1. ## Uniform Discontinuous Proof

Prove that f(x)=1/x is not uniformly continuous on (0, inf)?

First of all I wanted to make sure that the negation of:
for every epsilon > 0 there exists delta > 0 such that x,y element S and |x-y| => |f(x)-f(y)|< epsilon
is:
there exists epsilon > 0 such that for every delta > 0 there exists x, y element S such that |x-y|< delta and |f(x)-f(y)|>= epsilon

So here is how I started:
We can set epsilon = 1 so |1/x - 1/y|>= 1. Then |x-y|<delta so |x-y|=delta/2 which means y = x + delta/2. Then we can plug in to get
|1/x - 1/(x+delta/2)|>=1. Simplifying we get delta/2 >= x(x+delta/2).

Here I don't know what x I need to use so that the statement is true. Actually, I don't think my way is possible. I was wondering if there is another way to approach this but still use the definition.
Thanks

2. Scratch that, I actually have no idea how to do this problem.

3. Note that $\begin{array}{rcl}
f(x) = \frac{1}{x}\quad & \Rightarrow & \quad \left| {f\left( n \right) - f\left( {n + k} \right)} \right| \\
& = & \quad \left| {\frac{1}{n} - \frac{1}{{n + k}}} \right| \\
& = & k \\
\end{array}$
.

Consider the sequence $\left( {\frac{1}{n}} \right) \to 0$.

4. Let sn = 1/n for n element N. Then (sn) is cauchy in (0,inf). Since f(sn) = n, (f(sn)) is not cauchy. Therefore f cannot be uniformly continuous on (0, inf).

I think that proof works. I was just wondering how you would do a proof using the precise definition uniformly continuous and a contradiction

for each eps >0 there exists delt>0 s. t. s, y ele S and |x-y| => |f(x)-f(y)|<eps

THanks

5. Oh yeah by the way how do you get the math symbols like epsilon, delta, subset and stuff?

Thanks

6. Theorem: If $f$ is uniformly continous on a non-empty set $S$ then any Cauchy sequence $x_k$ in $S$ we have $f(x_k)$ is a Cauchy sequence.

So consider $x_k = \frac{1}{k}$ this is a Cauchy sequence in $(0,\infty)$ but $f(x_k) = k$ is not.