1 / 2)^{j + 1})\) limits to \(1\).
But there is no such any \(J^`\), because such any \(J^`\) would have at least \(1\) element, \(l \in J^`\), but as \(S_l\) did not cover \(S\), \(J^`\) would have at least another \(l' \in J^`\) such that \(l \lt l'\), and as far as \(S \subseteq \cup_{j \in J^`} S_j\) held, \(S \subseteq \cup_{j \in J^` \setminus \{l\}} S_j\) held, because as \(S_l \subseteq S_{l'}\), \(\cup_{j \in J^` \setminus \{l\}} S_j = \cup_{j \in J^`} S_j\).
As the example shows, even when \(J\) is countable infinite instead of uncountable infinite, \(J^`\) may not exist.
\(S'\) can be regarded to be the Euclidean topological space and \(\{S_j \subseteq S' \vert j \in J\}\) can be regarded to be an open cover of \(S\), but that does not change the conclusion at all that \(J^`\) does not exist.
