11.2 Compactness, Completeness, and Connectedness in R^n

11.2 Compactness, Completeness, and Connectedness in R^n

Compactness, completeness, and connectedness are the triad of topological concepts that govern analysis in Euclidean and metric spaces. The GRE Mathematics Subject Test evaluates both their theoretical equivalents and their standard counterexamples.

---

Compactness and the Heine-Borel Theorem

Open Coverings and Compactness

Let $(M, d)$ be a metric space and $K \subseteq M$.

• An open cover of $K$ is a family $\{U_\alpha\}{\alpha \in I}$ of open sets such that $K \subseteq \bigcup{\alpha \in I} U_\alpha$.
• A finite subcover is a finite subfamily $\{U_{\alpha_1}, \dots, U_{\alpha_k}\}$ with $K \subseteq \bigcup_{j=1}^k U_{\alpha_j}$.
• A set $K$ is compact if every open cover of $K$ admits a finite subcover.

The Heine-Borel Theorem in $\mathbb{R}^n$

In finite-dimensional Euclidean space $\mathbb{R}^n$, the Heine-Borel Theorem provides a concrete geometric characterization: $K \\subset \\mathbb{R}^n \\text{ is compact} \\iff K \\text{ is closed and bounded}.$

Failure of Heine-Borel in General Metric Spaces

In any metric space, compact sets are always closed and bounded. However, closed and bounded does NOT imply compact in general:

1. Discrete Metric on Infinite Sets: If $X$ is infinite and $d(x, y) = 1$ for $x \neq y$, $X$ is closed and bounded (diameter 1), but the open cover of singletons $\{\{x\}\}_{x \in X}$ has no finite subcover.
2. Infinite-Dimensional Spaces: In $\ell^2$, the closed unit ball $\bar{B}(0, 1)$ is closed and bounded, but the orthonormal basis vectors $e_n$ satisfy $d(e_n, e_m) = \sqrt{2}$ ($n \neq m$), admitting no convergent subsequence.

---

Sequential Compactness and Bolzano-Weierstrass

• Sequential Compactness: A metric space is sequentially compact if every sequence has a subsequence converging to a point in the space.
• Metric Equivalence: In all metric spaces, compactness is strictly equivalent to sequential compactness, and also to completeness plus total boundedness.
• Bolzano-Weierstrass Theorem: Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. Equivalently, every infinite bounded subset of $\mathbb{R}^n$ has an accumulation point.

---

Metric Completeness and Cantor's Intersection Theorem

• Cauchy Sequence: A sequence $(x_n)$ is Cauchy if for all $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that $d(x_n, x_m) < \varepsilon$ for all $n, m \ge N$. Every convergent sequence in a metric space is Cauchy.
• Complete Metric Space: A space $(M, d)$ is complete if every Cauchy sequence converges to a limit in $M$. $\mathbb{R}^n$ is complete; $\mathbb{Q}$ and $(0, 1)$ with $|x - y|$ are incomplete. Every compact metric space is complete.
• Cantor's Intersection Theorem: In a complete metric space, if $\{F_n\}{n=1}^\infty$ is a nested sequence of nonempty closed sets ($F{n+1} \subseteq F_n$) with $\lim_{n \to \infty} \operatorname{diam}(F_n) = 0$, then $\bigcap_{n=1}^\infty F_n = \{p\}$ contains exactly one point.

---

Connectedness and Path-Connectedness

• Connectedness: A metric space $M$ is disconnected if $M = U \cup V$ for nonempty, disjoint open sets $U, V$. Otherwise, $M$ is connected. Equivalently, the only clopen subsets of $M$ are $\emptyset$ and $M$.
• Subsets of $\mathbb{R}$: A subset of $\mathbb{R}$ is connected if and only if it is an interval.
• Path-Connectedness: A space $M$ is path-connected if any two points can be joined by a continuous path $\gamma: [0, 1] \to M$.
  • Path-connectedness implies connectedness.
  • The converse is false: The topologist's sine curve $T = \{(x, \sin(1/x)) : x \in (0, 1]\} \cup \{(0, y) : y \in [-1, 1]\}$ is connected and compact in $\mathbb{R}^2$, but not path-connected.
  • In $\mathbb{R}^n$, any open connected set is path-connected.

---

Comparison Table: Compactness, Completeness, and Connectedness

---

Step-by-Step Worked Problems

Problem 1: Compactness in Sequence Spaces

Let $S = \{x \in \ell^\infty : \|x\|_\infty \le 1\}$. Is $S$ compact?

Solution:

1. $S$ is closed and bounded in the supremum norm.
2. Consider sequence $x^{(n)} = e_n = (0, \dots, 0, 1, 0, \dots)$.
3. For $n \neq m$, $\|e_n - e_m\|_\infty = \max(1, 1) = 1$.
4. No subsequence can be Cauchy, so no subsequence converges.
5. $S$ is not sequentially compact, hence not compact.

Problem 2: Connectedness of Hyperbolic Level Sets

Determine whether $S = \{(x, y) \in \mathbb{R}^2 : x^2 - y^2 = 1\}$ is connected.

Solution:

1. The hyperbola has two branches: $x \ge 1$ and $x \le -1$.
2. Let $U = \{(x, y) : x > 0\}$ and $V = \{(x, y) : x < 0\}$. Both are open in $\mathbb{R}^2$.
3. $S \subset U \cup V$, with $S \cap U \neq \emptyset$ and $S \cap V \neq \emptyset$, while $(S \cap U) \cap (S \cap V) = \emptyset$.
4. $S$ is disconnected into two disjoint clopen components.

---

GRE Exam Traps & Pitfalls

Trap 1: Applying Heine-Borel Beyond $\mathbb{R}^n$ "Closed and bounded $\iff$ compact" is exclusively true in finite-dimensional Euclidean space. In function spaces and $\ell^p$, closed bounded sets are non-compact.

Trap 2: Concluding Path-Connectedness from Connectedness Connected does not imply path-connected. The topologist's sine curve is connected (as the closure of a connected graph) but fails path-connectedness.

Trap 3: Omitting Shrinking Diameters in Cantor's Theorem In complete spaces, nested closed sets must satisfy $\lim \operatorname{diam}(F_n) = 0$. Sets like $F_n = [n, \infty)$ are closed, nested, and nonempty, but have empty intersection.