Harrison Chapman

Math 317: Homework 7

Due: Friday, October 25, 2019

Exercises:

1. Let $$f$$ be uniformly continuous on a bounded set $$S$$. Prove that $$f$$ is bounded (on the set $$S$$).

Proof. Assume that $f$ is uniformly continuous on $S$ but suppose to the contrary that $f$ is unbounded on $S$. Then for all $n \in \mathbb N$, there exists as number $x_n$ so that $f(x_n) > n$. Use these to define the sequence $(x_n)$.

$S$ is bounded, so by the Bolzano-Weierstrauss theorem, there exists a convergent, hence Cauchy, subsequence $(x_{n_k})$. As $f$ is assumed to be uniformly continuous, this means that $(f(x_{n_k}))$ is itself a Cauchy, hence convergent, sequence. Say it converges to the number $y$.

But for all $n_k$, we have that $f(x_{n_k}) > n_k$. As the sequence $$(n_k)$$ diverges to $$+\infty$$, this implies that the sequence $$(f_{n_k})$$ cannot possibly converge to $y$, a contradiction. Hence $f$ must be bounded on $S$. $\Box$

2. Give an example of each of the following, or explain why one does not exist.

1. A function that is continuous on $$[a,b]$$ but not uniformly continuous.

2. A function that is continuous on $$(a,b)$$ but not bounded.

3. Two functions $$f$$ and $$g$$ so that neither $$f$$ nor $$g$$ is continuous at $$x_0 = 1$$ but $$fg$$ is.

1. Cannot exist; by theorem 19.2 all functions continuous on $$[a,b]$$ are uniformly continuous on $$[a,b]$$.

2. Consider for instance $$f(x) = 1/x$$ on $$(0,1)$$.

3. Consider for instance $$f(x) = 1$$ except $$f(1) = 0$$ and $$g(x) = 0$$ except $$g(1) = 10$$. Then $$(fg)(x) = 0$$ which is constant and hence continuous, while neither $$f$$ nor $$g$$ were at $$x=1$$.

3. Show that $$x = \cos x$$ for some $$x \in [0, \pi]$$.

Consider $$g(x) = \cos x - x$$; $$g(0) = 1$$ and $$g(\pi) = 1-\pi < 0$$. $$g(x)$$ is continuous, so by the Intermediate Value Theorem there exists $$c \in [0, \pi]$$ with $$g(c) = 0$$, i.e., $$\cos c = c$$.

4. Recall the definition of open set from homework 6.

1. If $$A$$ and $$B$$ are open sets, prove that $$A \cap B$$ and $$A \cup B$$ are open.

2. Prove that if $$(A_i)$$ is a sequence of open sets then $$B = \bigcup_{i=1}^{\infty} A_i = \{ x : x \in A_i \textrm{ for some } i \in \mathbb N \}$$ is also open.

3. Find a sequence of open sets $$(A_i)$$ for which the set $$\bigcap_{i=1}^\infty A_i = \{x : x \in A_i \textrm{ for all } i \in \mathbb N\}$$ is not open.

1. If $$x \in A \cap B$$, then $$x$$ is interior to $$A$$ and $$B$$, meaning there exists $$r > 0$$ with $$B_r(x) = (x-r, x+r)$$ a subset of both $$A$$ and $$B$$. This means that $$B_r(x) \subseteq A \cap B$$. So $$x$$ is interior; $$x$$ is arbitrary, so the set is open.

If $$x \in A \cup B$$, then without loss of generality $$x$$ is interior to $$A$$, meaning there exists $$r > 0$$ with $$B_r(x) = (x-r, x+r)$$ which is a subset of $$A$$. This means that $$B_r(x) \subseteq A \cup B$$. So $$x$$ is interior; $$x$$ is arbitrary, so the set is open.

2. If $$x \in \bigcup_{i=1}^\infty A_i$$, then $$x$$ is interior to $$A_k$$ for some $$k$$, meaning there exists $$r > 0$$ with $$B_r(x) = (x-r, x+r)$$ a subset of $$A_k$$. So every $$x’ \in B_r(x)$$ is in $$A_k$$ and hence also in $$\bigcup_{i=1}^\infty A_i$$. This means that $$B_r(x) \subseteq \bigcup_{i=1}^\infty A_i$$. So $$x$$ is interior; $$x$$ is arbitrary, so the set is open.

3. Consider for instance $$A_i = (-1/i, 1/i)$$. Then the only element of $$\bigcap_{i=1}^\infty A_i$$ is $$0$$ (all other reals are not in $$A_k$$ for some $$k$$ sufficiently large). So $$\bigcap_{i=1}^\infty A_i = \{0\}$$, which is not open by definition.