What bounds on cardinality have you found? For $d=1$ the answer would seem to be $2N-1$ by building a binary tree on subintervals, and I feel like a similar construction should work in the higher-$d$ case; e.g., by alternating dimensions we get $\{\{[i]\times[j]: 1\leq i,j\leq N\}$ $\cup\{[2i, 2i+1]\times[j]: 1\leq (2i), j\leq N\}$ $\cup\{[2i, 2i+1]\times[2j, 2j+1]\}: 1\leq(2i), (2j)\leq N\}$ $\cup\ldots\}$. (Note that for non-power-of-two $N$ you have to intersect these intervals with $[1,N]$ to keep from spilling over but that shouldn't affect things too much.)

@DanielAsimov Wouldn't the latter reading ('not homeomorphic to any of $\mathbb{R}$'s proper subsets') just be trivially false?

You might also want to keep an eye on math.stackexchange.com/questions/4937730/… (which at least as of yet has no answers) ; I didn't ask it there but I'm also curious as to whether this formula is fundamentally new; I'll be keeping an eye on this too.

@JoelDavidHamkins I think the biggest difference between square root and general rootfinding is the montonicity; it's very clear e.g. that $\sqrt{x^L}$ is in the left set for $\sqrt{x}$ and $\sqrt{x^R}$ is in the right set, because $x^R\gt x\implies \sqrt{x^R}\gt\sqrt{x}$. You might be able to elide this by using small enough neighborhoods to ensure that $f(x)$ is monotonic on that interval, but I suspect getting there is a little tricky in its own right.

Speaking of iteration, you might want to ponder on the definition of $1/y$ from Wikipedia again; it's qualitatively different from the definitions for $+$ and $\times$ in that it involves the options to previous approximations of $1/y$. Square root (which I found and will offer up in a new answer) has the same issue.

IIRC there's a definition for sqrt() in ONAG, though I wouldn't 100% trust it; I'll take a look when I have a chance.

Since the equation for your critical points is $e^z+2z+1=0$, you can solve it using the Lambert W function; see the first example at en.wikipedia.org/wiki/Lambert_W_function#Applications for how to do this.

Have you tried to find $F(p)$ explicitly as a function of $p$? It may be feasible to show that $F()$ is one to one and either strictly expanding or strictly contracting in some neighborhood of your fixed point, and that would ensure its existence.

Unless I'm very confused, this semigroup is 'trivially infinite'. any one-tile subset of a full tiling $\mathcal{T}$ is a member of $\mathcal{S(T)}$; so is any two-tile subset, etc. There's clearly at least one subset of each cardinality, so infinitely many elements of $\mathcal{S(T)}$. OTOH, since all members of $\mathcal{S(T)}$ are finite subsets of $\mathcal{T}$ there are only countably many of them.

One simple note: the regular tetrahedron cannot be cut into any number of regular tetrahedra, and generically 'most' tetrahedra can't be cut into any number of tetrahedra congruent to the original, by non-zero Dehn invariant. (Of course, this does nothing to solve OP's question since this involves a strong additional constraint, but I figured it was worth noting)

More broadly, as Gro-Tsen noted, 'how do I prove these conjectures that have been open for multiple decades' is not likely to garner much in the way of useful information. You could ask 'what strategies have been used to show that similar games are draws?', and TBH I suspect the first answer would be something along the lines of 'Go look at Winning Ways.'

Have you read either Winning Ways or the various Games of No Chance collections? They're perhaps the best starting point for this. AFAIK you may be able to show that 5-in-a-row is a win on a sufficiently large board 'conceptually', but I would expect that to be well explored already and (given the lack of results) to not readily yield to any techniques that we could offer you. Finding the smallest board on which it's a win seems almost guaranteed to be a product of long computer exploration if at all.

Fantastic question! Out of curiosity, have you looked at the statistics for $0-1$ polynomials (or, maybe better, polynomials with coefficients in $\{-1, 1\}$)? It would be interesting to see if the discrete case matches the continuous.

This is excellent information; thank you so very much!

I don't know if you were taking 'unique' as part of your definition, but it already follows from the rest of the definition: if there are two points $P$ and $Q$ then only one of them can bisect the chord passing through both of them.