uniform continuity of $e^{-\frac{1}{|z+1|}}$

Question

Is function $e^{-\frac{1}{|z+1|}}$ uniform continuous on set of complex numbers $z \neq -1$. I tried to prove by definition, or make an example that it is not( tried sequences around -1, such as $x_{n}=-1+\frac{1}{n}$ or $x_{n}=-1+\ln(\frac{1}{n}+1)$ but still nothing.

Answer 1

Consider the function$$\begin{array}{rccc}f\colon&\Bbb C&\longrightarrow&\Bbb R\\&z&\mapsto&\begin{cases}e^{-1/|z+1|}&\text{ if }z\ne-1\\0&\text{ if }z=1.\end{cases}\end{array}\tag1$$Then $f$ is an extension of your function. Therefore, if $f$ is uniformly continuous, then so is your function. Also, note that $f$ is continuous and that therefore the restriction of $f$ to any compact subset of $\Bbb C$ is uniformly continuous.

If $f$ was not uniformly continuous, then there would be a number $\varepsilon>0$ and there would be two sequences $(z_n)_{n\in\Bbb N}$ and $(w_n)_{n\in\Bbb N}$ of complex numbers such that$$(\forall n\in\Bbb N):|z_n-w_n|<\frac1n\quad\text{and that}\quad\bigl|f(z_n)-f(w_n)\bigr|\geqslant\varepsilon.$$ There are wo possibilites now:

1. The sequence $(z_n)_{n\in\Bbb N}$ is unbounded. Then we can assume WLOG that $\lim_{n\to\infty}|z_n|=\infty$. So, $\lim_{n\to\infty}|w_n|=\infty$ too. But then $\lim_{n\to\infty}f(z_n)=\lim_{n\to\infty}f(w_n)=0$ and therefore we cannot have $(1)$.
2. The sequence $(z_n)_{n\in\Bbb N}$ is bounded. Then there is some $R>0$ such that every $z_n$ belongs to $\overline{D_R(0)}$ and so $(\forall n\in\Bbb N):z_n,w_n\in\overline{D_{R+1}(0)}$. But $\overline{D_{R+1}(0)}$ is compact. So, the restriction of $f$ to $\overline{D_{R+1}(0)}$ is uniformly continuous and, again, we canot have $(1)$.