# Proof $f(x)=\frac{1}{x}$ is not uniformly continuous checking

I want to evaluate if $f(x)=\frac{1}{x}$ is uniformly continuous.

According to the definition:$\forall\epsilon>0\exists\delta>0\forall(x,x´)\in X\times X\:\:\:d_X(x,x´)<\delta\implies d_y(f(x),f(x´))<\epsilon$

$\delta>0,\:\:|x-c|<\delta$, taking $x=\frac{1}{\delta}$ and $c=x+\delta$. Considering $\epsilon>0$.

Then,

$|f(x)-f(c)|=|\frac{1}{x}-\frac{1}{c}|=|\frac{c-x}{xc}|\leqslant\frac{|c-x|}{xc}<\frac{\delta}{xc}=\frac{\delta}{\frac{1}{\delta^2}+1}>\frac{\delta}{\delta}=1>\epsilon$

Question:

Is my proof right?

1. Showing that $a < b > c$ shows does not show any relationship between $a$ and $c$. As such, you can't conclude that $|f(x) - f(c)| > \varepsilon$ from the above argument.
2. The $x$ and $c$ constructed satisfy $|x - c| = \delta$, not $|x - c| < \delta$. Might I suggest letting $c = x + \frac{\delta}{2}$ instead?
3. Point 1 can be circumvented by noting that $$\left|\frac{c - x}{xc}\right| = \frac{c - x}{xc} = \frac{\delta}{xc}.$$
4. I find the best way to disprove uniform continuity is to find a Cauchy sequence in the domain that doesn't map to a Cauchy sequence. For example, $(1/n)_{n=1}^\infty$ is Cauchy, but maps to $(n)_{n=1}^\infty$ which is not Cauchy.