description/proof of that intersection of non-increasing sequence of nonempty open or closed subsets does not necessarily contain point
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description 1
- 2: Proof 1
- 3: Structured Description 2
- 4: Proof 2
- 5: Note
Starting Context
- The reader knows a definition of closed subset of topological space.
- The reader knows a definition of sequence.
Target Context
- The reader will have a description and a proof of the proposition that the intersection of a non-increasing sequence of nonempty open or closed subsets does not necessarily contain a point.
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 1
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\subseteq \mathbb{N}\)
\(s\): \(: J \to \{\text{ the open subsets of } T\} \setminus \{\emptyset\}\), \(\in \{\text{ the sequences }\}\), such that \(\forall j, j' \in J \text{ such that } j \lt j' (s (j') \subseteq s (j))\)
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Statements:
not necessarily "\(\cap_{j \in J} s (j) \neq \emptyset\)"
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2: Proof 1
Whole Strategy: Step 1: see a counterexample.
Step 1:
Let us see a counterexample.
Let \(T := \mathbb{R}\), as the Euclidean topological space, \(J = \mathbb{N}\), and \(s (j) := (0, 1 / (j + 1)) \subseteq T\).
\(s\) satisfies the conditions, because for each \(j \in J\), \(s (j) \neq \emptyset\), and for each \(j, j' \in J\) such that \(j \lt j'\), \(s (j') = (0, 1 / (j' + 1)) \subseteq (0, 1 / (j + 1)) = s (j)\).
Let \(t \in T\) be any.
When \(t \le 0\), \(t \notin (0, 1) = s (0)\), so, \(t \notin \cap_{j \in J} s (j)\).
When \(0 \lt t\), there is a large enough \(j \in J\) such that \(1 / (j + 1) \lt t\), so, \(t \notin (0, 1 / (j + 1)) = s (j)\), so, \(t \notin \cap_{j \in J} s (j)\).
So, anyway, \(t \notin \cap_{j \in J} s (j)\).
So, \(\cap_{j \in J} s (j) = \emptyset\).
3: Structured Description 2
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\subseteq \mathbb{N}\)
\(s\): \(: J \to \{\text{ the closed subsets of } T\} \setminus \{\emptyset\}\), \(\in \{\text{ the sequences }\}\), such that \(\forall j, j' \in J \text{ such that } j \lt j' (s (j') \subseteq s (j))\)
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Statements:
not necessarily "\(\cap_{j \in J} s (j) \neq \emptyset\)"
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4: Proof 2
Whole Strategy: Step 1: see a counterexample.
Step 1:
Let us see a counterexample.
Let \(T := \mathbb{R}\), as the Euclidean topological space, \(J = \mathbb{N}\), and \(s (j) := (- \infty, - j] \subseteq T\).
\(s\) satisfies the conditions, because for each \(j \in J\), \(s (j) \neq \emptyset\), and for each \(j, j' \in J\) such that \(j \lt j'\), \(s (j') = (- \infty, - j'] \subseteq (- \infty, - j] = s (j)\).
Let \(t \in T\) be any.
There is a large enough \(j \in J\) such that \(- j \lt t\), so, \(t \notin (- \infty, - j] = s (j)\), so, \(t \notin \cap_{j \in J} s (j)\).
So, \(t \notin \cap_{j \in J} s (j)\).
So, \(\cap_{j \in J} s (j) = \emptyset\).
5: Note
When the codomain of \(s\) is not restricted to the set of the nonempty open subsets or the set of the nonempty closed subsets but to the set of the nonempty subsets, not necessarily "\(\cap_{j \in J} s (j) \neq \emptyset\)" even more, because Proof 1 and Proof 2 are some counterexamples.
The point of Description 2 is that \(T\) is not necessarily compact.
If \(T\) is compact, the intersection necessarily contains a point, by the proposition that any topological space is compact if and only if for its every collection of closed subsets for which the intersection of any finite members is not empty, the intersection of the collection is not empty.
