Verification of Proof: “$K$ is compact $\implies$ $\exists \inf K \in K$ and $\exists \sup K \in K$”

analysisconvergence-divergencegeneral-topologyreal-analysis

I hope life is going okay for everyone; I am hoping someone here can comment on aspects of my proof's accuracy, coherence, simplicity, readability, etc.. Anything really goes, I am trying to improve all fronts of mathematical intuition – please criticize as much of my work as you can. Note that the problem below is an exercise (3.3.1.) in Understanding Analysis ed.2 by S. Abbott.

Problem

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

Definitions

Compact Set: A set $K \subseteq \mathbb{R}$ is compact if every sequence in $K$ has a subsequence that converges to a limit that is also in $K$.

Closed Set: A set $F \subseteq R$ is closed if it contains its limit points.

Bounded Set: A set $A \subseteq \mathbb{R}$ is bounded if there exists $M > 0$ such that $|a|\leq M$ for all $a \in A$.

Theorem 2.5.2. – Subsequences of a convergent sequence converge to the same limit as the original sequence.

Theorem 3.3.4. – A set $K \subseteq \mathbb{R}$ is compact if and only if it is closed and bounded.

Lemma 1.3.8.

  1. Assume $s \in R$ is an upper bound for a set $A \subseteq \mathbb{R}$. Then, $s = \sup A$ if and only if, for every choice of $\epsilon > 0$, there exists an element $a \in A$ satisfying $s – \epsilon < a$.
  2. Assume $i \in R$ is a lower bound for a set $A \subseteq \mathbb{R}$. Then, $i = \inf A$ if and only if, for every choice of $\epsilon > 0$, there exists an element $a \in A$ satisfying $a < i + \epsilon $.

Axiom of Completeness: Every nonempty set of real numbers that is bounded above has a least upper bound and every nonempty set of real numbers that is bounded below has a greatest lower bound.

Limit Point: A point $x$ is a limit point of a set $A$ if every $\epsilon$-neighborhood $\mathcal{B}_{\epsilon}(x)$ of $x$ intersects the set $A$ at some point other than $x$.

My Proof

$\text{ }$ Suppose that a set $K \subseteq \mathbb{R}$ is compact and nonempty. Let $\alpha = \sup K$ for some $\alpha \in \mathbb{R}$. First we want to show $\alpha$ exists and $\alpha \in K$. By Theorem 3.3.4., $K$ is bounded, i.e. there exists $M > 0$ such that $-M \leq k \leq M$ for all $k \in K$. This implies that $\alpha$ exists by the Axiom of Completeness. Now let $\epsilon > 0$. By Lemma 1.3.8. we know that there exists an element $k \in K$ satisfying $\alpha – \epsilon < k$. In particular, we have $\alpha – \epsilon < k < \alpha + \epsilon$, which means that $k \in \mathcal{B}_{\epsilon}(\alpha)$. Moreover, $\alpha$ is a limit point of $K$ since $k \in \mathcal{B}_{\epsilon}(\alpha) \cap K$. Since $K$ is closed by Theorem 3.3.4., then we have $\alpha \in K$, as desired.

$\text{ }$ Let $\beta = \inf K$ for some $\beta \in \mathbb{R}$. What remains to be shown is that $\beta$ exists and $\beta \in K$. Since $K$ is bounded by Theorem 3.3.4., the Axiom of Completeness implies that $\beta$ exists. Let $\epsilon > 0$. By Lemma 1.3.8. we know that there exists an element $k \in K$ satisfying $\beta + \epsilon > k$. In particular, we have $\beta + \epsilon > k > \beta – \epsilon$, which means that $k \in \mathcal{B}_{\epsilon}(\beta)$. Since $k \in \mathcal{B}_{\epsilon}(\beta) \cap K$, we have that $\beta$ is a limit point of $K$. Thus, $\beta \in K$ since $K$ is closed.

Best Answer

The proof is fine. It's basically 3 observations:

  • $K$ is compact so $K$ is bounded so $\inf K$ and $\sup K$ exist.
  • Always $\sup K$ and $\inf K$ are limit points of $K$ (which is not strictly true, consider a finite set e.g, though they are adherence points of $K$).
  • $K$ compact implies is $K$ closed so it contains its limit points (or rather its adherent points).
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