Provide examples of the following. [No justification is required.]
a true implication whose converse is false,
a function \(f: \mathbb{R} \to \mathbb{R}\) which is continuous at \(7\), but not differentiable there,
a function \(f: \mathbb{R}^2 \to \mathbb{R}\) which is continuous in each variable separately, but is not continuous at \((0,0)\),
a bounded sequence which is not Cauchy,
a (real) power series whose domain of convergence is the closed interval \([0,2]\).
Solution
1. Part (a): A true implication whose converse is false:
1.1. Consider the statement for an integer \(n \in \mathbb{Z}\):\[ P(n) \implies Q(n), \quad \text{where } P(n) \text{ is } ``4 \mid n" \text{ and } Q(n) \text{ is } ``2 \mid n". \]If \(4 \mid n\), then \(n = 4k = 2(2k)\), so \(2 \mid n\), which is a true implication.
1.2. The converse \(Q(n) \implies P(n)\) is false, as witnessed by the counterexample \(n = 2\): \(2 \mid 2\) is true, but \(4 \mid 2\) is false.
2. Part (b): Continuous at \(7\) but not differentiable at \(7\):
2.1. Define \(f: \mathbb{R} \to \mathbb{R}\) by:\[ f(x) = |x - 7|. \]\(f\) is continuous on \(\mathbb{R}\) as the composition of continuous functions.
2.2. Compute the one-sided derivative difference quotients at \(x = 7\):\[ \lim_{h \to 0^+} \frac{f(7+h) - f(7)}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = 1, \qquad \lim_{h \to 0^-} \frac{f(7+h) - f(7)}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1. \]Because the left and right limits do not agree, \(f'(7)\) does not exist.
3. Part (c): Separately continuous on \(\mathbb{R}^2\) but not continuous at \((0, 0)\):
3.1. Define \(f: \mathbb{R}^2 \to \mathbb{R}\) by:\[ f(x, y) = \begin{cases} \dfrac{xy}{x^2 + y^2} & (x, y) \neq (0, 0), \\ 0 & (x, y) = (0, 0). \end{cases} \]
3.2. For any fixed \(y_0\), \(x \mapsto f(x, y_0)\) is continuous on \(\mathbb{R}\) (at \(x = 0\), \(f(x, 0) = 0\) for all \(x\), so \(\lim_{x\to 0} f(x, 0) = 0 = f(0, 0)\)).Symmetrically, for any fixed \(x_0\), \(y \mapsto f(x_0, y)\) is continuous on \(\mathbb{R}\).Thus \(f\) is separately continuous in each variable.
3.3. Along the line \(y = x\), for \(x \neq 0\):\[ f(x, x) = \frac{x^2}{x^2 + x^2} = \frac{1}{2}. \]Thus \(\lim_{x \to 0} f(x, x) = \frac{1}{2} \neq 0 = f(0, 0)\), so \(f\) is not continuous at \((0, 0)\).
4. Part (d): A bounded sequence that is not Cauchy:
4.1. Consider the sequence \((a_n)_{n=1}^\infty\) defined by \(a_n = (-1)^n\).Then \(|a_n| = 1 \le 1\) for all \(n\), so the sequence is bounded.
4.2. For any \(n \in \mathbb{N}\), \(|a_{n+1} - a_n| = |(-1)^{n+1} - (-1)^n| = 2\).Taking \(\varepsilon = 1\), there is no \(N\) such that \(|a_n - a_m| < 1\) for all \(n, m \ge N\).Thus \((a_n)\) is not Cauchy.
5. Part (e): A real power series whose domain of convergence is \([0, 2]\):
5.1. Define the power series centered at \(c = 1\):\[ S(x) = \sum_{n=1}^\infty \frac{(x - 1)^n}{n^2}. \]
5.2. The radius of convergence \(R\) satisfies:\[ \frac{1}{R} = \lim_{n \to \infty} \left| \frac{1}{n^2} \right|^{1/n} = 1 \implies R = 1. \]Thus the series converges absolutely for \(|x - 1| < 1\) (i.e. \(x \in (0, 2)\)) and diverges for \(|x - 1| > 1\).
5.3. At the endpoints:- For \(x = 2\): \(\sum_{n=1}^\infty \frac{1^n}{n^2} = \sum_{n=1}^\infty \frac{1}{n^2}\), which converges (a \(p\)-series with \(p = 2 > 1\)).- For \(x = 0\): \(\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\), which converges absolutely by comparison to \(\sum \frac{1}{n^2}\).Therefore the exact interval of convergence is the closed interval \([0, 2]\).
6. Conclusion:All five requested examples are defined and verified. Q.E.D.