Exercises

Star Issue Watch Follow

If \(p\) is a prime number, the \(p\)-adic metric on \(\mathbb{Z}\) is defined as

\[|a - b|_p = p^{-n},\]

where \(n\) is the largest integer such that \(p^n\) divides \(a - b\).

Exercise 1.1

Show that the sequence \(2015, 20015, 200015, 2000015, \ldots\) converges in the 2-adic metric on \(\mathbb{Z}\).

Solution

We show that the sequence converges to \(15\) in the 2-adic metric. Let \(a_n\) be the \(n\)th term of the sequence. Then \(a_n - 15 = 2 \times 10^{n + 2}.\) Note that \(2 \times 10^{n + 2}\) is divisible by \(2^{n + 3}.\) Therefore, \(|a_n - 15|_2 = 2^{-(n + 3)} \to 0\) as \(n \to \infty.\)

Exercise 1.2

Determine whether the following subsets \(A \subseteq \mathbb{R}^2\) are open, closed or neither, under the standard topology on \(\mathbb{R}^2.\)

1. \(A = \{(x, y) \in \mathbb{R}^2 \vert x < 0\} \cup \{(x, y) \in \mathbb{R}^2 \vert x > 0, y > 1 / x\}.\)

2. \(A = \{(x, \sin(1 / x)) \in \mathbb{R}^2 \vert x > 0\} \cup \{(0, y) \in \mathbb{R}^2 \vert y \in [-1, 1]\}.\)

3. \(A = \{(x, y) \in \mathbb{R}^2 \vert x \in \mathbb{Q} \text{ and } x = y^n \text{ for some } n \in \mathbb{N}\}.\)

Solution

Set 1: The set

\[A = \{(x, y) \in x < 0\} \cup \{(x, y) \in x > 0, y > 1 / x\}\]

is open, because it is the union of two open sets

\[\{(x, y) \in x < 0\} ~\text{ and }~ \{(x, y) \in x > 0, y > 1 / x\}.\]

Note that the set \(A\) cannot be closed, because the only sets that are both open and closed are the empty set and the entire space, and \(A\) is neither.

Set 2: First, note that the set \(S_z = \{(x, \sin(1/x)): x > z\}\) is closed in \(\mathbb{R}^2\) for any \(z > 0.\) Therefore, its complement is open and the set

\[T_z = S_z' \cap \{(x, y) \in \mathbb{R}^2 \vert x > z\}\]

is also open in \(\mathbb{R}^2.\) Then the set

\[U_z = T_z \cap \{(x, y) \in \mathbb{R}^2 \vert |y| > 1\} \cap \{(x, y) \in \mathbb{R}^2 \vert x \notin [0, z]\}\]

is also open because it is the intersection of open sets. Since \(A = \bigcup_{z > 0} U_z,\) the set \(A\) is open in \(\mathbb{R}^2.\) Again, note that the set \(A\) cannot be closed, by the same reasoning that was used to argue for set 1.

Set 3: The set \(A\) is not open because for any rational \(x \in \mathbb{Q},\) there exists an irrational \(x' \in \mathbb{R} \setminus \mathbb{Q}\) that is arbitrarily close to it, so for any \((x, y) \in A,\) there exists an \((x', y)\) that is aribtrarily close to it but is not in \(A,\) because \(x'\) is irrational. In addition, the set \(A\) is not closed because it does not contain all its limit points. For example, consider the point \((0, 1) \in \mathbb{R}^2.\) This is a limit point of the set \(A\): for any \(\epsilon > 0,\) there exists a rational \(q \in \mathbb{Q}\) such that \(0 < |q| < \epsilon / 2\) and a sufficiently large \(n \in \mathbb{N}\) such that \(|q^{1/n} - 1| < \epsilon / 2.\) Then \(|(q, q^{1/n}) - (0, 1)|_\infty = |q| + |q^{1/n} - 1| < \epsilon,\) from which it follows that \((0, 1)\) is a limit point of \(A\) that is not contained in \(A.\)

Exercise 1.3

Show that the maps \(f, g: \mathbb{R}^2 \to \mathbb{R}\) given by \(f(x, y) = x + y\) and \(g(x, y) = xy\) are continuous with respect to the usual topology on \(\mathbb{R}.\) Let \(X\) be \(\mathbb{R}\) equipped with the topology whose open sets are intervals of the form \((a, \infty).\) Are the maps \(f, g: X \times X \to X\) continuous?

Solution

First, we show that the maps \(f, g: \mathbb{R}^2 \to \mathbb{R}\) are continuous. For this first part of the question we consider open balls to be taken with respect to the \(\ell_1\) norm. Let \(U\) be an open set in \(\mathbb{R}\) and \(z = (x, y) \in f^{-1}(U).\) Then, note that, for all \((x', y') \in B_{\delta}(z),\) where \(\delta = \epsilon / 2,\) we have

\[|f(x', y') - f(x, y)| = |x' - x + y' - y| < |x' - x| + |y' - y| < \epsilon.\]

for all \((x', y') \in B_\delta(z).\) Therefore, \(f(B_{\delta}(z)) \subseteq U,\) which implies that \(B_{\delta}(z) \subseteq f^{-1}(U),\) so \(f\) is continuous.

Similarly, let \(U\) be an open set in \(\mathbb{R}\) and \(z = (x, y) \in g^{-1}(U).\) Then, note that for all \((x', y') \in B_{\delta}(z),\) where \(\delta = \sqrt{1 + \epsilon} - 1\) we have

\[\begin{split}\begin{align} |g(x', y') - g(x, y)| = |x'y' - xy| &= |(x + (x' - x))(y + (y' - y)) - xy| \\ &= |x(y' - y) + y(x' - x) + (x' - x)(y' - y)| \\ &< |x||y' - y| + |y||x' - x| + |x' - x||y' - y| \\ &< 2\delta + \epsilon^2 \\ &= \epsilon \\ \end{align}\end{split}\]

for all \((x', y') \in B_\delta(z).\) Therefore, \(g(B_{\delta}(z)) \subseteq U,\) which implies that \(B_{\delta}(z) \subseteq g^{-1}(U),\) so \(g\) is continuous.

In the second part of the exercise we replace \(\mathbb{R}^2\) with \(X \times X,\) where \(X = \mathbb{R}\) with the topology whose open sets are intervals of the form \((a, \infty).\)

First, we show that \(f\) is continuous under the product topology on \(X \times X.\) Let \(U\) be an open set in \(X\) and \(z = (x, y) \in f^{-1}(U).\) If \(U = \emptyset\) or \(U = X,\) then \(f^{-1}(U) = \emptyset\) or \(f^{-1}(U) = X \times X,\) respectively, which are both open. Suppose instead that \(U \neq \emptyset\) and \(U \neq X.\) Then \(U = (a, \infty)\) for some \(a \in \mathbb{R}.\) Then

\[f^{-1}(U) = \{(x, y) \in X \times X \vert x + y > a\} = \bigcup_{x \in \mathbb{R}} (x, \infty) \times (a - x, \infty),\]

Noting that the sets \((x, \infty) \times (a - x, \infty)\) are open in the product topology on \(X \times X,\) we see that \(f^{-1}(U)\) is open in \(X \times X,\) so \(f\) is continuous.

Next we show that \(g\) is not continuous under the product topology on \(X \times X.\) Let \(U = (-\infty, 0),\) which is open in \(X.\) Then

\[g^{-1}(U) = (-\infty, 0) \times (0, \infty) \cup (0, \infty) \times (-\infty, 0).\]

We will show that \(g^{-1}(U)\) is not open in \(X \times X.\) Suppose instead \(g^{-1}(U)\) is open in \(X \times X.\) Note that \(\mathbb{R}\) is open in \(X,\) so

\[G = g^{-1}(U) \cap (0, \infty) \times \mathbb{R} = (0, \infty) \times (-\infty, 0)\]

is also open in \(X \times X.\) However, this is a contradiction, because for any \(z = (x, y) \in G,\) there do not exist sets \(V_1 = (a, \infty) \times (b, \infty)\) and \(V_2 = (c, \infty) \times (d, \infty)\) such that \(x \in V_1\) and \(y \in V_2\) and \(V_1 \times V_2 \subseteq G,\) because for any \(b \in \mathbb{R}\) we have \((b, \infty) \not \subseteq (-\infty, 0).\) This is a contradiction, so \(g^{-1}(U)\) is not open in \(X \times X,\) and \(g\) is not continuous.

Exercise 1.4

Let \(C^1([0, 1])\) be the set of continuously differentiable functions on \([0, 1],\) that is differentiable functions whose first derivative is continuous. For \(f \in C^1([0, 1]),\) define

\[||f||_{1, 1} = \int_0^1 |f(x)| \, dx + \int_0^1 |f'(x)| \, dx.\]

Show that \(||\cdot||_{1, 1}\) is a norm on \(C^1([0, 1]).\) If a sequence \((f_n)\) converges with respect to this norm, show that it also converges with respect to the uniform norm. Give an example to show that the converse statement does not hold.

Solution

First part: First, we show that \(||\cdot||_{1, 1}\) is a norm on \(C^1([0, 1]),\) by showing that it satisfies the properties of a norm.

Positive definiteness: By the definition of \(||\cdot||_{1, 1},\) we have \(||f||_{1, 1} \geq 0,\) for all \(f \in C^1([0, 1]).\)

Non-degeneracy: Let \(f \in C^1([0, 1]).\) If \(f = 0,\) then \(||f||_{1, 1} = 0.\) Conversely, suppose that \(||f||_{1, 1} = 0.\) Then it must be the case that \(|f(x)|\) is identically zero, because if it were not, then the integral \(\int_0^1 |f(x)| \, dx\) would be positive since non-negative functions with zero integral are zero almost everywhere.

Homogeneity: Let \(f \in C^1([0, 1])\) and \(\lambda \in \mathbb{R}.\) Then

\[\begin{split}\begin{align} ||\lambda f||_{1, 1} &= \int_0^1 |\lambda f(x)| \, dx + \int_0^1 |\lambda f'(x)| \, dx \\ &= |\lambda| \int_0^1 |f(x)| \, dx + |\lambda| \int_0^1 |f'(x)| \, dx \\ &= |\lambda| ||f||_{1, 1}. \end{align}\end{split}\]

Triangle inequality: Let \(f, g \in C^1([0, 1]).\) Then

\[\begin{split}\begin{align} ||f + g||_{1, 1} &= \int_0^1 |f(x) + g(x)| \, dx + \int_0^1 |f'(x) + g'(x)| \, dx \\ &\leq \int_0^1 |f(x)| + |g(x)| \, dx + \int_0^1 |f'(x)| + |g'(x)| \, dx \\ &= \int_0^1 |f(x)| \, dx + \int_0^1 |f'(x)| \, dx + \int_0^1 |g(x)| \, dx + \int_0^1 |g'(x)| \, dx \\ &= ||f||_{1, 1} + ||g||_{1, 1}. \end{align}\end{split}\]

Therefore, \(||\cdot||_{1, 1}\) is a norm on \(C^1([0, 1]).\)

Second part: Here, we want to show that if a sequence \((f_n)\) in \(C^1([0, 1])\) converges to some limit \(f \in C^1([0, 1])\) with respect to the norm \(||\cdot||_{1, 1},\) then it also converges to \(f\) with respect to the uniform norm. Without loss of generality, we can assume that \(f = 0,\) because if \(f \neq 0,\) then we can consider the sequence \((f_n - f)\) instead.

Let \((f_n)\) be a sequence in \(C^1([0, 1])\) that converges with respect to the norm \(||\cdot||_{1, 1}.\) Let \(\epsilon > 0.\) Then there exists an \(N \in \mathbb{N}\) such that for all \(n \geq N,\) we have \(||f_n - f||_{1, 1} < \epsilon / 2.\) Now, for each \(f_n,\) with \(n \geq N,\) there must exist at least one \(x_n \in [0, 1]\) such that \(|f(x_n)| < \epsilon / 2,\) because if not, then the integral \(\int_0^1 |f(x_n)| \, dx\) would be at least \(\epsilon / 2,\) which would be a contradiction. Since \([0, 1]\) is a closed interval and each \(f_n\) is continuous, each \(f_n\) attains its maximum at some \(x_n' \in [0, 1].\) Therefore, we have

\[\begin{split}\begin{align} ||f_n||_{\infty} &= |f(x_n')| \\ &\leq |f(x_n)| + |f(x_n') - f(x_n)| \\ &< \epsilon / 2 + |f(x_n') - f(x_n)| \\ &= \epsilon / 2 + \left| \int_{x_n}^{x_n'} f'(x) \, dx \right| \\ &\leq \epsilon / 2 + \int_{x_n}^{x_n'} |f'(x)| \, dx \\ &\leq \epsilon / 2 + ||f_n||_{1, 1} \\ &< \epsilon. \end{align}\end{split}\]

for all \(n \geq N.\) Therefore \(||f_n||_{\infty} < \epsilon\) for all \(n \geq N,\) so \((f_n)\) converges to \(f\) with respect to the uniform norm.

The converse statement does not hold. A counterexample is the sequence of functions \((f_n)\) defined by \(f_n(x) = \sin(2 \pi n^2 x) / n\) for \(x \in [0, 1].\) This sequence converges to the zero function with respect to the uniform norm, because \(|f_n|\) is bounded by \(1 / n,\) which converges to zero. However, the sequence does not converge to the zero function with respect to the norm \(||\cdot||_{1, 1},\) because

\[\begin{split}\begin{align} ||f_n||_{1, 1} &= \int_0^1 \left| \frac{\sin(2 \pi n^2 x)}{n} \right| \, dx + \int_0^1 \left|2 \pi n \cos(2 \pi n^2 x) \right| \, dx \\ &> 2 \pi n \int_0^1 \left|\cos(2 \pi n^2 x) \right| \, dx. \end{align}\end{split}\]

Now, note that the integral on the right hand side is constant in \(n,\) so \(||f_n||_{1, 1}\) is unbounded as \(n \to \infty.\) Therefore, the sequence \((f_n)\) does not converge to the zero function with respect to the norm \(||\cdot||_{1, 1}.\)