Stepan Malkov

Week 4-6 Problem Set

Easy

  1. Show that the image of a connected set under a continuous map is connected.
    2. Solution:
    Suppose not. Then, there exist U,V \subset Y open such that f(A) \subset U \cup V, U \cap V = \varnothing. Then, by the topological definition of continuity, f^{-1}(U), f^{-1}(V) are open, f^{-1}(U) \cap f^{-1}(V) = \varnothing, and A \subset f^{-1}(U) \cap f^{-1}(V). But this implies A is disconnected, which is a contradiction.
  2. Show that if a sequence f_n: X \to Y of continuous functions converges uniformly to a function f: X \to Y, then f is continuous.
    3. Solution:
    Let \epsilon>0. Then, there exists an N and a \delta>0 such that \sup_{x \in X} d(f_n(x),f(x)) < \frac{\epsilon}{3} and d(f_n(x),f_n(y))<\frac{\epsilon}{3} whenever d(x,y)<\delta and n \geq N. Then, by the triangle inequality,

        \[d(f(x),f(y)) \leq d(f(x),f_n(x)) + d(f_n(x),f_n(y)) + d(f_n(y),f(y)) < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon\]

    whenever d(x,y) < \delta and n \geq N, showing that f is continuous.
  3. Provide an example of a sequence of continuous functions f_n that converges pointwise to some function f, but does not converge uniformly to f.
    4. Solution:
    Define f_n:[0,1] \to \mathbb{R} given by f_n(x) = x^n. Then, f_n(x) \to 0 for x<1 and f_n(1) \to 1, so f_n \to f pointwise, where f(x) = 0 for x<1 and 1 otherwise. But by the previous example, a uniformly convergent sequence of continuous functions converges to a continuous function, and since f is not continuous, f_n does not converge uniformly to f.
  4. Let f: X \to Y be a function from a connected metric space (X,d) to a metric space (Y,d_{disc}) with the discrete metric. Show that f is continuous if and only if it is constant.
    5. Solution:
    Note that a constant function is trivially continuous, since x_n \to x always implies f(x_n) \to f(x) since f(x_n) = f(x) for all n. Conversely, using the sequential definition of continuity, suppose f: X \to Y is continuous. Then, since the image of a connected set under a continuous map is connected, f(X) is connected. If f(X) had at least two distinct elements y_1,y_2, then U=\{y_1\}, V=f(X) \setminus \{y_1\} would be open sets (since every set in a discrete metric space is open) such that U \cup V = X, U \cap V = \varnothing, contradicting the fact that f(X) is connected. Thus, f(X)=\{y\}, i.e. f(x) = y for all x \in X, showing that f is constant. Consequently f: X \to Y is continuous if and only if it is constant.

Medium

  1. Show that the topologist’s sine curve (the set defined in Problem 1 on HW3) is connected but not path-connected.
    2. Solution:
    Let X be the topologist’s sine curve. Note that X= A \cap B, where B is an interval and is therefore path-connected. Moreover, A is path connected since the path \gamma:[0,1] \to A such that

        \[\gamma(t) = \left(t x_1 + (1-t)x_2, \sin\left(\frac{1}{t x_2+(1-t)x_1}\right)\right)\]

    is such that

        \[\gamma(0) = \left(x_1,\sin\left(\frac{1}{x_1}\right)\right),\gamma(1) = \left(x_2, \sin\left(\frac{1}{x_2}\right)\right).\]

    Thus, it suffices to show A and B are connected but not path-connected. Indeed, X is not path-connected, since there is no path from the point (0,1) to the point (1,0), for if there were a continuous path \gamma: [0,1] \to X with \gamma(0) = (1,0), \gamma(1) = (0,1), there would exist an x such that x_0 = \inf\{x \in [0,1]: \gamma(x) \in B\}. But then, there must exist a \delta such that d(\gamma(x_0),\gamma(x_0-x)<\epsilon whenever |x_0-x|<\delta, i.e. \gamma(x_0-x) \in B(\gamma(x_0),\epsilon) \cap A. But note that B(\gamma(x_0),\epsilon) \cap A is disconnected, with each component being of the form \{y=(x,\sin\left(\frac{1}{x}\right): \|y-x\| < \epsilon\}. Thus, no such path exists, showing that X is not path-connected.
    To show that X is connected, proceed by contradiction, and suppose U,V are open sets such that X \subset U \cup V, U \cap V = \varnothing. Without loss of generality, suppose U \cap A \not = \varnothing, V \cap B \not = \varnothing. Let u = \sup\{u_1: (u_1,u_2) \in U\} and v = \inf \{v_1: (v_1,v_2) \in V\}. if u \not = v, then for any u<a<v, the point (a, \sin\left(\frac{1}{a}\right)) \not \in U \cup V, which is a contradiction. If u = v>0, then x= (u,\sin\left(\frac{1}{u}\right)) is either in U or in V. Without loss of generality, suppose x \in V, and note that since V is open, there exist an open ball B(x, \epsilon) \subset V. But this implies u-\frac{\epsilon}{2} < u is an x-coordinate of a point in V, which contradicts the definition of v. Finally, if u = v = 0, one notes that A \subset V. Since U,V are open and U \cap V = \varnothing, U \cap \overline{V} = \varnothing. But X = \overline{A} \subset \overline{U}, so U = \varnothing, which is a contradiction. Thus, no such U,V exist, showing that the topologist’s sine curve is connected.
  2. Show that a continuous function on a compact metric space is uniformly continuous.
    3. Solution:
    Let X be a compact metric space and suppose f: X \to Y is continuous. Fix \epsilon>0. Since f is continuous, for every x \in X there exists a \delta_x>0 such that d(y,x)<\delta_x implies d(f(y),f(x))<\epsilon. Note that \bigcup_{x \in X} B(x,\frac{\delta_x}{2}) (we will justify the choice of radii later) is an open cover of X, so by compactness, it has a finite subcover B(x_1,\delta_1),...,B(x_n,\delta_{x_n}). Then for \delta = \min(\delta_1,...,\delta_n), whenever d(x,y)<\delta, y \in B(x, \delta) \subset B(x_i,\delta_i) for some i, so d(x,y)<\delta_i, implying by the triangle inequality that

        \[d(f(x),f(y)) \leq d(f(x),f(x_i)) + d(f(x_i),f(y))<\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.\]

    Since \delta is independent of x, we conclude that f is uniformly continuous.
  3. Let (x_n)_{n=1}^\infty be a sequence of real numbers that converges to 0, and suppose f: \mathbb{R} \to \mathbb{R} is uniformly continuous. Define a sequence (f_n)_{n=1}^\infty, where f_n(x) = f(x+x_n). Show that f_n converges uniformly to f.
    4. Solution:
    Since f is uniformy continuous, for every \epsilon>0, there exists \delta>0 such that |f(x)-f(y)|<\epsilon whenever |x-y|<\delta. In particular, note that |f(x) - f_n(x)| = |f(x) - f(x+x_n)|< \epsilon whenever |x+x_n-x| = |x_n|<\delta, which is guaranteed for large enough n since (x_n)_{n=1}^\infty converges to 0. Since the choice of \delta is independent of x, we thus conclude that f_n converges uniformly to f.

Challenging:

  1. Prove Dini’s theorem: let X be a compact metric space, and let f_n: X \to Y be a monotonically increasing sequence of functions, i.e. f_n(x) \leq f_{n+1}(x) for all x \in X and n \in \mathbb{N}, that converges pointwise to some function f: X \to Y. Furthermore, suppose f is continuous. Then, f_n converges uniformly to f.
    2. Solution:
    Proceed by contradiction. Note that without loss of generality one can set f(x) =0 for all x \in X. Then, since f_n does not converge uniformly to 0, there exists an \epsilon>0 and a sequence (x_n) \in X such that |f_n(x_n)|> \epsilon for all n. Since X is compact, (x_n) has a convergent subsequence (x_{n_k}) \to x \in X. Then, since f_n is continuous for all n, for any \epsilon>0 there exists a \delta>0 such that

        \[|f_{n_k}(x_{n_k}) - f_{n_k}(x)| < \frac{\epsilon}{2}\]

    whenever |x-x_{n_k}|< \delta. This shows that |f_{n_k}(x)| > \frac{\epsilon}{2} for n_k sufficiently large. But since f_n(x) is an increasing sequence, it follows that

        \[\lim_{n \to \infty} |f_n(x)| = \lim_{n_{k} \to \infty} |f_{n_k}(x)| = |f(x)|\geq \frac{\epsilon}{2},\]

    so f(x) \not = 0, and we thus arrive at a contradiction. Consequently, f_n must converge uniformly to f.
  2. Suppose f_n: [a,b] \to \mathbb{R} is a sequence of Riemann integrable functions that converges uniformly to a function f:[a,b] \to \mathbb{R}. Show that f is Riemann integrable and

        \[\int_a^b f_n(x)dx \to \int_a^b f(x)dx.\]

    3. Solution:
    By the equivalent conditions for Riemann integrability, it suffices to show that there exists a sequence of partitions \mathcal{P} such that

        \[\sum_{i=0}^{n-1} (\sup_{[x_i,x_{i+1}]} f(x) - \inf_{[x_i,x_{i+1}]} f(x)) (x_{i+1}-x_i) \to 0.\]

    But note that by the triangle inequality,

        \[|\sup f(x) - \inf f(x)| \leq |\sup f(x) - \sup f_n(x)| + |\sup f_n(x)-\inf f_n(x)| + |\inf f_n(x)-\inf f(x)|,\]

    where one can pick a sequence of equally spaced x_i with x_{i+1}-x_i<\delta (for example, x_i = \frac{i}{n} for \frac{1}{n}<\delta) such that |f(x)-f(y)|<\epsilon, i.e. |\sup_{[x_i,x_{i+1}]} f(x) - \inf_{[x_i,x_{i+1}]} f(x)| <\epsilon for x,y \in [x_i,x_{i+1}] (since f_n are continuous on a compact set, i.e. uniformly continuous). Picking n large enough such that |\sup f(x) - \sup f_n(x)|<\frac{\epsilon}{2},|\inf f(x) - \inf f_n(x)|\leq \frac{\epsilon}{2} and summing over all intervals yields

        \[|\sup f(x) - \inf f(x)| \leq \epsilon + \sup f_n(x) -\inf f_n(x),\]

    so summing over all intervals yields

        \begin{equation*} \begin{split} \sum_{i=0}^{n-1} (\sup_{[x_i,x_{i+1}]} f(x) & - \inf_{[x_i,x_{i+1}]} f(x)) (x_{i+1}-x_i) \\ & \leq \sum_{i=0}^{n-1} \epsilon (x_{i+1}-x_i) + \sum_{i=0}^{n-1} (\sup_{[x_i,x_{i+1}]} f_n(x) - \inf_{[x_i,x_{i+1}]} f_n(x)) (x_{i+1}-x_i)\\ & = \epsilon (b-a)+ \sum_{i=0}^{n-1} (\sup_{[x_i,x_{i+1}]} f_n(x) - \inf_{[x_i,x_{i+1}]} f_n(x)) (x_{i+1}-x_i). \end{split} \end{equation*}

    We now note that since f_n is Riemann integrable, there exists a sequence of partitions such that for any \epsilon > 0

        \[\sum_{i=0}^{n-1} (\sup_{[x_i,x_{i+1}]} f_n(x) - \inf_{[x_i,x_{i+1}]} f_n(x)) (x_{i+1}-x_i)< \epsilon.\]

    Taking a common refinement of this sequence with our partition, we get that

        \[\sum_{i=0}^{n-1} (\sup_{[x_i,x_{i+1}]} f(x) - \inf_{[x_i,x_{i+1}]} f(x)) (x_{i+1}-x_i) < \epsilon (b-a+1),\]

    which demonstrates that there exists a sequence of partitions that satisfies the condition for Riemann integrability. Thus, f is Riemann integrable. Finally, by uniform convergence, for sufficiently large n, |f(x)-f_n(x)| \leq \epsilon implies

        \[f_n(x) - \epsilon \leq f(x) \leq f_n(x) + \epsilon,\]

    and integrating each side on [a,b] yields

        \[\int_a^b f_n(x) dx - \epsilon (b-a) \leq \int_a^b f(x) dx \leq \int_a^b f_n(x) dx + \epsilon (b-a),\]

    so

        \[\left|\int_a^b f_n(x) dx -\int_a^b f(x) dx\right|\leq\epsilon(b-a).\]

    Letting \epsilon \to 0 shows that

        \[\int_a^b f_n(x) dx \to \int_a^b f(x) dx,\]

    which completes the proof.