Chapter 5 - Continuity
Section 5.1 - Continuous Functions
Definition. Let \(A \subseteq \mathbb{R}\), and \(f: A \rightarrow \mathbb{R}\). Then, if \(a \in A\), \(f\) is continuous at \(a\) if, given any \(\varepsilon > 0\), there exists some \(\delta > 0\) such that for all \(x \in A\),
Note that if \(a\) is an isolated point of \(A\), that is, not a cluster point, then \(a\) is automatically continuous.
If \(a\) is a cluster point of \(A\), then this definition collapses to the definition of \(\lim_{x \rightarrow a} f(x) = f(a)\).
Note that a function cannot be continuous at a point outside of its domain, even if the limit exists.
Definition. \(f\) is continuous on \(A\) if it is continuous at every point \(a \in A\).
Theorem. \(f\) is continuous if and only if for every sequence \((x_n)\) in \(A\) that converges to \(a\), the sequence \((f(x_n))\) converges to \(f(a)\).
Definition. Let \((S, d_S)\) and \((T, d_T)\) be metric spaces. A function \(f: S \rightarrow T\) is continuous at a point \(a \in S\) if given any \(\varepsilon > 0\), there exists some \(\delta > 0\) such that for all \(x \in S\),
Theorem. A function \(f: S \rightarrow T\) is continuous at a point \(a \in A\) if and only if given some neighborhood \(V(f(a)) \in B\), there exists some \(U(a) \in A\) such that \(f(U) \subseteq V\).
Section 5.2 - Combinations of continuous Functions
Theorem. Let \(f, g: A \rightarrow \mathbb{R}\) be continuous at \(a \in A\). Then,
- \(f + g\) and \(fg\) are continuous at \(a\)
- If \(g(x) \neq 0\) for all \(x \in A\), then \(\frac{f}{g}\) is continuous at \(a\).
As a consequence, every polynomial, rational, and basic trigonometric function are continuous on its domains.
Theorem. Lett \(A, B \subseteq \mathbb{R}\), such that \(f: A \rightarrow B\) and \(g: B \rightarrow \mathbb{R}\). Then, if \(c\) is a cluster point of \(A\) such that \(\lim_{x \rightarrow c} f(x) = L \in B\) and \(g\) is continuous at \(L\), then
Corollary. let \(A, B \subseteq \mathbb{R}\), with \(f: A \rightarrow B\) and \(g: B \rightarrow \mathbb{R}\). If \(f\) is continuous at \(a \in A\) and \(g\) is continuous at \(f(a) \in B\), then \(g(f(x))\) is continuous at \(a\).
Section 5.3 - continuous functions on Intervals
Theorem. Let \(S, T\) be metric spaces with \(A \subseteq S\) and \(f: A \rightarrow T\). If \(A\) is a compact subset of \(S\), then \(f(A)\) is a compact subset of \(T\).
Corollary. Let \(f: A \rightarrow \mathbb{R}\) be a continuous function, with \(A\) being a compact subset of metric space \(S\). Then, \(f(A)\) is closed and bounded. Moreover, there exists a \(p, q \in A\) such that \(f(p)\) and \(f(q)\) are the supremum and infimum of \(f(A)\).
Corollary. Maximum-Minimum Theorem. If \(I = [a, b]\) is a closed and bounded interval and \(f: I \rightarrow \mathbb{R}\) is continuous on \(I\), then \(f\) has an absolute minimum and maximum on \(I\).
Theorem. Let \(S, T\) be metric spaces and \(A \subseteq S\). Then, if \(f: A \rightarrow T\) is continuous on \(A\), and \(A\) is a connected subset of \(S\), then \(f(A)\) is a connected subset of \(T\).
Corollary. Suppose that \(I\) is an interval. Let \(f: I \rightarrow \mathbb{R}\) be continuous on \(I\). Then, \(f(I)\) is an interval.
Theorem. (Bolzano's) Intermediate Value Theorem. Suppose \(f: [a, b] \rightarrow \mathbb{R}\) is continuous on \([a, b]\) with \(a \neq b\). Then, given some \(k\) such that \(f(a) < k < f(b)\), there exists some \(c \in (a, b)\) such that \(k = f(c)\).
Definition. Let \(A \subseteq R\). Then, a function \(f: A \rightarrow \mathbb{R}\) is uniformly continuous if given any \(\varepsilon > 0\), there exists some \(\delta > 0\) depending only on \(\varepsilon\) such that for any \(x, y \in A\),
Note that if \(f\) is uniformly continuous, it must be continuous on \(A\).
Theorem. Let \(I = [a, b]\) be a closed and bounded interval. If \(f: I \rightarrow \mathbb{R}\) is continuous on \(I\), then \(f\) is uniformly continuous.
Remark. If \(S, T\) are metric spaces, \(K\) is a compact subset of \(S\), and \(f: K \rightarrow T\) is continuous on \(K\), then \(f\) is uniformly continuous.
Theorem. Suppose \(A \subseteq \mathbb{R}\) and \(f: A \rightarrow \mathbb{R}\) is uniformly continuous. Then, if \((x_n)\) is a Cauchy sequence in \(A\), \((f(x_n))\) is a Cauchy sequence in \(\mathbb{R}\).
Remark. Suppose \(S, T\) are metric spaces and \(f: S \rightarrow T\) is uniformly continuous. Then, if \((x_n)\) is a Cauchy sequence in \(S\), \((f(x_n))\) is a Cauchy sequence in \(T\).