Stepan Malkov

Week 1 Problem Set

Easy

  1. Let X=C([a,b]) be the space of continuous functions on the interval [a,b]. Show that the map d: X \times X \to [0,\infty) defined by

        \[d(f,g) = \max_{x \in [0,1]}[f(x)-g(x)]\]

    is a metric, thus making X into a metric space.

    2. Solution:
    Note that

        \[d(f,f) = \max_{x \in [0,1]} [f(x)-f(x)] = 0,\]

        \[d(f,g) = \max_{x \in [0,1]}[f(x)-g(x)]=\max_{x \in [0,1]} [g(x)-f(x)] = d(g,f),\]

    and

        \[d(f,g) = \max_{x \in [0,1]} |f(x)-h(x) + h(x) - g(x)| \leq \max_{x \in [0,1]} |f(x)-h(x)|+\max_{x \in [0,1]} |h(x)-g(x)| = d(f,h)+d(h,g)\]

    for all h \in C([a,b]). Thus, the metric axioms are satisfied for the map d, making d into a metric and X into a metric space.
  2. Let X = \{(x,y) \in \mathbb{R}^2: x,y \in \mathbb{Q}\} \subset \mathbb{R}^2, considered as a metric subspace. Find the boundary of X in \mathbb{R}^2.

    3. Solution:
    Since \mathbb{Q} is dense in \mathbb{R}, for any \epsilon > 0 and any r \in \mathbb{R} there exists a q \in \mathbb{Q} such that |r-q|<\epsilon. Thus, for r=(r_1,r_2) \in \mathbb{R}^2, pick q=(q_1, q_2) \in \mathbb{Q}^2 such that |r_1-q_1| < \frac{\epsilon}{\sqrt{2}}, |r_2-q_2|<\frac{\epsilon}{\sqrt{2}}. Then, d(r,q) < \epsilon, i.e. B(r,\epsilon) \cap X \not = \varnothing. Since r is arbitrary, it follows that every point is an adherent point of X, i.e. the boundary of X is \mathbb{R}^2.
  3. Let X be a metric space. Suppose B_1 = B(x_1,r_1) and B_2 = B(x_2,r_2) are open balls in X. Prove the following statement or find a counterexample: if the union B_1 \cup B_2 is an open ball, then either B_1 \subset B_2 or B_2 \subset B_1.

    4. Solution:
    Pick X = \mathbb{R} with the typical metric. Then, B(0,1) = (-1,1) and B(\frac{1}{2},1) = (-\frac{1}{2},\frac{3}{2}) are both open balls in \mathbb{R}, with B_1 \cup B_2 = (-1, \frac{3}{2}) = B(\frac{1}{4},\frac{5}{4}) also being an open ball, but it is not true that either B_1 \subset B_2 or B_2 \subset B_1.
  4. Let A \subset X be a subset of a metric space. Show that A = \overline{A} if and only if for every sequence (x_n)_{n=1}^\infty, x_n \in A, if \lim_{n \to \infty} x_n = x \in X, then x \in A.

    5. Solution:
    Suppose A = \overline{A}. Then for any sequence (x_n)_{n=1}^\infty such that \lim_{n \to \infty} x_n = x \in X exists, it follows that x \in \overline{A}, since for any \epsilon > 0, there exists an n such that d(x_n,x) < \epsilon, i.e. B(x,\epsilon) \cap A \not = \varnothing. Since x is an adherent point of A and A = \overline{A}, it follows that x \in \overline{A} implies x \in A. Conversely, suppose that for any sequence (x_n)_{n=1}^\infty, x_n \in A, if \lim_{n \to \infty} x_n = x \in A, then x \in A. Let y \in \overline{A}. Then, for \epsilon = \frac{1}{k} > 0, B(y,\epsilon) \cap A \not = \varnothing, i.e. there exists a sequence y_k, y_k \in A, such that d(y_n,y)< \frac{1}{k} for n \geq k, i.e. \lim_{k \to \infty} y_k = y. This implies that y \in A, i.e. \overline{A} \subseteq A. Since A \subseteq \overline{A}, it immediately follows that A = \overline{A}.

Intermediate

  1. Let A \subset X be a subset of a metric space. Show that A is closed in X if and only if for every open ball B(x,r) contained in A, \overline{B(x,r)} is contained in A.

    2. Note:
    The problem as originally stated was not correct. As a counter example, take A to be a single point in the discrete metric and r=1. Then, \overline{B(x,r)} \not \subset A but A is closed. For the “only if” implication, take A=\mathbb{Q} \subset \mathbb{R} with the usual metric. Since there are no open balls in \mathbb{Q}, the statement holds vacuously true, but \mathbb{Q} is not closed.
  2. Let Y = \mathcal{P}([0,1]) be the power set of [0,1], that is, the set of subsets of [0,1]. Let A,B \subset [0,1], and define d: Y \times Y \to [0, \infty) by d(A,B) = \sup_{a \in A, b \in B} |a-b|. Is d a metric?

    3. Solution:
    No, since for any A \in Y with at least two distinct elements \{x,y\},

        \[d(A,A) = \sup_{a \in A, b \in A} |a-b| \geq |x-y|>0,\]

    which contradicts the first axiom of a metric.
  3. If d is a metric, (x_n)_{n=1}^\infty is a sequence, and d(x_n,y) converges, is it true that \lim_{n \to \infty} x_n exists?

    4. Solution:
    No. As an example, take (x_n)_{n=1}^\infty = (-1)^n and y=1 in \mathbb{R}. Then d(x_n,y) = 1 for all n, but (x_n)_{n=1}^\infty does not converge to any real number \mathbb{R}.

Challenging

  1. A set that is as a countable intersection of open sets is called a G_\delta set, and a set that is a countable union of closed sets is called an F_\sigma set. Show that the irrationals are a G_\delta set.

    2. Solution:
    1. Since the rationals can be written as a countable union of closed sets \mathbb{Q} = \bigcup_{q \in \mathbb{Q}} \{q\}, by de Morgan’s laws, the irrationals can be written as a countable intersection of open sets

          \[\mathbb{Q}^c = \left(\bigcup_{q \in \mathbb{Q}} \{q\}\right)^c = \bigcap_{q \in \mathbb{Q}} \{q\}^c = \bigcap_{q \in \mathbb{Q}} \mathbb{R} \setminus \{q\},\]

      which is a G_\delta set since the complement of a closed set is an open set.
    2. A metric space (X,d) is called an ultrametric space if the condition

          \[d(x,z) \leq d(x,y) + d(y,z)\]

      is replaced by the stronger condition

          \[d(x,z) \leq \max(d(x,y),d(x,z))\]

      for all x,y,z \in X.
      1. Give an example of an ultrametric space.
      2. Show that every open ball is closed in X and every closed ball is open in X.

      3. Solution:
      1. An example of an ultrametric space is any set X with the discrete metric, which is easily verified to be a metric. It is an ultrametric since if x=z, d(x,z) = 0 \leq \max(d(x,y), d(y,z)) for any x,y,z \in X, and if x \not = z, then either y \not = x or y \not = z, i.e.

            \[d(x,z) = 1 \leq \max(d(x,y),d(x,z)).\]

        Thus, (X,d) is indeed an ultrametric space.
      2. Let B(x,r) be an open ball in X and let y_n \in B(x,r) be a sequence such that \lim_{n \to \infty} y_n = y \in X. Then,

            \[d(x,y) \leq \max(d(x,y_n),d(y_n,y)) = d(x,y_n) < r,\]

        (since d(y_n,y) \to 0 as n \to \infty), so y \in B(x,r). Thus, B(x,r) is closed. Similarly, let C(x,r) be a closed ball in X and let y \in C(x,r). Then, for z \in X such that d(y,z) < r,

            \[d(x,z) \leq \max(d(x,y),d(y,z)) < r,\]

        i.e. z \in B(x,r). Thus, C(x,r) is open.