TITLE: Question on Uniform Continuous Functions
QUESTION [2 upvotes]: I have a question that I need help. The question is:
Let $f$ be uniformly continuous on $A \subset \mathbb{R}$. Suppose there exists a constant $k$ such that $f(x) \geq k \gt 0$ for all $x$ in $A$. Prove that $\dfrac{1}{f}$ is also uniformly continuous on $A$. 
I was able to show that $\dfrac1f$ is continuous but I have difficulty showing whether $A$ is a closed interval. 
Just wondering if I'm on the right track for my proof? 
Any thoughts?

REPLY [1 votes]: You seem to try to use the theorem that a continuous function on a closed interval is uniformly continuous. This is not the way to go about this question since you really have no idea what the domain of definition of $f$ and there is no reason in the world why that would be a closed interval. 
Instead, follow the definition of uniform continuity. Start out with explicitly writing down what it would mean to show that $1/f$ is uniformly continuous. Then remember that $f$ is given to be uniformly continuous and try to use that in order to show what you want to show. Somewhere you will find it useful to know that $f(x)\ge k >0$ for all $x\in A$. That should indicate to you you are on the right track. 
Alternatively, the composition of uniformly continuous functions is uniformly continuous. So, (and this is a much shorter solution but it may obscure what is actually going on here) can you express the function $1/f(x)$ as the composition of two functions? make sure you use the condition that $f(x)\ge k >0$ for all $x\in A$ to properly define the other uniformly continuous function!
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