When I am reading the descriptions of 2-dimensional Poincare conjecture on Wikipedia , it menstions "a compact 2-dimensional surface without boundary". People told me that a 2-sphere will be an example of "a compact 2-dimensional surface without boundary". My question is, what is the topological space we are talking about? Is it $\mathbb R^3$ or the 2-sphere itself?
If the whole topological space is $\mathbb R^3$, then a 2-sphere is clearly bounded. It is closed because $\mathbb R^3 \backslash \{(x, y, z) \vert x^2 + y^2 + z^2 = 1\}$ is open. But in this case the 2-sphere contains the boundary, because the interior of a 2-sphere will be empty, and the closure will be 2-sphere itself. Hency the boundary will be 2-sphere itself.
If the whole topological space is the 2-sphere itself, then it is also bounded. It is closed because the empty set is open. Hence it is compact. The interior is 2-sphere itself, and the closure is also 2-sphere. So it is without a boundary.
So looks like we are using the 2-sphere as the whole topological space. Is my understanding correct?
To summarize my question, what will be the whole topological space in the context of the 2-dimensional Poincare conjecture?
Best Answer
You're mistaken, the two sphere does not contain a boundary because every point is an interior point. This mean it has no boundary points. Note that the two sphere is merely the surface of a sphere, no other points are included. The interior and exterior you're describing are part of $\mathbb{R}^3$ not the two sphere.