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The proof for Theorem 13.9 from "Introduction to Metric & Topological Spaces" by Wilson Sutherland. The definition for compact is that any open cover can be reduced to a finite open cover for a compact set.

https://books.google.dk/books?id=dxrpDAAAQBAJ&pg=PA129&lpg=PA129&dq=Wilson+A.+Sutherland+Compactness+of+closed+bounded+intervals&source=bl&ots=-M-q4YnqKS&sig=ACfU3U0nKuavUPNxcnTqBZYWVA7kdJjX1g&hl=de&sa=X&ved=2ahUKEwji0t7ZrIbnAhXqtYsKHXBuDRIQ6AEwAHoECAkQAQ#v=onepage&q&f=false

I understand the proof, but I don't see why we need x > a.

HF_
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  • $[a,x]=\emptyset$ if $x<a$. It's not necessary that $G$ be defined only for $x \geq a$, but it doesn't add anything to allow $x<a$. If you allowed $x<a$, then $G$ would also contain all of $(-\infty,a]$. However, we are only concerned with showing that $b \leq \sup G$, so this does not matter. – kccu Jan 15 '20 at 19:52
  • Right, but we have x>=a, since x $\in$ G – HF_ Jan 15 '20 at 19:54
  • For which $x$? I don't see what the problem is. – kccu Jan 15 '20 at 19:55
  • The x used to get the contradiction, I am wondering why it's necessary to point out that c > a so we can conclude that x > a. Why will x >= a not work? – HF_ Jan 15 '20 at 19:58
  • I don't think it makes any difference. As long as $x > c-\epsilon$, we have $[a,x]\cup (c-\epsilon, c+\epsilon/2] \supseteq [a,c+\epsilon/2]$. – kccu Jan 15 '20 at 20:05

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