Math 317: Homework 8

Due: Friday, November 1, 2019

Reading: Read sections 25, 28, 29.


  1. Give an example of each of the following:

    1. A function which is bounded on a set \(S\) but not uniformly continuous on \(S\).

    2. A function \(f\) and a point \(a\) for which \(\lim_{x \to a^{+}} f(x) \ne \lim_{x \to a^{-}} f(x)\) (and both limits exist).

    3. A function \(f\) and a point \(a\) for which \(\lim_{x \to a^{\mathbb Q}} f(x) \ne \lim_{x \to a^{\mathbb R \setminus \mathbb Q}} f(x)\) (and both limits exist).

    4. A nonzero power series \(g(x) = \sum_{n=0}^\infty {a_n x^n}\) that is defined for all \(x \in \mathbb R\).

    1. The step function \(f(x) = \lfloor x \rfloor\) is bounded, but not continuous on \([0,2]\) so not uniformly continuous.

      The function \(f(x) = \sin(1/x)\) is continuous and bounded on \((0,1)\) but not uniformly continuous.

    2. The step function \(f(x) = \lfloor x \rfloor\) and the point \(x = 1\); the limit from the left is 0 but the limit from the right is 1.

    3. Let \(f(x) = 1\) if \(x \in \mathbb Q\) or \(f(x) = 0\) otherwise. (This function is called the indicator function of the rationals \(\mathbb Q\)) Then the limit along the rationals at any point \(x = a\) is 1 while the limit along the irrationals at that same point is 0.

    4. We require any power series whose radius of convergence is \(+\infty\), for instance \(\sum \frac{x^n}{n^n}\).

  2. Answer each question with a “yes” or “no,” and a brief (1 sentence) justification.

    1. If \(f\) is continuous on the bounded set \(S\), is it uniformly continuous on \(S\)?

    2. Is the interval \([a, b]\) an open set?

    3. Is there is a power series that converges only on the interval \([-1, 1)\)?

    1. Not necessarily. For instance \(f(x) = 1/x\) is continuous on the bounded set \((0,1)\) but not unformly continuous. Also, the function \(f(x) = \sin(1/x)\) is continuous and bounded on the bounded set \((0,1)\) but not uniformly continuous.

    2. No, as neither the point \(a\) nor \(b\) is interior to \([a,b]\). For instance, with \(a\): Let \(\varepsilon > 0\) and consider the set \((a-\varepsilon, a+\varepsilon)\), which contains \(a - \varepsilon/2\). But \(a - \varepsilon/2 < a\) is not in \([a, b]\), so this set cannot be a subset of \([a,b]\). \(\varepsilon> 0\) was arbitrary, so \(a\) is not interior to \([a,b]\).

    3. Yes. For instance \(\sum \frac{x^n}{n}\).

  3. A real set \(S\) is compact if it contains all of its limit points. In other words, the set \(S\) is compact if, for all sequences \((x_n)\) in \(S\) with limits (including \(\pm \infty\)), \(\lim x_n \in S\). Let \(S\) be any real, compact set.

    1. Show that \(S\) is bounded.

    2. Show that \(\bar S = \mathbb R \setminus S\) is open.

    1. Suppose without loss of generality that \(S\) is unbounded above. Then for all \(n \in \mathbb N\) there exists \(x_n \in S\) with \(x_n > n\). So \((x_n) \to +\infty\), which means that as \(S\) is compact, \(+\infty \in S\). But \(S\) is a real set, so this is a contradiction. So \(S\) is bounded.

    2. Consider \(x \in \bar S\). Suppose \(\forall n \in \mathbb N\), that the interval \((x - 1/n, x+1/n)\) intersects the set \(S\) (otherwise, \(\bar S\) would be open!) in at least one point we will call \(x_n\). By the Squeeze lemma, \((x_n) \to x\), so \(x \in S\) and hence \(x \not \in \bar S\), but this is a contradiction. So \(\bar S\) is open.

      (Remark: If \(\mathbb R \setminus S\) is open, we say “\(S\) is closed”)

  4. Let \(S \subset \mathbb R\) be compact, and let \(f\) be a real-valued function that is continuous on \(S\). Show that \(f\) is uniformly continuous on \(S\). Suggestion: Follow the proof of Theorem 19.2 in the book.

    Assume \(f\) is not uniformly continuous on \(S\). Then there exists \(\varepsilon >0\) so that for each \(\delta > 0\) there exist \(x,y \in S\) so that \(\vert x-y\vert < \delta\) and yet \(\vert f(x) - f(y) \vert \ge \varepsilon\). Thus for each \(n \in \mathbb N\) there exist \(x_n, y_n \in S\) such that \(\vert x_n - y_n \vert < 1/n\) and yet \(\vert f(x_n) - f(y_n) \vert \ge \varepsilon\). By the Bolzano-Weierstrauss Theorem, as \(S\) is bounded by 3.a., \((x_n)\) has a convergent subsequence \((x_{n_k})\) converging to \(x_0\). Moreover as \(S\) is compact, \(x_0 \in S\). By the Squeeze lemma, notice also that the subsequence \((y_{n_k})\) must also converge to \(x_0\). Since \(f\) is continuous at \(x_0 \in S\), we have

    \[ f(x_0) = \lim_{k \to \infty} f(x_{n_k}) = \lim_{k \to \infty} f(y_{n_k}) \]

    So by limit laws,

    \[ \lim_{k \to \infty} \left[ f(x_{n_k}) - f(y_{n_k}) \right] = 0. \]

    But this contradicts that \(\vert f(x_{n_k}) - f(y_{n_k}) \vert \ge \varepsilon\) for all \(k\). So \(f\) must be uniformly continuous on \(S\).