Intro: I would to show that $f(x)=\frac{1}{x-3}$ is not uniformly continuous on $(3,\infty)$. I would like to use the theorem that states: If $f$ is uniformly continuous on a set $S$ and $(s_n)$ is a Cauchy sequence in $S$, then $f(s_n)$ is a Cauchy Sequence.
My work so far$\rightarrow$ Proof: To get a contradiction, assume $f(x)=\frac{1}{x-3}$ is uniformly continuous. The sequence $(s_n)$ defined by $s_n=e^{-x}+3$ lies in $(3,\infty)$ and converges to $3$, so $s_n$ is a Cauchy Sequence. By the assumption and the theorem above, $$f(s_n)=\frac{1}{(e^{-x}+3)-3}=e^{x}$$is a Cauchy Sequence. This is a contradiction because the sequence $(e^{x})$ diverges to $+\infty$, so it cannot be a Cauchy sequence; therefore, $f(x)$ is not uniformly continuous on $(3, \infty)$. $\square$
I am concerned that my chosen $s_n$ will not work since its domain is $(-\infty,\infty)$. Is it true that $s_n$ lies in $(3,\infty)$? That is, am I okay to state the bolded part of my proof? Clearly $(3,\infty)$ lies in $(-\infty,\infty)$.
Any help with this scenario would be greatly appreciated!
Thank you!