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 $\mathbb{R}^1$?

  • 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

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 $E\subset\cup_{k=1}^n B_k$.

  • $\begingroup$ 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

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?

  • $\begingroup$ @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

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