description/proof of that for linearly-ordered set, subset does not necessarily have supremum or infimum
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of linearly-ordered set.
- The reader knows a definition of supremum of subset of partially-ordered set.
- The reader knows a definition of infimum of subset of partially-ordered set.
Target Context
- The reader will have a description and a proof of the proposition that for a linearly-ordered set, a subset does not necessarily have any supremum or any infimum.
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, \(\lt'\)
\(S\): \(\subseteq S'\)
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Statements:
Not necessarily, \(\exists Sup (S)\)
\(\land\)
Not necessarily, \(\exists Inf (S)\)
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2: Note
It is easily guessed that a subset of a partially-ordered set has no supremum or infimum, but being a linearly-ordered set does not guarantee the existence of supremum or infimum of a subset.
3: Proof
Whole Strategy: Step 1: see an example that \(S\) has no supremum; Step 2: see an example that \(S\) has no infimum.
Step 1:
Let \(S' = \mathbb{Q}\), the rational numbers set, with the canonical linear ordering, \(\lt'\).
Let \(S = \{s \in S' \vert s \lt \sqrt{2}\}\): \(\sqrt{2} \notin \mathbb{Q}\), so, \(s \lt \sqrt{2}\) is not by the ordering for \(\mathbb{Q}\) but by the ordering for \(\mathbb{R}\) in which \(\mathbb{Q}\) is contained.
Then, \(S\) has no supremum, because while \(Ub (S) = \{s' \in S' \vert \sqrt{2} \lt s'\}\), \(Ub (S)\) has no minimum, because for each \(s' \in Ub (S)\), there is an \(s'' \in S'\) such that \(\sqrt{2} \lt s'' \lt s'\), which means that \(s'' \in Ub (S)\) and \(s'' \lt s'\).
Step 2:
Let \(S' = \mathbb{Q}\), the rational numbers set, with the canonical linear ordering, \(\lt'\).
Let \(S = \{s \in S' \vert \sqrt{2} \lt s\}\): \(\sqrt{2} \notin \mathbb{Q}\), so, \(\sqrt{2} \lt s\) is not by the ordering for \(\mathbb{Q}\) but by the ordering for \(\mathbb{R}\) in which \(\mathbb{Q}\) is contained.
Then, \(S\) has no infimum, because while \(Lb (S) = \{s' \in S' \vert s' \lt \sqrt{2}\}\), \(Lb (S)\) has no maximum, because for each \(s' \in Lb (S)\), there is an \(s'' \in S'\) such that \(s' \lt s'' \lt \sqrt{2}\), which means that \(s'' \in Lb (S)\) and \(s' \lt s''\).
