I'm studying singularities in complex analysis and came across the following function:
$$f(z):=\frac{1}{z\cos(\frac{1}{z})}$$
I already figured out that $f$ has a non-isolated singularity in $0$ and that there are poles in $$z_k:=\frac{2}{\pi(1+2k)}$$ for $k\in\mathbb{Z}$, tending towards $0$ for $|k|\to\infty$. I am now trying to find out if the domain $$D_f:=\mathbb{C}\setminus(\{z_k|k\in\mathbb{Z}\}\cup\{0\}),$$ is closed, open or neither.
I already came up with a proof that $D_f$ cannot be closed because if $z\in\{z_k|k\in\mathbb{Z}\}\cup\{0\}$, there is no $\epsilon>0$ such that $B(z_1,\epsilon)\subseteq \{z_k|k\in\mathbb{Z}\}\cup\{0\}$ due to $z_k\le z_1$ for all $k\in\mathbb{Z}$.
Can one help me out with the rest?
Thank you in advance.
Best Answer
${D_f} = {\Bbb C}\backslash \left( {\left\{ {{z_k}} \right\}_{ - \infty }^\infty \cup \left\{ 0 \right\}} \right)$ is in fact open.
I believe it is easier to see that $B = \left\{ {{z_k}} \right\}_{ - \infty }^\infty \cup \left\{ 0 \right\}$ is closed.
According to this:
Prove that a set is closed iff it contains all its accumulation points
it suffices to check that $B$ contains all of its accumulation points.
Well the set $B$ has only one accumulation point and it's $0$ and it's contained in $B$.
Thus $B$ is closed and $${D_f} = {\Bbb C}\backslash B$$ is open.
Alternatively we can show that the set $B$ is closed by showing that it contains all of its boundary points.
By the definition of a boundary point ($x$ is a boundary points of $B$ if every neighborhood of $x$ contains at least one element in $B$ and at least one element not in $B$) we conclude that every point in $B$ is a boundary point. Thus $B$ is closed.