Does this work for #7? (and question) (Spoilers for #6):
I did #6 using 2D Sperner’s lemma and closedeness. Imagine the the destination points are colored [as in #5, which was a nice hint] by where they are relative to their source points—split the possible difference vectors into a colored circle as in #5 [pick the center to be a fourth color so you can notice if you ever sample a fixed point directly, but if fixed points are rare this shouldn’t matter], and take samples to make it look like 2d Sperner’s lemma, in which there must be at least one interior tri-colored patch. Define a limit of zooming in that moves you towards the tri-colored patch, apply closedness to say the center (fixed) point is included, much like how we were encouraged to do #2 with 1D Sperner’s lemma.
To do #7, it seems like you just need to show that there’s a continuous bijection that preserves whether a point is interior or on the edge, from any convex compact subset of R^2 to any other. And there is indeed a recipe to do this—it’s like you imagine sweeping a line across the two shapes, at rates such that they finish in equal time. Apply a 1D transformation (affine will do) at each point in time to make the two cross sections match up and there you are. This uses the property of convexity, even though it seems like you should be able to strengthen this theorem to work for simply connected compact subsets (if not—why not?).
EDIT: (It turns out that I think you can construct pathological shapes with uncountable numbers of edges for which a simple linear sweep fails no matter the angle, because you’re not allowed to sweep over an edge of one shape while sweeping over a vertex of the other. But if we allow the angle to vary slightly with parametric ‘time’, I don’t think there’s any possible counterexample, because you can always find a way to start and end at a vertex.)
Then once you’ve mapped your subset to a triangle, you use #6. But.
This doesn’t use the hint! And the hints have been so good and educational everywhere I’ve used them. So what am I missing about the hint?
Does this work for #7? (and question) (Spoilers for #6):
I did #6 using 2D Sperner’s lemma and closedeness. Imagine the the destination points are colored [as in #5, which was a nice hint] by where they are relative to their source points—split the possible difference vectors into a colored circle as in #5 [pick the center to be a fourth color so you can notice if you ever sample a fixed point directly, but if fixed points are rare this shouldn’t matter], and take samples to make it look like 2d Sperner’s lemma, in which there must be at least one interior tri-colored patch. Define a limit of zooming in that moves you towards the tri-colored patch, apply closedness to say the center (fixed) point is included, much like how we were encouraged to do #2 with 1D Sperner’s lemma.
To do #7, it seems like you just need to show that there’s a continuous bijection that preserves whether a point is interior or on the edge, from any convex compact subset of R^2 to any other. And there is indeed a recipe to do this—it’s like you imagine sweeping a line across the two shapes, at rates such that they finish in equal time. Apply a 1D transformation (affine will do) at each point in time to make the two cross sections match up and there you are. This uses the property of convexity, even though it seems like you should be able to strengthen this theorem to work for simply connected compact subsets (if not—why not?).
EDIT: (It turns out that I think you can construct pathological shapes with uncountable numbers of edges for which a simple linear sweep fails no matter the angle, because you’re not allowed to sweep over an edge of one shape while sweeping over a vertex of the other. But if we allow the angle to vary slightly with parametric ‘time’, I don’t think there’s any possible counterexample, because you can always find a way to start and end at a vertex.)
Then once you’ve mapped your subset to a triangle, you use #6. But.
This doesn’t use the hint! And the hints have been so good and educational everywhere I’ve used them. So what am I missing about the hint?