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This is an idea for a proof I had of a theorem in Understanding Analysis by Stephen Abbott. I'm unsure of if it's valid:

Assume that any open cover of K $\subseteq \mathbb{R}$ has a finite subcover. Then for any open cover $\{O_{\lambda} | \lambda \in \Lambda\}$, there is some $\lambda_{1}, \lambda_{2}, \dots, \lambda_{k}$ such that $\bigcup_{i=1}^kO_{\lambda_{i}}=K$. From all of the $O_{\lambda_{i}}$s, take $O_i$ with smallest infimum and $O_j$ with largest supremum. Then for all $k \in K$, $|k| \le$ max(inf $O_i$, sup $O_j$). Hence K is bounded.

Would this proof work or am I misunderstanding the meaning of open covers/finite subcovers? Thank you so much for any help!

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    $\begingroup$ What do you mean by the smallest infimum? Are you talking about sets in $\mathbb{R}$? $\endgroup$ Commented Jan 17 at 15:49
  • $\begingroup$ Yes, apologies or not specifying. All of these sets are in the reals. I meant the smallest infimum of those sets in the finite subcover. $\endgroup$
    – AyanNair
    Commented Jan 17 at 15:53
  • $\begingroup$ @mertunsal not true, you can’t rule out the possibility that $O_i$ is unbounded. You can, however, assume without loss of generality that each $O_\lambda$ is bounded. Of course, this latter assumption is the heart of the proof, and should be explicitly justified $\endgroup$
    – Andrew
    Commented Jan 17 at 16:13
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    $\begingroup$ @mertunsal no that’s not true, you can always take the open cover to be $\mathbb R$ $\endgroup$
    – Andrew
    Commented Jan 17 at 16:23
  • $\begingroup$ @Andrew I see how the proof will not follow through without this assumption, but I'm not sure I fully see why I can assume that all sets in the finite subcover are bounded. $\endgroup$
    – AyanNair
    Commented Jan 17 at 16:25

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The approach given here doesn't quite work. The problem is that you have no control over an arbitrary open cover of a set. For example, what if the open cover is $\mathbb{R}$? or the collection of half-lines $(-n,\infty)$ with $n\in\mathbb{N}$? These kinds of covers cannot be used to show that a set is bounded.

As such, I think that the approach here is missing something important about compactness (that is, the property that any open cover must have a finite subcover). The game here should not be to start with an arbitrary open cover and then try to somehow deduce that the set is bounded—rather, the game should be to carefully construct a cover in some way that the fact that it has a finite subcover becomes useful.

The following argument seems to get the job done:

Let $K \subseteq \mathbb{R}$ have the property that every open cover of $K$ has an open subcover. For each natural number $n$, let $$ O_n = (-n,n). $$ That is, $O_n$ is the open ball of radius $n$ centered at zero. Since $$ \mathbb{R} = \bigcup_{n=1}^{\infty} O_n, $$ it must be the case that the $O_n$ cover $K$. But every cover of $K$ has a finite subcover, so there must be a finite set $A \subseteq \mathbb{N}$ such that $$ K \subseteq \bigcup_{n \in A} O_n . $$ Choose $N = \max\{A\}$ (the maximum exists because $A$ is finite) and observe that the $O_n$ are nested (that is, if $m < n$, then $O_m \subseteq O_n$). This implies $K \subseteq O_N$, which means that if $x\in K$, then $|x|<N$. Therefore $K$ is bounded.

Note that this argument is given to be very general. What has actually been shown here is that if $(X,d)$ is any metric space and $K \subset X$ is compact (i.e. it has the property that any open cover has a finite subcover), then $K$ is bounded. The only modification which needs to be made is to choose $O_n = B(x,n)$ (the ball of radius $n$ centered at $x$) where $x$ is some fixed point in $X$.

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