Is D = $\{ (x,y,z) \in \mathbb{R}^3 \vert x^2 + y^2 +z^2 = 1, 0 < z < 1 \}$ homeomorphic to $S^1$ and to $\mathbb{R}^2\setminus \{(0,0)\}$

Question

Is D = $\{ (x,y,z) \in \mathbb{R}^3 \vert x^2 + y^2 +z^2 = 1, 0 < z < 1 \}$ homeomorphic to $S^1$ and to $\mathbb{R}^2\setminus \{(0,0)\}$?

I define $S^1 = \{(x,y) \in \mathbb{R}^2 \vert x^2 + y^2 = 1 \}$. I think that D is not compact because is limited but not closed, is that right? If it is, then it cannot be homeomorphic to $S^1$ because $S^1$ is indeed compact.

On the other hand I think it is homeomorphic to $\mathbb{R}^2 \setminus \{(0,0)\}$ but I do not know how to prove it.

Any tips?

Answer 1

You are right about $D$ not being homeomorphic to $S^1$ (and about the reason why).

On the other hand$$\begin{array}{rccc}f\colon&D&\longrightarrow&D\bigl((0,0),1)\setminus\{0\}\\&(x,y,z)&\longrightarrow&(x,y)\end{array}$$is a homeomorphism and I suppose that you know how to prove that $D\bigl((0,0),1)\setminus\{0\}$ and $\mathbb{R}^2\setminus\{(0,0)\}$ are homeomorphic.