# Real Analysis

Continuous Functions

# Continuity

Let \(X\) and \(Y\) be metric spaces, \(E \subset X\), \(p \in E\), \(f : E \rightarrow Y\). Say function \(f\) is *continuous at \(p\)* if for any \(\epsilon > 0\), there exists \(\delta > 0\) such that for any \(x \in E\), \(d(x, p) < \delta\) implies \(d(f(x), f(p)) < \epsilon\).

**Theorem.** If \(p\) is a limit point of \(E\) then \(f\) is continuous at \(p\) if and only if \(\lim\limits_{x \rightarrow p} f(x) = f(p)\). Also, if \(x_n\) is a convergent sequence then \(f\) is continuous if and only if \(\lim\limits_{n \rightarrow \infty} f(x_n) = f\left(\lim\limits_{n \rightarrow \infty} x_n\right)\).

**Corollary.** Given \(Y \subset \mathbb{R}\), sums and products of continuous functions are continuous; quotients \(\frac{f}{q}\) are continuous when \(g \neq 0\).

**Corollary.** \(f, g : X \rightarrow \mathbb{R}^k\) such that \(f = (f_1, f_2, \dots, f_k)\).

- \(f\) is continuous if and only if each \(f_i\) is continuous.
- \(f + g\) and \(f \cdot g\) are continuous.

**Theorem.** \(f : X \rightarrow Y\) is continuous if and only if for any open set \(U \subset Y\), \(f^{-1}(U)\) is open in \(X\).

*Proof.*

(\(\Rightarrow\)): Given \(U\) which is open in \(Y\), consider \(x \in f^{-1}(U)\). We’ll show \(x\) is an interior point. Since \(U\) is open and \(f(x) \in U\), there exists an open ball \(B_\epsilon(f(x)) \subset U\). By continuity of \(f\), there exists an open ball \(B_\delta(x)\) that maps into \(B_\epsilon(f(x)) \subset U\). This means \(B_\delta(x) \subset f^{-1}(U)\). So \(x\) is an interior point of \(f^{-1}(U)\).

(\(\Leftarrow\)): Fix arbitrary \(x \in X\) and \(\epsilon > 0\). Consider \(B_\epsilon(f(x))\). So, \(x \in f^{-1}(B_\epsilon(f(x)))\), which is open by assumption. So \(p\) is an interior point. Hence, there exists an open ball \(B_\delta(x) \subset f^{-1}(B_\epsilon(f(x)))\). This \(\delta\) has the required property. So \(f\) is continuous. QED.

## Consequences

**Theorem.** If \(f : X \rightarrow Y\), \(g : Y \rightarrow Z\), \(f\) and \(g\) are continuous then \(g \circ f\) is continuous.

*Proof.* Given \(U\) open in \(Z\), \(g^{-1}(U)\) is open in \(Y\), by continuity of \(g\). So \(f^{-1}(g^{-1}(U))\) is open in \(X\), by continuity of \(f\). But \(f^{-1}(g^{-1}(U)) = (g \circ f)^{-1}(U)\). QED.

**Theorem.** \(f : X \rightarrow Y\) is continuous if and only if for any closed set \(K \subset Y\), \(f^{-1}(K)\) is closed in \(X\).

**Theorem.** If \(f : X \rightarrow Y\), \(f\) is continuous, and \(X\) is compact, then \(f(X)\) is compact.

*Proof.* Consider any open cover \(\{V_\alpha\}\) of \(f(X)\). Let \(U_\alpha = f^{-1}(V_\alpha)\). By compactness of \(X\), there exists a finite subcover \(\{U_1, \dots, U_n\}\). Then the corresponding \(\{V_1, \dots V_n\}\) is a finite subcover for \(f(X)\), as desired. QED.

**Corollary.** If \(f : X \rightarrow \mathbb{R}^k\) and \(X\) is compact then \(f(X)\) is closed and bounded.

**Corollary.** If \(f : X \rightarrow \mathbb{R}\) and \(X\) is compact then \(f\) achieves its maximum and minimum.

**Theorem.** If \(f : X \rightarrow Y\), \(f\) is a bijection and is continuous, and \(X\) is compact then \(f^{-1}\) is continuous.

*Proof.* Suppose \(U\) is open in \(X\). So, \(U^c\) is closed. Since \(U^c\) is closed in a compact set \(X\), \(U^c\) is compact. Then its image \(f(U^c)\) compact. Since \(f(U^c)\) is compact, it is also closed. Hence \(f(U)\) is open. Thus, \(f^{-1}\) is continuous. QED.

**Theorem.** If \(f : X \rightarrow Y\) is continuous, \(E\) is a connected subset of \(X\), then \(f(E)\) is connected.

*Proof (by contradiction).* Suppose \(f(E)\) is not connected. Then \(f(E) = A \cup B\), a separation (i.e. \(A \neq \emptyset\), \(B \neq \emptyset\), and \(\overline{A} \cap B = A \cap \overline{B} = \emptyset\)). Notice \(K_A = f^{-1}(\overline{A})\) and \(K_B = f^{-1}(\overline{B})\) are closed since \(f\) is continuous. Also notice \(E_1 = f^{-1}(A) \cap E\) and \(E_2 = f^{-1}(B) \cap E\) are disjoint and non-empty. Since \(E_1 \subset K_A\), \(\overline{E_1} \subset K_A\). Similarly, \(\overline{E_2} \subset K_B\). Since \(K_A = f^{-1}(\overline{A})\), \(E_2 \subset f^{-1}(B)\), and \(\overline{A} \cap B = \emptyset\), \(K_A \cap E_2 = \emptyset\). Similarly, \(K_B \cap E_1 = \emptyset\). Hence, \(E\) is separated, which is a contradiction. QED.

**Theorem (Intermediate Value).** Given a continuous function \(f : [a, b] \rightarrow \mathbb{R}\), where \(f(a) < c < f(b)\), then there exists \(x \in (a, b)\) such that \(f(x) = c\).

*Proof (by contradiction).* Notice \([a, b]\) is connected. So, \(f([a, b])\) is connected. But if \(c\) is not achieved, then \(((c, +\infty) \cap f([a, b])) \cup ((-\infty, c) \cap f([a, b]))\) is a separation of \(f([a, b])\), a contradiction. QED.

The converse of the intermediate value theorem is false. Consider the topologist’s sine curve \[\begin{align} f(x) = \begin{cases} 0 & \text{if } x = 0 \\ \sin\left(\frac{1}{x}\right) & \text{otherwise} \end{cases} \end{align}\] The topologist’s sine curve is not continuous at \(0\) but it satisfies the intermediate value property.

# Uniform Continuity

A function \(f : X \rightarrow Y\) is *uniform continuous* on \(X\) if for every \(\epsilon > 0\), there exists \(\delta > 0\) such that for every \(x\) and \(p\) in \(X\), \(d(x, p) < \delta\) implies \(d(f(x), f(p)) < \epsilon\).

If \(K\) is closed, define \(d(x, K) := \inf\limits_{y \in K} d(x, y)\). \(d(x, K)\) is a continuous function of \(x\).

**Lebesgue Covering Lemma.** If \(\{U_\alpha\}\) is an open cover of a compact metric space \(X\) then there exists \(\delta > 0\) (called the *Lebesgue number of the cover*) such that for every \(x \in X\), \(B_\delta(x)\) is contained in some \(U_\widehat{\alpha}\) of the cover.

*Proof.* Since \(X\) is compact, there exists a finite subcover \(\{U_{\alpha_i}\}_{i = 1}^n\). Define \(f(x) = \frac{1}{n} \sum\limits_{i = 1}^n d(x, U_{\alpha_i})\). \(f\) is a continuous function on a compact set, so it must attain its minimum value \(\delta\). Hence, if \(f(x) \geq \delta\) then at least one of \(d(x, U_{\alpha_i}) \geq \delta\). So, for this \(i\), \(B_\delta(x) \subset U_{\alpha_i}\). Notice \(\delta > 0\) because \(f(x) > 0\) at each \(x\) due to \(U_{\alpha_i}\). QED.

**Theorem.** Given \(f : X \rightarrow Y\) and \(X\) compact, \(f\) is *uniformly continuous* on \(X\).