Uniform continuity and bounded

1800bigk
Messages
42
Reaction score
0
Prove that if f is uniformly continuous on a bounded set S then f is bounded on S.

Our book says uniform continuity on an interval implies regular continuity on the interval, and in the previous chapter we proved that if a function is continuous on some closed interval then it is bounded. Is that how I should prove this, or am I missing something? It seems to easy this way and that usually means I am overlooking something.


tia
 
Physics news on Phys.org
You're overlooking the fact that not all bounded sets are intervals.

A possible way to prove the result, is to use that a uniformly continuous function maps Cauchy sequences to Cauchy sequences. If you assume that f is unbounded on S, this will lead to a contradiction.
 
I was just getting ready to say just because S is bounded doesn't mean S is closed and I need S to be closed to go the easy way. I supposed f is uniformly continuous and unbounded then I said since S is bounded then there is a convergent sequence {Xn} in S by bolzano and since {Xn} converges then {Xn} is cauchy. Since f is uniformly continuous on S then f(Xn) is cauchy so f is bounded, contradiction
 
That's not correct, but on the right path. You don't actually have a contradiction (unbounded sets can (and always do!) contain Cauchy sequences). You need to place more requirements on x_n.
 
what kind of requirements do I need? Do I need to say something about Xn converging to an element in S? Do I need to use subsequences?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top