Real Analysis Uniform Continuity Question help If $f$ is uniformly continuous on a bounded subset $A$ of $\mathbb{R}$, then $f$ is bounded on $A$.
How would i start my proof? Could anybody give me a hint? I
 A: Hint for one approach:
If $\delta>0$, how many balls of radius $\delta$ are needed to cover $A$?

 If $f$ is uniformly continuous, we can find a $\delta>0$ such that $|f(x)-f(y)| < 1$ as long as $|x-y| < \delta$ and $x,y \in A$. It takes at most a finite number of sets of the form $\{y | |y-k \delta| < \delta \}$ to cover $A$. Call them $B_i$, with $i=1,...n$. Pick any point $x_i \in B_i$. Then show that $f$ is bounded by $\max(f(x_1),...,f(x_n))+1$.

Hint for another approach:
Can you extend $f$ to $\bar{A}$?

 If $f$ is uniformly continuous on $A$, it can be extended to a continuous function $\bar{f}$ on $\bar{A}$, the closure of $A$. Note that $\bar{A}$ is compact, hence $\bar{f}$ has a $\min$, $\max$. Hence $f$ is bounded.

A: Another approach to the problem: suppose that $f(A)$ is unbounded


*

*Since $f(A)$ is unbounded, there must be a sequence $\{x_n\}$ such that $\lim_{n \to \infty} f(x_n)$ is $\infty$ or $-\infty$

*Note that bounded sets in $\mathbb{R}$ are totally bounded; that is, every sequence in $A$ will have a Cauchy subsequence

*Note that $f$ is uniformly continuous, which means it maps Cauchy sequences to Cauchy sequences.


Putting these three together, you can derive a contradiction and force the conclusion that $f(A)$ is bounded.
Copper's first approach is the more common one, and if I recall correctly it pops up in Rudin somewhere.
