Uniformly continuous and bounded

[math2006]

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

Since f is uniformly continuous, take $$\epsilon = m, \exists \delta > 0$$
such that
$$|f(x)-f(p)| < \epsilon$$
whenever $$|x-p|<\delta$$ and $$x,p \in A$$
Now we have
$$|f(x)| < m + |f(p)|$$

Obviously i should show $$b + |f(p)|$$ is bounded, but no idea how.
Could someone help me? thx

 

[AKG]

Obviously i should show $$b + |f(p)|$$ is bounded, but no idea how.

What is b? And why would you want to show that b + |f(p)| is bounded? You mean bounded as a function of p? Well that's no easier than showing directly that f is bounded, so this doesn't seem to be a worthwhile approach.

The idea is simple. Fix $\epsilon > 0$. There exists $\delta > 0$ such that, well, you know. Since A is bounded, you can choose a FINITE number of points x₁, ..., x_n in A such that the intervals of radius $\delta$ about the x_i cover A. The total variation of f on these intervals is at most $2\epsilon$, so the total variation of f over all of A is at most $2n\epsilon$.

Oh, and if this was homework, it should have been in the homework section.