The subsets you partition in are necessarily non-measurable sets. That makes them very strange. Any reasonble set that you can come up with is measurable. Even the set of points with rational coordinates is measurable. Of course you cannot partition a gold cube into two pieces, one with the points that have rational coordinates, and the other with points that have irrational coordinates. Non-measurable sets are even weirder.

This is only an indication that arbitrary subsets of R^3 are not a good way to model three dimensional space.

It's not that the cuts are strange, it's that they involve infinitely small details. You can't cut physical atoms like that, and so the construction fails in the real world.

Then his article just leaves me more confused than before. I was already familiar with the traditional resolution of BT as "well, that's just an artifact if the weirdness when you have uncountably many points".

But now the author insists that "oh no, you get the same paradox with finite pieces". And yet on every probe of that point, it comes back to an issue of infinities. So what's wrong with the traditional explanation? And how does this article justify a "finite version" of the partition.

Having finitely many pieces is different from finitely many operations you would need to cut the pieces in real life.

For example, there is a curve that cuts the plane into just two pieces, but the line itself has infinitely small resolution [1] and hence you can't cut it. In fact, there is no segment of this line, no matter how small, that you can cut in a finite amount of time!

[1]: http://en.wikipedia.org/wiki/Weierstrass_function

The article doesn't try to explain how the construction works. The construction is done with finitely many pieces (I think five pieces is the limit), but those pieces have infinitesimally small details.

I don't understand what you think the "traditional resolution" might be. In the Banach-Tarski theorem you are partitioning a 3-dimensional solid ball into finitely many pieces. Because the ball has uncountably many points, those pieces will have uncountably many points.[0]

Does that help?

[0] Actually that only shows that at least one of the pieces must have uncountably many points, but in the theorem we find that at least four pieces must have uncountably many points.