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One could look at this question to find theorems which are true in low dimensions but false in high dimensions. The highest dimension involved seems to be $65$. So $\mathbb R^{65}$ is similar to all higher dimensional spaces in the sense that each property mentioned in that post either holds for all of them or fails for all of them
@GerhardPaseman The problem is to find a tile that can almost tile both of them. In particular such that the uncovered area is as small as possible relative to the area of the tile. You can look at the discussion page linked to from there if you want to see more detail.
Amazing! It sounds like he's well ahead of us. I was actually interested in the general conjecture from the start (a solution would let me kill this tricky problem), but I thought that making the problem more concrete would make it more interesting for a wider audience.
In fact this answer shows that if a triangle of side $n$ can be tiled then so can a triangle of side $n+15$. This because a triangle of side $n+15$ can be broken down into a triangle of side $n$, the above trapezium, and a parallelogram with sides $15$ and $n-15$. I just found a tiling of a trapezium whose two equal sides are length $20$ (image), so a tiling of the triangle with side $n$ also gives a tiling of the triangle with side $n+20$. So the side lengths we know we can tile are $\{45,60,65,75\}$ and all higher multiples of $5$.
Thanks for this! These insights could probably be combined with Timothy Chow's comment above to yield a fast computer program. I might implement this when I have the time.
I suspect you might want to strengthen your continuity condition even more. As it stands you can still have the following not-very-continuous-looking situation: However $F$ has been defined on the interval $[0,\tau]$ you can pick any $b$ with a map $m:F(\tau)\to b$ and define $F(t)=b$, $F(\tau\leq t)=m$ for $t>\tau$, with $F(t\leq t')=\mathrm{id}_B$. Your definition imposes an analogue of lower semicontinuity, so maybe you also want to add the dual criterion that $F$ commutes with limits from the right.
@SteveHuntsman Yes, one can use a binary proposition for each possible orientation and position of the tile, and then form the SAT formula that asserts that each small tile in the grid is covered by at least one tile and at most one tile. But I'm not sure that a general SAT solver would be quicker than Burr Tools which uses an algorithm specifically designed for the exact cover problem. Maybe it would be though. I've heard that modern SAT solvers are ferociously fast.
If we write $\mu$ as a weighted average of some other distributions $\nu_i$, then the distribution formed by the weighted average of $\tilde{\nu}_i$ has this property. One could probably state this as Jensen's inequality on $X$ where $X$ is some random variable in a joint distribution with some other random variable $Y$ which we are conditioning on. I guess $\mathbb E\left(f\left(X\right)\right)\geq\mathbb E\left(f\left(\mathbb E\left(X|Y\right)\right)\right)$.
