Unit 2: Topology of the Real Line: Open Sets, Compactness & Heine-Borel

1. The Standard Metric on the Real Line

The topological structure of $\mathbb{R}$ is induced by the Euclidean metric (distance function):

This distance function satisfies the three axioms of a metric:

1. Positivity & Identity of Indiscernibles: $d(x, y) \ge 0$, and $d(x, y) = 0 \iff x = y$.

2. Symmetry: $d(x, y) = d(y, x)$.

3. Triangle Inequality: $d(x, z) \le d(x, y) + d(y, z)$.

Definition 2.1 ($\epsilon$-Neighborhoods and Open Balls): For any point $x_0 \in \mathbb{R}$ and radius $\epsilon > 0$, the $\epsilon$-neighborhood (or open ball) centered at $x_0$ is defined as:

The deleted $\epsilon$-neighborhood is:

2. Open Sets and Their Fundamental Properties

Definition 2.2 (Open Set): A subset $U \subseteq \mathbb{R}$ is called an open set if every point of $U$ is an interior point:

Theorem 2.1 (Topological Properties of Open Sets in $\mathbb{R}$):

1. The empty set $\emptyset$ and the whole space $\mathbb{R}$ are open sets.
2. The union of an arbitrary (finite or infinite) collection of open sets is open:

1. The intersection of any finite collection of open sets is open:

Rigorous Proof of Finite Intersection Property:

Let $x \in \bigcap_{i=1}^n U_i$. Then $x \in U_i$ for each $i \in \{1, 2, \dots, n\}$. Since each $U_i$ is open, there exist positive radii $\epsilon_i > 0$ such that $V_{\epsilon_i}(x) \subseteq U_i$. Define:

Since this is a minimum of finitely many strictly positive numbers, $\epsilon > 0$. Now, for any $y \in V_\epsilon(x)$, $|y - x| < \epsilon \le \epsilon_i$ for all $i \in \{1, \dots, n\}$. Hence $y \in V_{\epsilon_i}(x) \subseteq U_i$ for every $i$, which implies $y \in \bigcap_{i=1}^n U_i$. Thus $V_\epsilon(x) \subseteq \bigcap_{i=1}^n U_i$, proving that the finite intersection is open. $\blacksquare$

Remark (Counterexample for Infinite Intersections): An arbitrary intersection of open sets is not necessarily open. For example, let $U_n = \left(-\frac{1}{n}, \frac{1}{n}\right)$ for $n \in \mathbb{N}$. Each $U_n$ is an open interval. However:

The single-point set $\{0\}$ is closed, not open!

3. Limit Points, Isolated Points & Closed Sets

Definition 2.3 (Limit Point / Accumulation Point): Let $A \subseteq \mathbb{R}$. A point $p \in \mathbb{R}$ is a limit point (or accumulation point) of $A$ if every deleted neighborhood of $p$ contains at least one point of $A$:

Note that $p$ need not be an element of $A$. The set of all limit points of $A$ is called the derived set of $A$, denoted $A'$.

Definition 2.4 (Isolated Point): A point $a \in A$ is an isolated point of $A$ if there exists $\epsilon > 0$ such that $V_\epsilon(a) \cap A = \{a\}$. That is, $a \in A$ but $a \notin A'$.

Definition 2.5 (Closed Set): A subset $F \subseteq \mathbb{R}$ is closed if its complement $\mathbb{R} \setminus F = F^c$ is open.

Theorem 2.2 (Characterization of Closed Sets via Limit Points): A subset $F \subseteq \mathbb{R}$ is closed if and only if it contains all of its limit points:

Line-by-Line Proof:

$(\implies)$ Assume $F$ is closed, so $F^c$ is open. Let $p \in F'$ be a limit point of $F$. Suppose for contradiction that $p \notin F$. Then $p \in F^c$. Since $F^c$ is open, there exists $\epsilon > 0$ such that $V_\epsilon(p) \subseteq F^c$. This implies $V_\epsilon(p) \cap F = \emptyset$. Consequently, $V_\epsilon^*(p) \cap F = \emptyset$, which contradicts the definition of $p$ being a limit point of $F$! Therefore, $p \in F$, showing $F' \subseteq F$.

$(\impliedby)$ Assume $F' \subseteq F$. We must show that $F^c$ is open. Let $x \in F^c$. Since $x \notin F$ and $F' \subseteq F$, $x \notin F'$. Because $x$ is not a limit point of $F$, there exists some $\epsilon > 0$ such that:

Since $x \notin F$, this implies $V_\epsilon(x) \cap F = \emptyset$, which means $V_\epsilon(x) \subseteq F^c$. Thus every $x \in F^c$ is an interior point of $F^c$, so $F^c$ is open and $F$ is closed. $\blacksquare$

1. Interior, Exterior, and Boundary

Definition 2.6 (Interior, Exterior, Boundary): Let $A \subseteq \mathbb{R}$.

1. The interior of $A$, denoted $\operatorname{int}(A)$ or $A^\circ$, is the set of all interior points of $A$:

It is the largest open set contained within $A$, and is equal to the union of all open sets contained in $A$.

1. The exterior of $A$, denoted $\operatorname{ext}(A)$, is the interior of its complement: $\operatorname{ext}(A) = \operatorname{int}(A^c)$.
2. The boundary of $A$, denoted $\partial A$ or $\operatorname{bd}(A)$, is the set of points whose neighborhoods intersect both $A$ and $A^c$:

Theorem 2.3 (Partition of the Real Line): For any subset $A \subseteq \mathbb{R}$, the ambient space decomposes into the disjoint union:

2. Topological Closure and Kuratowski Axioms

Definition 2.7 (Closure): The closure of $A$, denoted $\overline{A}$ or $\operatorname{cl}(A)$, is defined as the union of $A$ and its derived set:

Theorem 2.4 (Properties of Closure):

1. $\overline{A}$ is closed.
2. $\overline{A}$ is the smallest closed set containing $A$:

1. $A$ is closed if and only if $A = \overline{A}$.
2. $\overline{A} = \operatorname{int}(A) \cup \partial A$.
3. A point $x \in \overline{A}$ if and only if every neighborhood $V_\epsilon(x)$ contains at least one point of $A$:

3. Dense Sets and Nowhere Dense Sets

Definition 2.8 (Dense and Nowhere Dense Subsets):

1. A subset $D \subseteq \mathbb{R}$ is dense in $\mathbb{R}$ if $\overline{D} = \mathbb{R}$.

Equivalently, every non-empty open interval $(a, b)$ contains a point of $D$.

1. A subset $E \subseteq \mathbb{R}$ is nowhere dense (or rare) if the interior of its closure is empty:

Classical Examples:

• The rationals $\mathbb{Q}$ and irrationals $\mathbb{R} \setminus \mathbb{Q}$ are both dense in $\mathbb{R}$: $\overline{\mathbb{Q}} = \mathbb{R}$ and $\overline{\mathbb{R} \setminus \mathbb{Q}} = \mathbb{R}$.
• The integers $\mathbb{Z}$ have $\overline{\mathbb{Z}} = \mathbb{Z}$ and $\operatorname{int}(\mathbb{Z}) = \emptyset$, so $\mathbb{Z}$ is nowhere dense in $\mathbb{R}$.
• The Cantor ternary set $\mathcal{C}$ is closed and contains no intervals: $\operatorname{int}(\mathcal{C}) = \emptyset$, making it nowhere dense.

1. Infinite Sets in Bounded Regions

A fundamental question asked by Bernard Bolzano and Karl Weierstrass was: what conditions guarantee that a set has at least one accumulation point? Finite sets clearly cannot have accumulation points, since any finite set can be isolated by taking $\epsilon$ smaller than the minimal distance between any two distinct points. What happens when an infinite set is confined to a bounded domain?

Theorem 2.5 (Bolzano-Weierstrass Theorem for Sets): Every bounded, infinite subset of $\mathbb{R}$ has at least one limit point in $\mathbb{R}$.

Complete Rigorous Proof (Bisection Method & Completeness):

Let $S \subset \mathbb{R}$ be an infinite and bounded set. Since $S$ is bounded, there exists a closed bounded interval $I_0 = [a_0, b_0]$ such that $S \subseteq I_0$. Let $L_0 = b_0 - a_0 > 0$ be the length of $I_0$.

We construct a nested sequence of closed intervals $\{I_n\}_{n=0}^\infty$ by induction:

1. Divide $I_0 = [a_0, b_0]$ at its midpoint $m_0 = \frac{a_0 + b_0}{2}$ into two equal subintervals:

1. Since $S \cap I_0$ is infinite, at least one of these two subintervals must contain infinitely many points of $S$ (if both were finite, their union would be finite, a contradiction).
2. Select one subinterval containing infinitely many points of $S$, and denote it $I_1 = [a_1, b_1]$. Its length is $L_1 = \frac{L_0}{2}$.
3. Continuing this bisection inductively, for each $n \in \mathbb{N}$, we obtain a closed interval $I_n = [a_n, b_n]$ satisfying:

• $I_{n} \subseteq I_{n-1}$.
• $I_n$ contains infinitely many points of $S$.
• The length of $I_n$ is $b_n - a_n = \frac{b_0 - a_0}{2^n}$.

Now, observe the sequence of left endpoints $\{a_n\}_{n=0}^\infty$:

• Since $I_n \subseteq I_{n-1}$, we have $a_{n-1} \le a_n \le b_n \le b_0$ for all $n$.
• Thus, the set $A = \{a_n : n \ge 0\}$ is non-empty and bounded above by $b_0$.

By the Completeness Axiom, $A$ has a supremum:

Similarly, $p = \inf \{b_n : n \ge 0\}$, and by the Nested Interval Property:

We claim that $p$ is a limit point of $S$. Let $\epsilon > 0$ be arbitrary. By the Archimedean property, choose $N \in \mathbb{N}$ large enough such that:

Since $p \in I_N = [a_N, b_N]$ and the length of $I_N$ is strictly less than $\epsilon$, for any $x \in I_N$:

Because $I_N$ contains infinitely many points of $S$, the neighborhood $V_\epsilon(p)$ contains infinitely many points of $S$. In particular, $V_\epsilon^*(p) \cap S$ contains infinitely many points, so it is certainly non-empty. Since $\epsilon > 0$ was arbitrary, $p$ is a limit point of $S$. $\blacksquare$

1. Open Covers and Topological Compactness

In classical Euclidean analysis, compactness is the ultimate finiteness property. It allows mathematicians to pass from local properties (which hold in small $\epsilon$-neighborhoods of points) to global properties (which hold uniformly over an entire set).

Definition 2.9 (Open Cover and Subcover): Let $K \subseteq \mathbb{R}$.

1. A collection of open sets $\mathcal{U} = \{ U_\alpha \}_{\alpha \in \Lambda}$ is an open cover of $K$ if:

1. A subcover is a sub-collection $\mathcal{U}' \subseteq \mathcal{U}$ that still covers $K$:

1. If $\mathcal{U}'$ contains only finitely many sets, it is a finite subcover.

Definition 2.10 (Compact Set): A subset $K \subseteq \mathbb{R}$ is compact if every open cover of $K$ has a finite subcover.

2. The Heine-Borel Theorem

Theorem 2.6 (Heine-Borel Theorem): A subset $K \subseteq \mathbb{R}$ is compact if and only if $K$ is closed and bounded.

Line-by-Line Proof:

$(\implies)$ Compactness implies Closed and Bounded.

1. $K$ is bounded:

For each $n \in \mathbb{N}$, define the open interval $U_n = (-n, n)$. Then $\bigcup_{n=1}^\infty U_n = \mathbb{R} \supset K$. Thus $\{U_n\}_{n=1}^\infty$ is an open cover of $K$. Since $K$ is compact, there exists a finite subcover:

Let $M = \max \{n_1, n_2, \dots, n_k\}$. Then $K \subseteq (-M, M)$, which proves that $K$ is bounded.

2. $K$ is closed:

We show $K^c = \mathbb{R} \setminus K$ is open. Let $y \in K^c$. For each $x \in K$, since $x \ne y$, let $r_x = \frac{|x - y|}{2} > 0$. Define the disjoint open neighborhoods:

Notice that $V_x \cap W_x = \emptyset$. The family $\{V_x\}_{x \in K}$ forms an open cover of $K$: $K \subseteq \bigcup_{x \in K} V_x$. Since $K$ is compact, there exists a finite subcover:

Define $W = \bigcap_{i=1}^m W_{x_i}$. Since $W$ is a finite intersection of open balls centered at $y$, $W$ is an open neighborhood of $y$. Moreover, for each $i \in \{1, \dots, m\}$, $W \cap V_{x_i} \subseteq W_{x_i} \cap V_{x_i} = \emptyset$. Therefore, $W \cap \left(\bigcup_{i=1}^m V_{x_i}\right) = \emptyset \implies W \cap K = \emptyset$. Hence $W \subseteq K^c$, proving that $K^c$ is open. Thus $K$ is closed.

$(\impliedby)$ Closed and Bounded implies Compact. Let $K \subseteq \mathbb{R}$ be closed and bounded. Since $K$ is bounded, there exists a closed bounded interval $[a, b]$ such that $K \subseteq [a, b]$. We first prove that the closed interval $[a, b]$ is compact (this is the core lemma):

Lemma: Every closed interval $[a, b]$ is compact. Proof of Lemma: Let $\mathcal{U} = \{U_\alpha\}$ be an open cover of $[a, b]$. Define the set:

• $a \in S$ because $a \in [a, b]$, so $a \in U_{\alpha_0}$ for some $\alpha_0$, hence $[a, a] = \{a\} \subseteq U_{\alpha_0}$. Thus $S \ne \emptyset$.
• $S$ is bounded above by $b$.

By the Completeness Axiom, $c = \sup S$ exists and $a \le c \le b$. Since $c \in [a, b]$, $c \in U_{\alpha_c}$ for some open set $U_{\alpha_c} \in \mathcal{U}$. Since $U_{\alpha_c}$ is open, there is $\epsilon > 0$ such that $(c - \epsilon, c + \epsilon) \subseteq U_{\alpha_c}$. By the definition of supremum, there exists $x_1 \in S$ with $c - \epsilon < x_1 \le c$. Since $x_1 \in S$, $[a, x_1]$ is covered by finitely many sets $U_1, \dots, U_k \in \mathcal{U}$. Then $[a, x_1] \cup (c - \epsilon, c + \epsilon)$ is covered by $\{U_1, \dots, U_k, U_{\alpha_c}\}$. Thus, any point $x \in [a, c + \epsilon) \cap [a, b]$ has $[a, x]$ covered by finitely many sets. If $c < b$, we could choose $x \in (c, c + \epsilon) \cap [a, b]$, contradicting that $c = \sup S$. Hence $c = b$. Furthermore, $b \in S$ because $[a, b]$ is covered by $\{U_1, \dots, U_k, U_{\alpha_c}\}$. Thus $[a, b]$ is compact.

Finally, since $K \subseteq [a, b]$ is closed, $K^c$ is open. If $\mathcal{U}$ is an open cover of $K$, then $\mathcal{U} \cup \{K^c\}$ is an open cover of $[a, b]$. Since $[a, b]$ is compact, there exists a finite subcover of $[a, b]$ from $\mathcal{U} \cup \{K^c\}$. Discarding $K^c$ leaves a finite sub-collection of $\mathcal{U}$ that still covers $K$. Therefore, $K$ is compact. $\blacksquare$

1. Topological Connectedness and Separations

Definition 2.11 (Separated Sets and Connectedness):

1. Two non-empty subsets $A, B \subset \mathbb{R}$ are separated if:

1. A subset $E \subseteq \mathbb{R}$ is disconnected (or separated) if it can be written as the union of two non-empty separated sets:

1. A subset $E \subseteq \mathbb{R}$ is connected if it is not disconnected.

Remark (Clopen Sets): An equivalent definition states that a subset $E$ is connected if the only subsets of $E$ that are both open and closed in the subspace topology of $E$ are $\emptyset$ and $E$ itself. In $\mathbb{R}$, the only sets that are both open and closed (clopen) are $\emptyset$ and $\mathbb{R}$.

2. Characterization of Connected Subsets of $\mathbb{R}$

The geometry of the real line imposes a very strict condition on connectedness: a subset of $\mathbb{R}$ is connected if and only if it has no "gaps" — that is, if it is an interval!

Theorem 2.7 (Connected Subsets of $\mathbb{R}$ are Intervals): A subset $E \subseteq \mathbb{R}$ is connected if and only if $E$ is an interval. That is, $E$ has the intermediate point property:

Complete Proof:

$(\impliedby)$ We prove that if $E$ satisfies the intermediate point property, then $E$ is connected. Assume for contradiction that $E$ is disconnected. Then $E = A \cup B$ where $A, B$ are non-empty and separated ($\overline{A} \cap B = \emptyset = A \cap \overline{B}$). Choose $a \in A$ and $b \in B$. Without loss of generality, assume $a < b$. Since $E$ has the intermediate point property, $[a, b] \subseteq E$. Define the set:

Since $a \in S$, $S \ne \emptyset$. Moreover, $S$ is bounded above by $b$. By the Completeness Axiom, $c = \sup S$ exists, and $a \le c \le b$. Thus $c \in [a, b] \subseteq E = A \cup B$. Therefore, either $c \in A$ or $c \in B$.

• Case 1: $c \in A$.

Since $\overline{A} \cap B = \emptyset$, $c \notin \overline{B}$. Thus $c \ne b$, so $c < b$. Since $c \notin \overline{B}$, there exists $\epsilon > 0$ such that $(c - \epsilon, c + \epsilon) \cap B = \emptyset$. Can choose $\epsilon < b - c$, so $(c, c + \epsilon) \subseteq [a, b]$. Since $(c, c + \epsilon) \cap B = \emptyset$ and $[a, b] \subseteq A \cup B$, we must have $(c, c + \epsilon) \subseteq A$. This contradicts that $c = \sup S$, because any point in $(c, c + \epsilon)$ is an element of $S$ strictly greater than $c$!

• Case 2: $c \in B$.

Since $A \cap \overline{B} = \emptyset$, $c \notin \overline{A}$. Thus $c \ne a$ (since $a \in A$). Since $c \notin \overline{A}$, there exists $\epsilon > 0$ such that $(c - \epsilon, c + \epsilon) \cap A = \emptyset$. This means that for all $x \in (c - \epsilon, c]$, $x \notin A$. Hence $c - \epsilon$ is an upper bound for $S = A \cap [a, b]$, which strictly contradicts that $c = \sup S$!

In both cases we reach a contradiction. Hence $E$ must be connected.

$(\implies)$ Suppose $E$ is connected. Assume for contradiction that $E$ is not an interval. Then there exist $x, y \in E$ with $x < y$, and some point $z \in (x, y)$ such that $z \notin E$. Define:

Since $x \in A$ and $y \in B$, both $A$ and $B$ are non-empty, and $E = A \cup B$. Furthermore, $\overline{A} \subseteq (-\infty, z]$ so $\overline{A} \cap B \subseteq (-\infty, z] \cap (z, \infty) = \emptyset$. Similarly, $A \cap \overline{B} \subseteq (-\infty, z) \cap [z, \infty) = \emptyset$. Thus $A$ and $B$ are separated, so $E$ is disconnected, contradicting the premise. Therefore, $E$ must be an interval. $\blacksquare$