description/proof of that for topological space, sequence of preimages of natural-numbers-closed-upper-bounds intervals under exhaustion function is exhaustion of space by compact subsets
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of exhaustion function on topological space.
- The reader knows a definition of exhaustion of topological space by compact subsets.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, the sequence of the preimages of the natural-numbers-closed-upper-bounds intervals under any exhaustion function is an exhaustion of the space by compact subsets.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(f\): \(: T \to \mathbb{R}\), \(\in \{\text{ the exhaustion functions on } T\}\)
\(s\): \(: \mathbb{N} \setminus \{0\} \to \{\text{ the compact subsets of } T\}, j \mapsto f^{-1} ((- \infty, j])\)
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Statements:
\(s \in \{\text{ the exhaustions of } T \text{ by compact subsets }\}\)
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2: Natural Language Description
For any topological space, \(T\), the Euclidean topological space, \(\mathbb{R}\), and any exhaustion function on \(T\), \(f: T \to \mathbb{R}\), the sequence, \(s: \mathbb{N} \setminus \{0\} \to \{\text{ the compact subsets of } T\}, j \mapsto f^{-1} ((- \infty, j])\), is an exhaustion of \(T\) by compact subsets.
3: Proof
Whole Strategy: Step 1: see that \(T = \cup_{j \in \mathbb{N} \setminus \{0\}} s (j)\); Step 2: see that \(s (j) \subseteq {s (j + 1)}^\circ\).
Step 1:
Let us see that \(T = \cup_{j \in \mathbb{N} \setminus \{0\}} s (j)\).
For each \(p \in T\), \(f (p) \le n\) for an \(n \in \mathbb{N} \setminus \{0\}\). So, \(p \in f^{-1} ((- \infty, n]) = s (n)\). So, \(T = \cup_{j \in \mathbb{N} \setminus \{0\}} s (j)\).
Step 2:
Let us see that \(s (j) \subseteq {s (j + 1)}^\circ\).
Step 2 Strategy: Step 2-1: see that it suffices to show that for each \(p \in s (j)\), there is an open neighborhood of \(p\) contained in \(s (j + 1)\); Step 2-2: prove the claim.
Obviously, \(s (j) \subseteq s (j + 1)\).
Step 2-1:
It suffices to show that for each \(p \in s (j)\), there is an open neighborhood of \(p\), \(U_p \subseteq T\), such that \(U_p \subseteq s (j + 1)\), because then, \(s (j) \subseteq \cup_{p \in s (j)} U_p \subseteq s (j + 1)\), while \(\cup_{p \in s (j)} U_p\) will be an open subset contained in \(s (j + 1)\), which will mean that \(\cup_{p \in s (j)} U_p \subseteq {s (j + 1)}^\circ\).
Step 2-2:
Let us show \(U_p\).
\(f (p) \in (- \infty, j]\). Let us take any \(\epsilon \lt 1\) and the open ball, \(B_{f (p), \epsilon} = (f (p) - \epsilon, f (p) + \epsilon) \subseteq \mathbb{R}\). \(B_{f (p), \epsilon} \subseteq (- \infty, j + 1]\). As \(f\) is continuous, there is an open neighborhood of \(p\), \(U_p \subseteq T\), such that \(f (U_p) \subseteq B_{f (p), \epsilon}\). So, \(f (U_p) \subseteq B_{f (p), \epsilon} \subseteq (- \infty, j + 1]\). That means that \(U_p \subseteq s (j + 1)\).
