Unit 5: Continuous Real Functions: Epsilon-Delta Rigor, Compactness & Intermediate Values

1. The Cauchy-Weierstrass $\epsilon$-$\delta$ Limit Definition

Definition 5.1 (Limit of a Function): Let $E \subseteq \mathbb{R}$, $f: E \to \mathbb{R}$, and let $c \in \mathbb{R}$ be a limit point of $E$. A real number $L \in \mathbb{R}$ is the limit of $f(x)$ as $x$ approaches $c$, denoted:

if for every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x \in E$:

In terms of topological neighborhoods: $x \in V_\delta^*(c) \cap E \implies f(x) \in V_\epsilon(L)$.

2. Heine's Sequential Characterization of Limits

Karl Weierstrass and Eduard Heine discovered that limits of functions can be completely characterized through limits of sequences, linking differential calculus with sequential analysis.

Theorem 5.1 (Heine's Sequential Criterion for Functional Limits): Let $f: E \to \mathbb{R}$ and let $c$ be a limit point of $E$. Then:

if and only if for every sequence $(x_n)_{n=1}^\infty$ in $E \setminus \{c\}$ that converges to $c$:

Complete Rigorous Proof:

$(\implies)$ Assume $\lim_{x \to c} f(x) = L$. Let $\epsilon > 0$. By Definition 5.1, there exists $\delta > 0$ such that:

Now let $(x_n)$ be any sequence in $E \setminus \{c\}$ such that $x_n \to c$. By Definition 3.2 of sequence convergence, corresponding to this $\delta > 0$, there exists an index $N \in \mathbb{N}$ such that for all $n \ge N$:

Since $x_n \ne c$, we have $0 < |x_n - c| < \delta$ for all $n \ge N$. Therefore, for all $n \ge N$:

This proves that $f(x_n) \to L$ as $n \to \infty$.

$(\impliedby)$ We prove the contrapositive: if $\lim_{x \to c} f(x) \ne L$, then there exists a sequence $(x_n)$ in $E \setminus \{c\}$ with $x_n \to c$ such that $f(x_n) \not\to L$. Assume $\lim_{x \to c} f(x) \ne L$. Negating Definition 5.1:

For each $n \in \mathbb{N}$, choose $\delta = \frac{1}{n} > 0$. By the negated condition, there exists a point $x_n \in E$ such that:

Since $0 < |x_n - c| < \frac{1}{n}$, by the Squeeze Theorem, $x_n \to c$ with $x_n \ne c$. However, since $|f(x_n) - L| \ge \epsilon_0$ for all $n$, the sequence $f(x_n)$ cannot converge to $L$. This establishes the contrapositive, completing the proof. $\blacksquare$

Application (Proving Non-Existence of Limits): To prove that $\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$ does not exist: Pick $x_n = \frac{1}{2\pi n} \to 0$, so $\sin(1/x_n) = \sin(2\pi n) = 0 \to 0$. Pick $y_n = \frac{1}{2\pi n + \pi/2} \to 0$, so $\sin(1/y_n) = \sin(2\pi n + \pi/2) = 1 \to 1$. Since two sequences converging to 0 yield different functional limits, $\lim_{x \to 0} \sin(1/x)$ does not exist!

1. Continuity at a Point and on Sets

Definition 5.2 (Continuity at a Point): Let $f: E \to \mathbb{R}$ and $c \in E$. The function $f$ is continuous at $c$ if for every $\epsilon > 0$, there exists $\delta > 0$ such that for all $x \in E$:

If $c$ is a limit point of $E$, continuity at $c$ is equivalent to:

If $c$ is an isolated point of $E$, $f$ is automatically continuous at $c$. The function $f$ is continuous on $E$ if it is continuous at every point $c \in E$.

2. Topological Characterization: Preimages of Open Sets

In modern topology, continuity is defined entirely through the geometry of open sets, freeing the concept from metric coordinates.

Theorem 5.2 (Topological Continuity Theorem): A function $f: \mathbb{R} \to \mathbb{R}$ is continuous on $\mathbb{R}$ if and only if for every open set $V \subseteq \mathbb{R}$, the preimage $f^{-1}(V)$ is open in $\mathbb{R}$:

Complete Line-by-Line Proof:

$(\implies)$ Assume $f$ is continuous on $\mathbb{R}$. Let $V \subseteq \mathbb{R}$ be an open set. To show $f^{-1}(V)$ is open, let $x_0 \in f^{-1}(V)$. This means $y_0 = f(x_0) \in V$. Since $V$ is open, there exists an $\epsilon > 0$ such that the ball $B(y_0, \epsilon) = (y_0 - \epsilon, y_0 + \epsilon) \subseteq V$. Since $f$ is continuous at $x_0$, corresponding to this $\epsilon > 0$, there exists $\delta > 0$ such that:

In terms of balls:

Therefore, $B(x_0, \delta) \subseteq f^{-1}(V)$. Thus, every point of $f^{-1}(V)$ is an interior point, so $f^{-1}(V)$ is open.

$(\impliedby)$ Assume $f^{-1}(V)$ is open for every open set $V \subseteq \mathbb{R}$. Let $x_0 \in \mathbb{R}$ and let $\epsilon > 0$. The ball $V = B(f(x_0), \epsilon) = (f(x_0) - \epsilon, f(x_0) + \epsilon)$ is an open set. By hypothesis, $f^{-1}(V)$ is an open subset of $\mathbb{R}$. Notice that $x_0 \in f^{-1}(V)$ because $f(x_0) \in V$. Since $f^{-1}(V)$ is open, there exists $\delta > 0$ such that:

This means that for all $x$ with $|x - x_0| < \delta$, $x \in f^{-1}(V) \implies f(x) \in V = B(f(x_0), \epsilon) \implies |f(x) - f(x_0)| < \epsilon$. Therefore, $f$ is continuous at $x_0$. Since $x_0$ was arbitrary, $f$ is continuous on $\mathbb{R}$. $\blacksquare$

3. Classification of Discontinuities

Let $f$ be defined on an interval and let $c$ be an interior point. Let $f(c^+) = \lim_{x \to c^+} f(x)$ and $f(c^-) = \lim_{x \to c^-} f(x)$.

1. Removable Discontinuity: Both one-sided limits exist and are equal ($f(c^+) = f(c^-)$), but do not equal $f(c)$ (or $f(c)$ is undefined).

2. Jump Discontinuity (Discontinuity of the First Kind): Both one-sided limits exist as finite numbers, but $f(c^+) \ne f(c^-)$.

3. Essential Discontinuity (Discontinuity of the Second Kind): At least one of the one-sided limits does not exist (or is infinite), e.g., $\sin(1/x)$ at $x=0$.

1. Preservation of Compactness

Theorem 5.3 (Compactness Preservation Theorem): Let $K \subset \mathbb{R}$ be compact, and let $f: K \to \mathbb{R}$ be continuous. Then the image $f(K) = \{f(x) : x \in K\}$ is compact.

Proof:

Let $\{V_\alpha\}_{\alpha \in \Lambda}$ be an open cover of $f(K)$:

Taking preimages:

By Theorem 5.2, each $f^{-1}(V_\alpha)$ is open in the subspace topology of $K$. Thus $\{f^{-1}(V_\alpha)\}$ is an open cover of $K$. Since $K$ is compact, there exists a finite subcover:

Applying $f$ to both sides:

Thus, $\{V_{\alpha_1}, \dots, V_{\alpha_m}\}$ is a finite subcover of $f(K)$. Therefore, $f(K)$ is compact. $\blacksquare$

2. The Weierstrass Extreme Value Theorem

Theorem 5.4 (Weierstrass Extreme Value Theorem - EVT): If $f: [a, b] \to \mathbb{R}$ is continuous on a closed bounded interval $[a, b]$, then:

1. $f$ is bounded on $[a, b]$.
2. $f$ attains both its maximum and minimum values on $[a, b]$:

Complete Line-by-Line Proof:

By the Heine-Borel Theorem, $[a, b]$ is compact. By Theorem 5.3, the image $f([a, b]) \subset \mathbb{R}$ is compact. By the Heine-Borel Theorem again, $f([a, b])$ must be closed and bounded.

1. Since $f([a, b])$ is bounded, $f$ is bounded on $[a, b]$.
2. Let $M = \sup f([a, b])$ and $m = \inf f([a, b])$.

By the properties of supremum and infimum, $M$ and $m$ are limit points (or isolated points) of $f([a, b])$. In either case, $M, m \in \overline{f([a, b])}$. Since $f([a, b])$ is closed, $\overline{f([a, b])} = f([a, b])$. Therefore:

This means there exist points $x_{\max}, x_{\min} \in [a, b]$ such that:

Consequently, for all $x \in [a, b]$, $f(x_{\min}) \le f(x) \le f(x_{\max})$. $\blacksquare$

1. Preservation of Connectedness

Theorem 5.5 (Connectedness Preservation Theorem): Let $E \subseteq \mathbb{R}$ be connected, and let $f: E \to \mathbb{R}$ be continuous. Then the image $f(E)$ is connected.

Proof:

We prove this by contradiction. Suppose $f(E)$ is disconnected. Then there exist non-empty separated sets $A, B$ such that $f(E) = A \cup B$, with $\overline{A} \cap B = \emptyset$ and $A \cap \overline{B} = \emptyset$. Define:

Since $A \ne \emptyset$ and $B \ne \emptyset$, $U$ and $V$ are non-empty subsets of $E$, and $E = U \cup V$. Furthermore, by continuity, $f(\overline{U}) \subseteq \overline{f(U)} = \overline{A}$. Since $\overline{A} \cap B = \emptyset$, we have $f(\overline{U}) \cap B = \emptyset$, which implies $\overline{U} \cap V = \emptyset$. Similarly, $U \cap \overline{V} = \emptyset$. Thus $U$ and $V$ form a separation of $E$, which contradicts the hypothesis that $E$ is connected! Therefore, $f(E)$ is connected. $\blacksquare$

2. Bolzano's Intermediate Value Theorem

Theorem 5.6 (Bolzano's Intermediate Value Theorem - IVT): Let $f: [a, b] \to \mathbb{R}$ be continuous. If $u$ is any number strictly between $f(a)$ and $f(b)$, then there exists at least one $c \in (a, b)$ such that:

Elegant Topological Proof:

The interval $[a, b]$ is connected (Theorem 2.7). By Theorem 5.5, the image $f([a, b])$ is a connected subset of $\mathbb{R}$. By Theorem 2.7, connected subsets of $\mathbb{R}$ are intervals! Since $f(a) \in f([a, b])$ and $f(b) \in f([a, b])$, the image interval contains the entire segment between $f(a)$ and $f(b)$. Since $u$ lies between $f(a)$ and $f(b)$, $u \in f([a, b])$. Therefore, there exists $c \in [a, b]$ such that $f(c) = u$. Since $u \ne f(a)$ and $u \ne f(b)$, $c \in (a, b)$. $\blacksquare$

1. Pointwise vs. Uniform Continuity

In standard continuity at a point $c$, the choice of $\delta$ depends on both $\epsilon$ AND the location $c$: $\delta(\epsilon, c)$. For example, for $f(x) = 1/x$ on $(0, 1)$, as $x \to 0$, the function steepens dramatically, forcing $\delta \to 0$ for a fixed $\epsilon$.

Definition 5.3 (Uniform Continuity): A function $f: E \to \mathbb{R}$ is uniformly continuous on $E$ if for every $\epsilon > 0$, there exists a $\delta > 0$ independent of the points such that:

2. The Heine-Cantor Theorem

Theorem 5.7 (Heine-Cantor Theorem): Let $K \subset \mathbb{R}$ be compact. If $f: K \to \mathbb{R}$ is continuous on $K$, then $f$ is uniformly continuous on $K$.

Complete Proof via Open Covers:

Let $\epsilon > 0$ be given. Since $f$ is continuous at each point $x \in K$, there exists a $\delta_x > 0$ such that:

For each $x \in K$, consider the open ball centered at $x$ with half-radius:

The family $\{U_x\}_{x \in K}$ forms an open cover of $K$: $K \subseteq \bigcup_{x \in K} U_x$. Since $K$ is compact, there exists a finite subcover:

Define $\delta$ to be the minimum of these finitely many positive half-radii:

Now, let $p, q \in K$ be any two points such that $|p - q| < \delta$. Since the finite subcover covers $K$, $p \in U_{x_k}$ for some $k \in \{1, 2, \dots, m\}$. This means:

Now, using the Triangle Inequality to estimate the distance between $q$ and $x_k$:

Since $|q - x_k| < \delta_{x_k}$, it follows by definition of $\delta_{x_k}$ that:

Finally, applying the Triangle Inequality to $|f(p) - f(q)|$:

Since $\delta$ depends solely on $\epsilon$ and the compact set $K$, $f$ is uniformly continuous on $K$. $\blacksquare$

3. Lipschitz Continuity

Definition 5.4 (Lipschitz Continuity): A function $f: E \to \mathbb{R}$ is Lipschitz continuous if there exists a constant $L > 0$ such that:

Proposition: Every Lipschitz function is uniformly continuous (choose $\delta = \epsilon / L$).