0
$\begingroup$

Here is Prob. 8, Chap. 4 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Let $f$ be a real uniformly continuous function on the bounded set $E$ in $\mathbb{R}^1$. Prove that $f$ is bounded on $E$.

Show that the conclusion is false if boundedness of $E$ is omitted from the hypothesis.

My effort:

Since $f$ is uniformly continuous on $E$, we can find a real number $\delta > 0$ such that $$ \vert f(x) - f(y) \vert < 1$$ for all points $x, y \in E$ for which $\vert x-y \vert < \delta$.

Let $a = \inf E$ and $b = \sup E$. Now let $N$ be a natural number such that $$N > \frac{b-a}{\delta}.$$ Then $$0 \leq \frac{b-a}{N} < \delta.$$

Now for each $k \in \{ 1, \ldots, N \}$, let $$I_k = \left[ a + \frac{(b-a)(k-1)}{N} , \ a + \frac{(b-a)k}{N} \right].$$ Then the length of $I_k$ is exactly $\frac{b-a}{N}$, and therefore we can conclude that, for any points $x, y \in E \cap I_k$, the distance $$\vert f(x) - f(y) \vert < 1.$$ Moreover, if $E \cap I_k$ is non-empty, then we take $x_k$ to be an arbitrary but fixed point in $E \cap I_k$, for each $k$, and note that, for any point $x \in E \cap I_k$, the following holds. $$ \vert f(x) \vert \leq \left\vert f(x) - f\left(x_k \right) \right\vert + \left\vert f\left( x_k \right) \right\vert < 1 + \left\vert f\left( x_k \right) \right\vert.$$

So if $x \in E$, then $x \in I_j$ for some $j$, and so $x \in E \cap I_j$ for that same $j$. Therefore we have $$ \vert f(x) \vert < 1 + \left\vert f\left( x_j \right) \right\vert \leq 1 + \max_{k=1}^N \left\vert f\left( x_k \right) \right\vert = 1 + M,$$ where $M$ is any real number such that $$ M > \max_{k=1}^N \left\vert f\left( x_k \right) \right\vert,$$ showing that $f$ is bounded on $E$.

Is this proof correct? If so, then can we generalize this result to a uniformly continuous mapping of a bounded subset of a given metric space into a metric space? If not, then where have I erred?

Can we find any real uniformly continuous function other than the identity function on $\mathbb{R}^1$ which is unbounded on any unbounded subset of $\mathbb{R}^1$?

$\endgroup$
  • 1
    $\begingroup$ Can we find any real uniformly continuous function other than the identity function on $\mathbb R^1$ which is unbounded on any unbounded subset of $\mathbb R^1$? Consider affine transformations of the identity function. $\endgroup$ – Theoretical Economist Jan 16 '17 at 22:46
2
$\begingroup$

Example: Endow $\Bbb R$ with the metric $d_*(x,y):=\min(|x-y|,1)$. Then (i) $\Bbb R$ is $d_*$-bounded, (ii) the identity function $f$ is $d_*$-uniformly continuous, but (iii) $f$ is not bounded.

The metric space $(\Bbb R,d_*)$ lacks a property, enjoyed by $(\Bbb R,d)$ where $d$ is the usual metric on $\Bbb R$, that is crucial to your argument: If $E\subset\Bbb R$ is $d$-bounded and $\delta>0$ is fixed, then there is a finite collection $B_1,B_2,\ldots,B_n$ of $d$-balls such that $E\subset\cup_{k=1}^n B_k$.

$\endgroup$
  • $\begingroup$ granted that $d_*$ makes $\Bbb R$ bounded, but what is the codomain for the identity function which you claim is uniformly continuous? $\endgroup$ – Saaqib Mahmood Mar 28 '17 at 19:40
  • 1
    $\begingroup$ You can take the co-domain to be $(\Bbb R, d^*)$ or $(\Bbb R,d)$. $\endgroup$ – John Dawkins Mar 28 '17 at 22:37
1
$\begingroup$

I think I've just managed to come up with another argument to prove the above result. Here's how the proof goes!!

Suppose $E \subset \mathbb{R}$, suppose $E$ is bounded, and suppose $f \colon E \to \mathbb{R}$ is a uniformly continuous function.

If $f(E)$ were unbounded, then we could find a point $x_1 \in E$ such that $f\left( x_1 \right) > 1$. Now assuming that $x_n \in E$ has been found, where $n \in \mathbb{N}$, if $f(E)$ were unbounded, then we could find a point $x_{n+1} \in E$ such that $f \left( x_{n+1} \right) > f\left( x_n \right) + 1$.

Thus we have a sequence $\left( x_n \right)_{n\in\mathbb{N}}$ in $E$ such that $$f\left( x_{n+1} \right) > f\left(x_n\right) + 1, \ \mbox{ and } \ f\left(x_n\right) > n \ \mbox{ for all } n \in \mathbb{N}.$$ Now by Theorem 3.6 (b) in Baby Rudin, the sequence $\left( x_n \right)_{n\in\mathbb{N}}$ in the bounded set $E \subset \mathbb{R}^1$ has a convergent subsequence, say, $\left( x_{\varphi(n)} \right)_{n\in\mathbb{N}}$, where $\varphi \colon \mathbb{N} \to \mathbb{N}$ is a strictly increasing function.

Now as $\left( x_{\varphi(n)} \right)_{n\in\mathbb{N}}$ is a Cauchy sequence in $E$ and as $f$ is uniformly continuous on $E$, so the image sequence $\left( f \left(x_{\varphi(n)}\right) \right)_{n\in\mathbb{N}}$ is a Cauchy sequence in the usual metric space $\mathbb{R}$, by Prob. 11, Chap. 4 in Baby Rudin. And, since $\mathbb{R}$ is a complete metric space, the subsequence $\left( f \left(x_{\varphi(n)}\right) \right)_{n\in\mathbb{N}}$ converges in $\mathbb{R}$.

But $\left( f \left(x_{\varphi(n)}\right) \right)_{n\in\mathbb{N}}$ is a subsequence of the sequence $ \left( f \left(x_n\right) \right)_{n\in\mathbb{N}}$, and therefore, for every natural number $n$, we have $$f \left( x_{\varphi(n)} \right) > \varphi(n) \geq n,$$ which contradicts the fact that the sequence $\left( f \left(x_{\varphi(n)}\right) \right)_{n\in\mathbb{N}}$ converges in $\mathbb{R}$. Hence $f(E)$ must be bounded.

Is the above proof correct? If so, then is my presentation of this proof good enough too? If not, then where do problems lie?

$\endgroup$
  • $\begingroup$ @Omnomnomnom can you please have a look at the answer I've just posted? What do you find it like? $\endgroup$ – Saaqib Mahmood Mar 28 '17 at 19:10
  • $\begingroup$ @JohnDawkins can you please go through my answer? I've attempted to give an alternative proof. Do you find it good enough? $\endgroup$ – Saaqib Mahmood Mar 28 '17 at 19:11
  • $\begingroup$ @TheoreticalEconomist I would be grateful if you could kindly take time reviewing my proof in the post I've just added. That would be so kind of you!! $\endgroup$ – Saaqib Mahmood Mar 28 '17 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.