Suppose that [a, b] ≤R is a closed and bounded interval, and let {G a E A} (where A is any index set) be a collection of open sets such that [a, b] ≤ U G and XR be any closed and bounded subset of R and we have s as the supremum of x € [a, b] such that the required result holds if [a, b] is replaced by [a, x] where some finite subcollection of the G covers [a, x]) and s> a, and s < b is impossible. Proves > a, and s a and s < b on la bl
Suppose that [a, b] ≤R is a closed and bounded interval, and let {G a E A} (where A is any index set) be a collection of open sets such that [a, b] ≤ U G and XR be any closed and bounded subset of R and we have s as the supremum of x € [a, b] such that the required result holds if [a, b] is replaced by [a, x] where some finite subcollection of the G covers [a, x]) and s> a, and s < b is impossible. Proves > a, and s a and s < b on la bl
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Suppose that \([a, b] \subseteq \mathbb{R}\) is a closed and bounded interval, and let \(\{G_\alpha : \alpha \in A\}\) (where \(A\) is any index set) be a collection of open sets such that \([a, b] \subseteq \bigcup_{\alpha \in A} G_\alpha\) and \(X \subseteq \mathbb{R}\) be any closed and bounded subset of \(\mathbb{R}\) and we have \(s\) as the supremum of \(x \in [a, b]\) such that the required result holds if \([a, b]\) is replaced by \([a, x]\) where some finite subcollection of the \(G_\alpha\) covers \([a, x]\) and \(s > a\), and \(s < b\) is impossible. Prove \(s > a\), and \(s < b\) is impossible holds given any collection \(\{G_\alpha : \alpha \in A\}\) of open sets such that \(X \subseteq \bigcup_{\alpha \in A} G_\alpha\) and there is some finite subset \(\{\alpha_1, \alpha_2, \ldots, \alpha_n\} \subseteq A\) such that \([a, b] \subseteq \bigcup_{i=1}^{n} G_{\alpha_i}\). Let
\(a = \inf X\) and \(b = \sup X\) show that \([a, b]\setminus X\) is an open set and apply the result that \(s > a\) and \(s < b\) on \([a, b]\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37e8ed93-7bef-4409-89ed-52264f64a27e%2Ffb09408f-c6be-4f65-90fc-63d1211e8465%2Fgdzqe_processed.png&w=3840&q=75)
Transcribed Image Text:Suppose that \([a, b] \subseteq \mathbb{R}\) is a closed and bounded interval, and let \(\{G_\alpha : \alpha \in A\}\) (where \(A\) is any index set) be a collection of open sets such that \([a, b] \subseteq \bigcup_{\alpha \in A} G_\alpha\) and \(X \subseteq \mathbb{R}\) be any closed and bounded subset of \(\mathbb{R}\) and we have \(s\) as the supremum of \(x \in [a, b]\) such that the required result holds if \([a, b]\) is replaced by \([a, x]\) where some finite subcollection of the \(G_\alpha\) covers \([a, x]\) and \(s > a\), and \(s < b\) is impossible. Prove \(s > a\), and \(s < b\) is impossible holds given any collection \(\{G_\alpha : \alpha \in A\}\) of open sets such that \(X \subseteq \bigcup_{\alpha \in A} G_\alpha\) and there is some finite subset \(\{\alpha_1, \alpha_2, \ldots, \alpha_n\} \subseteq A\) such that \([a, b] \subseteq \bigcup_{i=1}^{n} G_{\alpha_i}\). Let
\(a = \inf X\) and \(b = \sup X\) show that \([a, b]\setminus X\) is an open set and apply the result that \(s > a\) and \(s < b\) on \([a, b]\).
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