# Harrison Chapman

## Math 317: Homework 8

Due: Friday, November 1, 2019

Exercises:

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 $$S$$.