I'm trying to do the following exercise
1) Show that $\mathbb{C} \setminus \ (\{1/n \mid n \in \mathbb{N}\} \cup \{0\})$, $\mathbb{C}\setminus \mathbb{N}$ and $\mathbb{C}\setminus \mathbb{Z}$ are homeomorphic
2) Show that $\mathbb{C} \setminus \ \{1/n \mid n \in \mathbb{N}\}$ is not homeomorphic to the others
My attempt is:
1)
In the first space, holes accumulate at zero; in the second at infinity, so I thought to exchange zero with infinity with $z \to 1/z$. It is its own inverse, both are continuuos and the holes fit.
In the second space, holes go straight to the infinity; in the third they go in both direction, so I thought to use a "spiral" with $z \to e^{i\pi|z|}z$ and the inverse $z \to e^{-i\pi|z|}z$. This really goes to $\mathbb{C} \setminus \{(-1)^nn \mid n \in \mathbb{N}\}$, but I think this is homeomorphic to $\mathbb{C}\setminus \mathbb{Z}$ stretching the real axes with
$a + ib \to \begin{cases}1/2(a-1) + ib &a \leq -1\\ a + ib &-1 \leq a \leq 0\\ 1/2a + ib &a \geq 0\end{cases}$
2)
My idea here is that if they are homeomorphic it will exist a point corresponding to zero. But this has a neighborhood simply connected. Its image is homeomorphic so it's simple connected too, but it will be also contain many holes because holes accumulates at zero, so it's not simply connected.
Is 1) correct? And how can I formalize 2) ?
Thanks in advance
To formalize the simply-conectedness argument in part 2 (if you are not happy with it as is), you could consider all continuous injective maps $\overline{\Bbb D}\to X$ with $f(0)=z_0\in X$. If $X$ is any of the spaces of part 1 and $z_0$ any point of $X$, such maps clearly exist. If $X=\Bbb C\setminus\{\,\frac1n\mid n\in\Bbb N\,\}$ and $z_0=0$, such a map would have to contain $B(0,r)$ with $r=\min\{\,f(z)\mid z\in S^1\,\}$ in its image, but then it also contains $\frac1n$ for $n>\frac1r$.