Proof. If \(f\) is continuous at \(p\) and \(\varepsilon >0\), there is a \(\delta >0\) with \(|f(x) - f(p)|<\varepsilon \) for all \(x\in E\) with \(|x-p|<\delta \), in particular for those with \(0<|x-p|<\delta \). Hence \(\lim _{x\rightarrow p} f(x) = f(p)\).

Conversely, if \(\lim _{x\rightarrow p} f(x) = f(p)\) and \(\varepsilon >0\), there is a \(\delta >0\) with \(|f(x) - f(p)| < \varepsilon \) for all \(x \in E\) with \(0<|x - p| < \delta \). The inequality also holds trivially at \(x=p\). Hence \(f\) is continuous at \(p\). □

Proof. Let \(f\) be continuous at \(p\), and let \(\varepsilon > 0\). Then there is a \(\delta > 0\) such that \begin {equation} \label {eq:seqcont} |f(x) - f(p)| < \varepsilon \quad \text {whenever } x\in E \text { and } |x - p| < \delta . \end {equation} Now let \(\{x_n\}\) be a sequence in \(E\) with \(\lim _{n\rightarrow \infty } x_n = p\). For this \(\delta \) there is an \(N\in \mathbb {N}\) such that \begin {equation} \label {eq:seqconv} |x_n - p| < \delta \quad \text {for all } n > N. \end {equation} Putting \(x= x_n\) in (2.4) and using (2.5), we get \(|f(x_n) - f(p)| < \varepsilon \) for all \(n > N\). Hence \(\lim _{n\rightarrow \infty } f(x_n) = f(p)\).

Conversely, suppose \(f\) is not continuous at \(p\). Then there is an \(\varepsilon > 0\) such that for every \(\delta > 0\) there is an \(x\in E\) with \(|x - p| < \delta \) but \(|f(x) - f(p)|\geq \varepsilon \). Taking \(\delta = \frac {1}{n}\), \(n\in \mathbb {N}\), we obtain for each \(n\) a point \(x_n\in E\) with \begin {equation} \label {eq:seqneg} |x_n - p| < \frac {1}{n} \qquad \text {and}\qquad |f(x_n) - f(p)|\geq \varepsilon . \end {equation} The first inequality in (2.6) gives \(\lim _{n\rightarrow \infty } x_n = p\), while the second shows that \(f(x_n)\) does not converge to \(f(p)\). So if \(f(x_n)\longrightarrow f(p)\) for every such sequence, \(f\) must be continuous at \(p\). □

Proof. The sequence \(x_n = \frac {1}{n}\) lies in \([0,1]\) and converges to \(0\), but \(f(x_n) = n\) does not converge to \(f(0) = 0\). By the Discontinuity Criterion, \(f\) is discontinuous at \(0\). (No choice of \(f(0)\) would make \(f\) continuous there, since \(f(x_n)\) does not converge at all.) □

Proof. Let \(\{x_n\}\) be any sequence in \(E\) converging to \(p\). By the Sequential Criterion, \(f(x_n)\longrightarrow f(p)\) and \(g(x_n)\longrightarrow g(p)\). By the rules for limits of sequences, \begin {gather*} f(x_n)\pm g(x_n)\longrightarrow f(p)\pm g(p),\qquad \alpha f(x_n)\longrightarrow \alpha f(p),\\ f(x_n)g(x_n)\longrightarrow f(p)g(p), \end {gather*} and, if \(g(p)\neq 0\) (so that \(g(x)\neq 0\) for \(x\) near \(p\), by Theorem 2.3.3 below), \(\frac {f(x_n)}{g(x_n)}\longrightarrow \frac {f(p)}{g(p)}\). By the Sequential Criterion again, each function is continuous at \(p\). □

Proof. Let \(p\in E\) and \(\varepsilon > 0\). There exists \(\delta > 0\) such that \(|f(x) - f(p)| < \varepsilon \) whenever \(x\in E\) and \(|x - p| < \delta \). For such \(x\), \[\bigl ||f(x)| - |f(p)|\bigr | \leq |f(x) - f(p)| < \varepsilon .\] Therefore \(|f|\) is continuous at \(p\).

The converse is false. For example, the function \[f(x) = \begin {cases} 1, & x \text { rational},\\ -1, & x \text { irrational}, \end {cases}\] is discontinuous everywhere, but \(|f|= 1\) is constant and hence continuous everywhere. □

Proof. Let \(\varepsilon > 0\) be given. Since \(g\) is continuous at \(f(p)\), there exists \(\eta > 0\) such that \(|g(y) - g(f(p))| < \varepsilon \) whenever \(y \in f(E)\) and \(|y - f(p)| < \eta \). Since \(f\) is continuous at \(p\), there exists \(\delta > 0\) such that \(|f(x) - f(p)| < \eta \) whenever \(x\in E\) and \(|x - p| < \delta \). Hence \[|h(x) - h(p)| = |g(f(x)) - g(f(p))| < \varepsilon \qquad \text {whenever } x\in E \text { and } |x - p| < \delta .\] Therefore \(h\) is continuous at \(p\). □

Proof. Suppose \(f\) is continuous on \(X\) and \(O\) is open in \(Y\). Let \(p\in f^{-1}(O)\), so that \(f(p)\in O\). Since \(O\) is open in \(Y\), there exists \(\varepsilon > 0\) such that \(y \in O\) whenever \(y\in Y\) and \(|y - f(p)| < \varepsilon \). Since \(f\) is continuous at \(p\), there exists \(\delta > 0\) such that \(|f(x) - f(p)| < \varepsilon \) whenever \(x\in X\) and \(|x - p| < \delta \). For such \(x\), \(f(x)\in O\), that is, \(x \in f^{-1}(O)\). Hence \(f^{-1}(O)\) is open in \(X\).

Conversely, suppose \(f^{-1}(O)\) is open in \(X\) for every set \(O\) open in \(Y\). Fix \(p\in X\) and \(\varepsilon > 0\), and let \(O = \{y\in Y : |y - f(p)| < \varepsilon \}\), which is open in \(Y\). Since \(p\in f^{-1}(O)\) and this set is open in \(X\), there exists \(\delta > 0\) such that \(x\in f^{-1}(O)\) whenever \(x\in X\) and \(|x - p|< \delta \). For such \(x\), \(|f(x) - f(p)| < \varepsilon \). Therefore \(f\) is continuous at \(p\), and since \(p\) was arbitrary, \(f\) is continuous on \(X\). □

Proof. This follows from the theorem, since a set is closed in \(Y\) if and only if its complement in \(Y\) is open in \(Y\), and \(f^{-1}(Y\setminus C) = X\setminus f^{-1}(C)\) for every \(C\subseteq Y\). □

Solution. At \(c>0\): for \(x\geq 0\), \[|\sqrt {x} - \sqrt {c}| = \frac {|x-c|}{\sqrt {x}+\sqrt {c}}\leq \frac {|x-c|}{\sqrt {c}},\] so \(\delta = \varepsilon \sqrt {c}\) works. At \(c = 0\): \(|\sqrt {x}-0|<\varepsilon \) whenever \(0\leq x<\varepsilon ^2\), so \(\delta = \varepsilon ^2\) works. Notice that the \(\delta \) at \(c>0\) shrinks as \(c\longrightarrow 0\); we return to this under uniform continuity.

Solution. At \(0\): \(|f(x) - f(0)|\leq |x|\) for every \(x\), so \(\delta = \varepsilon \) works.

Let \(c\neq 0\). Choose irrational numbers \(x_n\longrightarrow c\) and rational numbers \(y_n\longrightarrow c\) (both exist because every interval contains rationals and irrationals). Then \(f(x_n) = 0\longrightarrow 0\) and \(f(y_n) = y_n\longrightarrow c\neq 0\). The two limits differ, so at least one of them differs from \(f(c)\), and by the Discontinuity Criterion \(f\) is discontinuous at \(c\).

Solution. Rational points. If \(c = \frac {p}{q}\), then \(f(c) = \frac {1}{q}>0\), but irrational \(x_n\longrightarrow c\) give \(f(x_n) = 0\). So \(f\) is discontinuous at \(c\).

Irrational points. Let \(c\) be irrational and \(\varepsilon >0\). Choose \(N\) with \(\frac {1}{N}<\varepsilon \). In \((0,1)\) there are only finitely many rationals \(\frac {p}{q}\) with \(q\leq N\), and none of them equals \(c\). Let \(\delta >0\) be smaller than the distance from \(c\) to each of them. If \(|x-c|<\delta \), then either \(x\) is irrational and \(f(x) = 0\), or \(x = \frac {p}{q}\) with \(q>N\) and \(f(x) = \frac {1}{q}<\varepsilon \). Either way \(|f(x) - f(c)| = f(x)<\varepsilon \).