Exercises

Exercises#

Exercise 1.1

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

Exercise 1.2

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

\(A = \{(x, y) \in \mathbb{R}^2 \vert x < 0\} \cup \{(x, y) \in \mathbb{R}^2 \vert x > 0, y > 1 / x\}.\)
\(A = \{(x, \sin(1 / x)) \in \mathbb{R}^2 \vert x > 0\} \cup \{(0, y) \in \mathbb{R}^2 \vert y \in [-1, 1]\}.\)
\(A = \{(x, y) \in \mathbb{R}^2 \vert x \in \mathbb{Q} \text{ and } x = y^n \text{ for some } n \in \mathbb{N}\}.\)

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 \(||\cdot||_{1, 1}.\)

Exercise 1.5

Let \(d: X \times X \to \mathbb{R}\) be a function which satisfies all the axioms for a metric except for the requirement that \(d(x, y) = 0 \implies x = y.\) For \(x, y \in X,\) define \(x \sim y\) if \(d(x, y) = 0.\) Show that \(\sim\) is an equivalence relation on \(X,\) and that \(d\) induces a metric on the quotient topology \(X / \sim.\)

Exercise 1.6

Find a closed \(A_1 \subseteq \mathbb{R}\) (with the standard topology) such that \(\overline{\text{Int}(A_1)} \neq A_1\) and an open \(A_2 \subseteq \mathbb{R}\) such that \(\text{Int}(\overline{A_2}) \neq \text{Int}(A_2).\)

Exercise 1.7

Let \(f: X \to Y\) be a map between topological spaces. Show that \(f\) is continuous if and only if \(f(\overline{A}) \subseteq \overline{f(A)}\) for all \(A \subseteq X.\) Deduce that if \(f\) is surjective and continuous, the image of a dense set in \(X\) is dense in \(Y.\)

Exercise 1.8

Suppose \(X\) is a topological space and \(Z \subseteq Y \subseteq X.\) If \(Y\) is dense in \(X\) and \(Z\) is dense in \(Y,\) must \(Z\) be dense in \(X\)?

Exercise 1.9

Define a topology on \(\mathbb{R}\) by declaring the closed subsets to be those which are (1) closed in the usual topology and (2) either bounded or all of \(\mathbb{R}.\) Show that this is a topology, that all points of \(\mathbb{R}\) are closed with respect to it, but that the topology is not Hausdorff.

Exercise 1.10

The diagonal in \(X \times X\) is the set \(\Delta = \{(x, x) \in X \times X | x \in X \}.\) If \(X\) is a Haussdorff space, show that \(\Delta\) is a closed subset of \(X \times X.\)

Exercise 1.11

Exhibit a countable basis for the usual topology of \(\mathbb{R}.\)