700: For Topological Space, Sequence of Preimages of Natural-Numbers-Closed-Upper-Bounds Intervals Under Exhaustion Function Is Exhaustion of Space by Compact Subsets

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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

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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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.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $\in \{\text{ the topological spaces }\}$
$\mathbb{R}$: $=\text{ the Euclidean topological space }$
$f$: $:T\to \mathbb{R}$, $\in \{\text{ the exhaustion functions on }T\}$
$s$: $:\mathbb{N}\setminus \{0\}\to \{\text{ the compact subsets of }T\},j\mapsto f^{-1}((-\mathrm{\infty },j])$
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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, $s:\mathbb{N}\setminus \{0\}\to \{\text{ the compact subsets of }T\},j\mapsto f^{-1}((-\mathrm{\infty },j])$, is an exhaustion of $T$ by compact subsets.

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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, $p\in f^{-1}((-\mathrm{\infty },n])=s(n)$. 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 $U_p$.

$f(p)\in (-\mathrm{\infty },j]$. Let us take any $ϵ<1$ and the open ball, $B_{f(p),ϵ}=(f(p)-ϵ,f(p)+ϵ)\subseteq \mathbb{R}$. $B_{f(p),ϵ}\subseteq (-\mathrm{\infty },j+1]$. As $f$ is continuous, there is an open neighborhood of $p$, $U_p\subseteq T$, such that $f(U_p)\subseteq B_{f(p),ϵ}$. So, $f(U_p)\subseteq B_{f(p),ϵ}\subseteq (-\mathrm{\infty },j+1]$. That means that $U_p\subseteq s(j+1)$.

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References

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