# A uniformly continuous function maps bounded set to bounded sets

I am trying to prove the following:

If $$A\subset\mathbb R$$ is bounded and $$\,f:A\to \mathbb R\,$$ is uniformly continuous, then $$f[A]$$ is bounded.

Could you check my proof?

Let $$A \subseteq [-K,K]\subseteq \mathbb R$$. Let $$\varepsilon = 1$$. If $$f$$ is uniformly continuous there is $$\delta$$ wuth $$|x-y| <\delta$$ imply that $$|f(x)-f(y)|<1$$ for all $$x,y\in A$$. Let $$a \in A$$. Then because $$A$$ is bounded there is a finite number of balls $$B(a_n,\delta)$$ that cover $$A$$. Let the number be $$N$$. Then $$f(a)-N \le f(x) \le f(a) + N$$ for all $$x\in A$$.

• It's not correct, you need to consider the values $f(a_n)$ for all $n$. If $A$ were assumed to be an interval, it would be correct, but without that assumption, it's not. Consider $A = \{-1,1\}$. Then two balls suffice, but $\lvert f(1) - f(-1)\rvert$ can be arbitrarily large. – Daniel Fischer Jan 20 '14 at 11:37
• @DanielFischer $|f(1) - f(-1)|$ cannot be arbitrarily large, since $f$ is fixed. – 5xum Jan 20 '14 at 11:38
• @5xum It is a fixed number once you know $f$. But if all you know is that $f \colon \{-1,1\} \to \mathbb{R}$ is uniformly continuous (which is vacuously true), then every non-negative real number is possible for $\lvert f(1) - f(-1)\rvert$. – Daniel Fischer Jan 20 '14 at 11:41
• @DanielFischer Yes, but if for a given $f$, you want to prove that $f(A)$ is bounded (which is what OP is doing), you first fix $f$. For any function $f$, $f({-1,1}) is in fact a bounded set. – 5xum Jan 20 '14 at 11:44 • @5xum Yes, it is bounded. But$f(A)$need not be contained in$[f(a)-N,f(a)+N]$, where$N$is the number of$\delta$-balls required to cover$A$, and$a\in A$. That is the point. But if we take$m = \min \{ f(a_n\}$and$M = \max \{ f(a_n)\}$, then$f(A) \subset [m-1,M+1]$. – Daniel Fischer Jan 20 '14 at 11:48 ## 2 Answers Assume that$f$is unbounded, and$\sup_{x\in A} f(x)=\infty$. (The case$\inf_{x\in A} f(x)=-\infty$can be treated in the same way.) Then, there is a sequence$\{x_n\}\subset A$, such that$f(x_n)\to\infty$. We can pick a subsequence$\{y_n\}$of$\{x_n\}$, such that$f(y_{n+1})-f(y_n)>1$, for all$n\in\mathbb N$. Since$f$is uniformly continuous, there exists a$\delta>0$, such that, for all$x,y\in A$, $$|x-y|<\delta\quad\Longrightarrow\quad |\,f(x)-f(y)|<1.\tag{1}$$ But as$A\subset\mathbb R$is bounded, then$\{y_n\}$has a convergent subsequence$z_n\to z\in \overline{A}$. In fact, we may pick the subsequence$\{z_n\}$, so that$|z_m-z_n|<\delta$, for all$m,n\in\mathbb N$, which implies that, for all$m,n\in\mathbb N$, with$m\ne n$, we have $$|z_m-z_n|<\delta, \quad\text{while}\quad |f(z_m)-f(z_n)|>1,$$ which contradicts$(1)$. Note. Since$f$is uniformly continuous, then$f$extends continuously to$\overline{A}$. This is done in the following way. If$\{x_n\}\subset A$is Cauchy, then so is$\{f(x_n)\}$, due to the uniform continuity of$f$. Hence,$f(x)$can be extended continuously (and uniquely) to$\overline{A}$. But$\overline{A}$is compact, since$A$is bounded. The extension of$f$shall be bounded, as it is continuous on a compact set, and thus$f$is bounded on$A$. • The proof is only okay if$A$is assumed connected. For disconnected$A$, you can't bound$f(A)$by one value of$f$and the number of balls needed to cover$A\$. – Daniel Fischer Jan 20 '14 at 11:50
• @DanielFischer: Correct. See updated version. – Yiorgos S. Smyrlis Jan 20 '14 at 12:38

My proof is super long but the person that used sequences made me realized there's a much easier proof. Here is the long one anyways because I thought it was interesting:

Let $$s=\sup A$$ and $$t=\inf A$$. There is $$\delta>0$$ such that if $$|x-y|<\delta, x,y\in A \Rightarrow |f(x)-f(y)|<1$$. Suppose that $$f(A)$$ is unbounded. Then $$s>t\Rightarrow \exists n$$ such that $$\displaystyle \frac{\delta}{s-t}>\frac{1}{n}\Rightarrow t+n\delta>s$$. We will prove that there are $$n$$ points of $$A$$ whose union of $$\delta-$$neighborhoods cover $$A$$.

Choose $$x_{1}\in A\Rightarrow \exists x_{2}\in A$$ such that $$f(x_{2})>f(x_{1})+1\Rightarrow |x_{2}-x_{1}|\geq \delta$$. Similarly, $$\exists x_{3}\in A$$ such that $$f(x_{3})>f(x_{2})+1$$ and notice that $$|x_{3}-x_{1}|\geq \delta$$ and $$|x_{3}-x_{2}|\geq \delta$$. Suppose that we have chosen $$x_{1},\ldots,x_{n}\in A$$ and assume, without loss of generality, $$x_{1}.

Take $$x\in A$$ and suppose that $$|x_{i}-x|\geq \delta,i=1,\ldots,n$$. If $$x_{i}-x\geq \delta \Rightarrow x_{n}-x=x_{1}-x+x_{2}-x_{1}+\cdots+x_{n}-x_{n-1}\geq x_{1}-x+\delta(n-1)\geq n\delta \Rightarrow x_{n}>x+n\delta$$. But $$x+n\delta \geq t+n\delta>s$$ and $$x_{n}\in A$$. Then we have a contradiction. If $$x-x_{i}<-\delta\Rightarrow \delta repeat the argument. Hence $$A\displaystyle \subset \bigcup_{i=1}^{n}(x_{i}-\delta,x_{i}+\delta)$$. But $$|x_{i}-x|<\delta$$ implies $$f(x) for some $$i\Rightarrow f(x)<\max \left\{f(x_{i})\right\}+1$$ contrary to $$f(A)$$ being unbounded.