Exercise 3.3.1 in Stephen Abott's book poses the following question.

Show that if $K$ is compact and non-empty, then $\sup K$ and $\inf K$ both exist and are elements of $K$.

I attempted to write a simple proof, and would like someone to verify, if the proof is technically correct and rigorous.

My Attempt.

Let $K$ be compact and non-empty. By the theorem on the characterization of compactness, $K$ is closed and bounded. Since $K$ is a bounded subset of $\mathbf{R}$, by AoC, $K$ has a supremum and an infimum. Let $s = \sup K$.

There is a trichotomy. Either (i) $s \in int(K)$ (ii) $s \in \partial K$ (iii) $s \in ext(K)$.

(i) Suppose $s \in int(K)$. Then, there exists an $\epsilon$ such that the open interval $(s - \epsilon,s + \epsilon)$ is contained in $K$. Consider $s + \epsilon/2 \in (s, s+\epsilon) \cap K$. $s + \epsilon/2 > s$. But, this implies that $s$ is not an upper bound for $K$, which is a contradiction. Hence, $s \notin int(K)$.

(ii) Suppose $s \in ext(K)$. Then, there exists an $\epsilon > 0$ such that the open interval $(s - \epsilon,s + \epsilon)$ is contained in $K^C$. Consider $s - \epsilon/2 \in K \cap (s - \epsilon,s)$. $s - \epsilon/2 \ge x$ for all $x \in K$, and $s - \epsilon/2 < s$. So, $s$ is not an upper bound for $K$, which is a contradiction. Hence, $s \notin ext(K)$.

(iii) If $s \in \partial K$, because $K$ is a closed set, $\partial K \subseteq K$. So, $s \in K$.

A similar argument can be made for $\inf K$.

This closes the proof.

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    $\begingroup$ Sounds good to me. Another approach is to use the fact that if $s = \mathbf{sup} (X)$ then $s = \lim_n x_n$ for some $(x_n) \subset X$. In particular, when $X$ is closed we have $s \in X$. $\endgroup$ – guidoar Jan 24 at 8:51
  • $\begingroup$ @guidoar, I wasn't entirely sure about that approach. For example, consider $K = [0,1] \cap \{2\}$. This is a closed and bounded set. So, it is compact. $\sup K = 2$, which is not a limit point. $\endgroup$ – Quasar Jan 24 at 8:54
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    $\begingroup$ $2$ is still a limit point, it's just not an -accumulation point-. $\endgroup$ – John Samples Jan 24 at 8:55
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    $\begingroup$ The sequence in that case is the constant sequence $2$, assuming you meant $[0,1] \cup \{2\}$ $\endgroup$ – guidoar Jan 24 at 8:55
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    $\begingroup$ Proof of my claim: since $s$ is a supremum, for each $n \geq 1$ the exists $x_n \in S$ such that $s-1/n \leq x_n$. Then $0 \leq s-x_n \leq 1/n$. Now use take limit and use the "sandwhich" theorem (not that we may very well have $x_n = s$ for some or all $n$). $\endgroup$ – guidoar Jan 24 at 8:57

Everything's fine except #2. Just because it's not in the set doesn't mean it isn't an upper bound; indeed, if $s > k$ for every $k \in K$ then it's still an upper bound. Also, it may be that $s$ is between components of $K$, or it could be less than every element of $K$ . . . being in the exterior of $K$ doesn't give you any info about if it's an upper/lower bound or neither.

The argument should look more like "$K$ is compact so closed, so ext($K$) is open, so $s$ is contained in an interval that doesn't intersect $K$." Then if $s$ is supposed to be the sup, you should have elements (not necessarily distinct, as per your comment) of $K$ converging to $s$ from below. But that's impossible, since the interval you constructed doesn't intersect $K$.

Make sense?

  • $\begingroup$ Hi, I agree that just because $s$ belongs $K^C$, doesn't mean it's an upper bound. But, we've already assume $s = \sup K$, so isn't it like implicit that $s > k$ for all $k \in K$? $\endgroup$ – Quasar Jan 24 at 9:07
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    $\begingroup$ Well that's sort of what you're trying to show, tho ofc it might be $\geq$ rather than just >. If you are using the set-theoretic sup, what you'll wanna do is note that any other element in that interval not intersecting $K$ is ALSO greater, and it contains elements less than $s$. So $s$ can't be the limsup, since there are lesser numbers than $s$ that are also a sup. $\endgroup$ – John Samples Jan 24 at 9:10
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    $\begingroup$ I think the issue John raises is that you need to justify your contradiction: yes, we have that $s \geq k$, but the contradiction is supposed to stem from the fact that $s$ is no longer the best upper bound. I.e. we should justify why $s- \varepsilon /2$ is still an upper bound. But I guess that's alright since the whole interval $(s-\varepsilon/2,s)$ is in the complement of $K$ and by definition of $s$ so is $(s,+\infty)$. Hence $K \subset (s-\varepsilon/2,+\infty)^c$ and $s-\varepsilon/2$ is an upper bound, a contradiction. $\endgroup$ – guidoar Jan 24 at 9:11

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