I'm trying to figure out whether $S^2 := \{ v \in \mathbb R^3, |v| = 1 \} $ is homeomorphic to $D^2 := \{ v \in \mathbb R^2, |v| \leq 1 \}$, the closed unit disk. Intuitively I would say no, since $S^2$ has empty interior as a subset of $\mathbb R^3$ but $D^2$ doesn't as a subset of $\mathbb R^2$. However, I couldn't find anything about the boundary or interior of homeomorphic sets.
Here's an actual attempt to a line of reasoning : a homeomorphism maps open sets to open sets, however $S^2$ does not have any open subset, while $D^2$ does (such as any open ball of radius smaller than 1 centered at the origin), so the two are not homeomorphic.
The part I'm unsure about is that technically, $D^2$ has empty interior as a subset of $\mathbb R^3$ as well. I'm not sure how to address this "is $D^2$ a subset of $\mathbb R^2$ or $\mathbb R^3$" problem.
Best Answer
I'm assuming you don't know about homotopy groups otherwise this problem would be easy. Here is a proof using a result that might show up in a first topology class.
The Borsuk-Ulam theorem states that any for any map $f: S^n \to \mathbb{R}^n$ there exists $x\in\mathbb{R}^n$ so that $f(x) =f(-x)$. Using $n=2$ and replacing the plane with the disk (since the disk is a subset of the plane), the theorem still holds; hence, there are no injective maps from the 2-sphere to the disk, so they cannot be homeomorphic.