11.1 The Real Field and Metric Space Topology

11.1 The Real Field and Metric Space Topology

Introductory real analysis formalizes the calculus of the real line. On the GRE Mathematics Subject Test, questions emphasize metric space topology, completeness axioms, topological set classifications, and the canonical Cantor set.

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The Real Numbers as a Complete Ordered Field

The real numbers $\mathbb{R}$ form a complete ordered field. While $\mathbb{Q}$ is also an ordered field, $\mathbb{R}$ is uniquely distinguished by the Least Upper Bound (Completeness) Axiom:

• Supremum ($\sup S$): For a nonempty set $S \subset \mathbb{R}$ bounded above, $s = \sup S$ satisfies $x \le s$ for all $x \in S$, and for every $\varepsilon > 0$, there exists $x \in S$ with $x > s - \varepsilon$.
• Infimum ($\inf S$): Dually, $m = \inf S$ satisfies $x \ge m$ for all $x \in S$, and for every $\varepsilon > 0$, there exists $x \in S$ with $x < m + \varepsilon$.
• Completeness: Every nonempty subset bounded above in $\mathbb{R}$ has a supremum in $\mathbb{R}$. In contrast, $\{q \in \mathbb{Q} : q^2 < 2\}$ has no supremum in $\mathbb{Q}$, proving $\mathbb{Q}$ is incomplete.

Archimedean Property and Density

• Archimedean Property: For any $x \in \mathbb{R}$, there exists $n \in \mathbb{N}$ with $n > x$. Equivalently, for any $\varepsilon > 0$, there exists $n \in \mathbb{N}$ with $1/n < \varepsilon$.
• Density of $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$: Between any two distinct reals $a < b$, there exists a rational $q$ and an irrational $r$ such that $a < q < r < b$.

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Metric Spaces and Topological Classifications

A metric space $(M, d)$ is a set $M$ with distance $d: M \times M \to [0, \infty)$ satisfying non-negativity ($d(x, y) \ge 0$, with equality iff $x = y$), symmetry ($d(x, y) = d(y, x)$), and the triangle inequality ($d(x, z) \le d(x, y) + d(y, z)$). Standard metrics on $\mathbb{R}^n$ include Euclidean $d_2$, taxicab $d_1$, and supremum $d_\infty$. In the discrete metric ($d(x, y) = 1$ for $x \neq y$), all subsets are clopen.

Neighborhoods and Set Topology

For open ball $B(x, r) = \{y \in M : d(x, y) < r\}$:

• Open Set: Every point has an open ball contained in $U$. Arbitrary unions and finite intersections of open sets are open.
• Closed Set: Complement $M \setminus F$ is open. Contains all its limit points.
• Interior ($E^\circ$): The union of all open subsets of $E$; largest open set in $E$.
• Limit Point ($E'$): $x \in M$ such that every punctured ball $(B(x, r) \setminus \{x\}) \cap E \neq \emptyset$.
• Closure ($\bar{E}$): $\bar{E} = E \cup E'$; smallest closed set containing $E$.
• Boundary ($\partial E$): $\partial E = \bar{E} \setminus E^\circ = \bar{E} \cap \overline{M \setminus E}$.
• Dense Set: $\bar{E} = M$. A space with a countable dense subset is separable (e.g., $\mathbb{Q} \subset \mathbb{R}$).

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The Cantor Ternary Set

Constructed by iteratively removing open middle thirds from $C_0 = [0, 1]$, the Cantor set $\mathcal{C} = \bigcap_{k=1}^\infty C_k$ has five core GRE properties:

1. Compactness: Closed and bounded in $\mathbb{R}$, hence compact.
2. Lebesgue Measure Zero: Total length removed is $\sum_{k=1}^\infty \frac{2^{k-1}}{3^k} = 1$, leaving $\lambda(\mathcal{C}) = 0$.
3. Uncountable: Base-3 expansions with digits $\{0, 2\}$ biject with $\{0, 1\}^\mathbb{N}$, so $|\mathcal{C}| = 2^{\aleph_0} = \mathfrak{c}$.
4. Empty Interior: $\operatorname{int}(\mathcal{C}) = \emptyset$; contains no open intervals.
5. Perfect: Closed with no isolated points ($\mathcal{C}' = \mathcal{C}$).

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Comparison Table: Subsets of $\mathbb{R}$

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Step-by-Step Worked Problems

Problem 1: Accumulation Points and Boundary

Find $\sup S$, $S'$, and $\partial S$ for $S = \{(-1)^n + 1/n : n \in \mathbb{Z}^+\} \cup (2, 4)$.

Solution:

1. Sequence terms: Even terms $1 + 1/n \in (1, 3/2]$; odd terms $-1 + 1/n \in (-1, 0]$.
2. Interval $(2, 4)$ yields $\sup S = 4$.
3. Limit points $S'$: Even terms converge to $1$; odd terms to $-1$. The interval $(2, 4)$ gives $[2, 4]$. Thus $S' = \{-1, 1\} \cup [2, 4]$.
4. Boundary $\partial S$: All sequence points are isolated, so each is in $\partial S$. Limit points $\{-1, 1\} \subset \partial S$. The interval contributes $\{2, 4\}$. Thus $\partial S = S_{seq} \cup \{-1, 1, 2, 4\}$.

Problem 2: Topological Distributivity Failure

Show that $\overline{A \cap B} \neq \bar{A} \cap \bar{B}$ in general.

Solution:

1. Let $A = (0, 1)$ and $B = (1, 2)$ in $\mathbb{R}$.
2. $A \cap B = \emptyset$, so $\overline{A \cap B} = \emptyset$.
3. Closures are $\bar{A} = [0, 1]$ and $\bar{B} = [1, 2]$, so $\bar{A} \cap \bar{B} = \{1\}$.
4. Since $\emptyset \neq \{1\}$, the closure of an intersection does not equal the intersection of closures.

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GRE Exam Traps & Pitfalls

Trap 1: "Open" and "Closed" Are Not Antonyms Sets need not be either open or closed. Subsets like $[0, 1)$ and $\mathbb{Q}$ are neither open nor closed, while $\emptyset$ and $\mathbb{R}$ are clopen in $\mathbb{R}$.

Trap 2: Assuming Boundary Points Belong to the Set Boundary points belong to $E$ iff $E$ contains them. For $E = (0, 1)$, $\partial E = \{0, 1\}$, but $\partial E \cap E = \emptyset$. A set is closed iff $\partial E \subseteq E$.

Trap 3: Equating Measure Zero with Countability The Cantor set has measure zero but is uncountable ($|\mathcal{C}| = 2^{\aleph_0}$). Never deduce countability from measure zero.