Given a set in (n+1) Euclidean space homeomorphic to the n-sphere, how do you show the existence of a simplicial complex also homeomorphic to the n-sphere arbitrarily close to it, if this is possible? I can find a proof for a diffeomorphism, but I was hoping to go the extra step.
(EDIT: What I mean by "the existence of a simplicial complex... arbitrarily close" is that, for every positive $\epsilon$, the existence of a complex every point of which is within $\epsilon$ of some point in the original set.)
Here's my initial thought: start with a homeomorphism from the boundary of the (n+1)-simplex (since this is homeomorphic to the sphere). Break the boundary down by barycentric subdivision, and use the convex hulls of the images of the vertices. By uniform continuity (since the simplex's boundary is compact), you can get arbitrarily close by this method, but the problem is that it isn't, in general, injective. It would be enough to show that it was injective for infinite steps, so that you could just skip over those that aren't. My hope was that you could maybe show that it only fails at finite steps (I don't at all know this is true), and I was thinking I might be able to get a contradiction by taking a representative pair of simplices in the domain at each step whose images under this new function meet and, assuming there are infinite steps, using compactness to pass the pairs of midpoints to a convergent subsequence, whose limit must be identical, since otherwise, eventually, simplices too small to meet would meet. The problem is I can't think of a problem with them converging to an identical pair. Is this tactic a fool's errand? Is there a similar tactic that might get me there, maybe altering the function when it fails to be injective?
I'd prefer a constructive proof if possible.