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I understand that the definition of compactness varies. I am wanting to prove the set $[a,b]$ is compact the following way just like Julien did here for an answer. This proof is also similarly set up on p. 42 of Royden (second edition) for Real Analysis. There are many proofs of this result which all use the same trick which I am having a difficult time with.

I am going to start to the proof and explain where I am getting hung up.

$\textbf{Proof:}$ Let $[a,b]$ be a set of real numbers where we will assume $a<b$. Let $\lbrace U_i\rbrace_{i\in I}$ be an arbitrary collection of open sets that is an arbitrary open cover of $[a, b]$. Then, by definition, we know $[a,b]\subseteq \bigcup_{i\in I} U_i$. Now, we need to show there exists a finite collection $\lbrace U_1, U_2, ..., U_k\rbrace$ (i.e. a finite collection whose objects are open sets in the set $\lbrace U_i\rbrace_{i\in I}$) such that $[a,b]\subseteq \bigcup_{j=1}^kU_j$.

What I am confused on now is why does everyone define the set $S=\lbrace x\in [a,b]|[a,x]\subseteq \bigcup_{\text{C is a finite subset of I}} \lbrace U_c \rbrace_{c\in C}\rbrace$ after this part?

I do not see how the set $S$ here resembles a finite collection of open sets $\lbrace U_1, U_2, ..., U_k\rbrace$ like I was trying to show with the whole subset thing. Why do we want to show $LUB(S)=b$ (i.e. the least upper bound of the set $S$) do? I get lost at this part in the proof as to what we are trying to show.

W. G.
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    $S$ Is all the endpoints of intervals (starting at $a$) that can be covered by finitely many open sets. If $b\in S$, then $[a,b]$ can be covered by finitely many open sets! – Hempelicious Aug 06 '18 at 17:17
  • Do you mean that we are trying to show $[a,b]\subseteq \bigcup S$? I understand your first sentence, but the second sentence I am unsure what you mean. I do not see how you go from the first sentence to the second sentence. – W. G. Aug 06 '18 at 17:23
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    You are trying to show $b$ is in $S$. You know $[a,b]$ is connected, and $S\subset [a,b]$ is not empty, since $a \in S$. All you got to do is show that $S$ is open and closed, hence, it has to be all $[0,1]$. Showing that $b$ is in $S$ suffices, because the set $S\subset [0,1]$ is the set of all $x \in [0,1]$, such that [0,x] has a finite subcover of ${U_i}_{i \in I}$ – Bajo Fondo Aug 06 '18 at 17:25
  • I bet that is the part that I am missing. Why do we want to show the set is both open and closed here? The connectivity thing I did not know was needed for this step. Is there a lemma or something that I need to prove first to get to there? – W. G. Aug 06 '18 at 17:32
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    By definition, elements of $S$ are those endpoints where $[a,x]$ can be finitely covered. If $b\in S$, then by definition of S, $[a,b]$ can be finitely covered. – Steve D Aug 06 '18 at 17:38
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    The argument is a bit confusing until you realise what it going on. The set $S$ gives the collection of all $x \in [a,b]$ such that $[a,x]$ is covered by a finite number of the open sets. The goal then is to show that $b \in S$. – copper.hat Aug 06 '18 at 17:41
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    Wait a second. I think I get part of it. We want to show $LUB(S)=b\in S$. Thus, $[a,b]\subseteq \bigcup_{\text{C is a finite subset of I}} \lbrace U_c \rbrace_{c\in C}$ for some finite set $C$ looking at the set builder notation of $S$. How do we show the $LUB(S)\in S$ though? – W. G. Aug 06 '18 at 18:02
  • Note that if $x \in S$ and $x<b$, since there is some open set containing $x$, then we must have $x+\epsilon \in S$ for some $\epsilon >0$. Hence we must have $\sup S =b$. Similarly, since there is some open set containing $b$ we must have $b \in S$. It is a fairly subtle argument. – copper.hat Aug 06 '18 at 18:28

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I think the idea behind this proof starts with "Okay, obviously I can finitely cover $[a,a]$. Can I go a step further? Yes, because there exists $\epsilon>0$ ($\epsilon \leq b-a$, to stay within the interval) such that the finite cover just chosen still works for $[a,a+\epsilon]$". Then you wonder how far to the right you can go with this reasoning and that's when your set $S$ comes into play. If $\sup S <b$, apply the procedure just described and get a contradiction. So $\sup S=b$. Now choose $\epsilon>0$ sufficiently small such that $[b-\epsilon,b]$ can be finitely covered, therefore $b\in S$, since $b-\epsilon \in S$.

SEBASTIAN VARGAS LOAIZA
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The general idea of this proof is to try to cover $[a,b]$ iteratively. To do this, start by choosing an open set $U_0$ which covers $a$. The uncovered portion of $[a,b]$ will have a smallest point, $a_1$. Next, find an open set $U_1$ which covers $a_1$, rinse and repeat.

The problem is that you might get stuck at some point. Namely, the infinite sequence $U_0,U_1,\dots,$ of open sets could get smaller and smaller in such a way that do not cover all of $[a,b]$ in total. If this happens, there will be a smallest uncovered point, $a'$. Find an open set $U'$ in the cover which contains $u'$. Since $U'$ will extend a little bit the left of $a'$, you can show you it will cover all but finitely many sets in the list $U_0,U_1,U_2,\dots$. This allows you to get back down to a finite list of open sets, and then restart the iterative procedure.

This is a mouthful to state. The $S$ trick is a way of describing this whole strategy in one fell swoop. The interval $[a,x]$ in the definition of $S$ represents the current progress you have made in trying to cover all of $S$. Proving that $S$ contains its upper bound amounts to the strategy in the second paragraph; if you can cover $[a,x]$ with finitely many intervals for all $x<s$, then you can cover $[a,s]$ as well. Then, showing that $\sup S$ cannot be less than $b$ is covered by the first paragraph; if you can cover $[a,s]$, then you can cover $[a,s+\epsilon]$ for some $\epsilon>0$.

Mike Earnest
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Consider the definition of $S$. The statement that some finite subset of $\{U_i\}_{i\in I}$ covers $[a,b]$ is exactly the statement that $b\in S.$

We show (i). $\sup S$ exists. (ii). $\sup S\in S.$ (iii). $\sup S\not \in [a,b). $ And conclude that $b=\sup S\in S.$

DanielWainfleet
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  • A set of sets does not need to written with an index set in the style ${U_i}{i\in I}.$ we can just say "Let $V$ be an open cover of $[a,b]$". And instead of $\cup{i\in }U_i$ we can just write $\cup V$. Extra notation tends to result in notational errors such as your ${U_1,..., U_k}$ which should be ${U_{i(1)},...,U_{i(k)}}$... You can just say we want to show there exists finite $W\subset V$ with $[a.b]\subset \cup W.$ – DanielWainfleet Aug 06 '18 at 21:00
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    I was thinking about saying for some $k\in \mathbb{W}$ to be explicit for notation to mean the set could be empty. I agree with you I could have avoided the whole index set thing. Why are you writing the $i(1), ..., i(k)$ like a function just out of curiosity? The union could be empty too if that is the confusion. I am just curious what you meant by my notation being off. – W. G. Aug 07 '18 at 18:49
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    If you already have the set $V= {U_i}_{i\in I}$ then ${U_1,...,U_k}$ is a specific unique subset of $V$ if ${1,...,k}\subset I,$ while if ${1,...,k} \not \subset I$ then ${U_1,...,U_k}$ doesn't exist. – DanielWainfleet Aug 07 '18 at 20:02
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Obviously {a} could be covered by one of those open sets, namely $U$.

Then this open set includes an open interval centered at $a$ so we can say some closed interval $[a,x_1]$ is covered by that open set $U$.

That means $[a,x_1]$ is covered by finitely many open sets.

If $x_1=b$ we are done otherwise we like to push the right end of [a,x] as far as it goes while it is still covered by finitely many of the open sets.

That is how the supremum comes to the picture and we show that the supremum can not be less than $b$

Mohammad Riazi-Kermani
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