forlesen wrote:If this is correct, we would seem to end up with for any uncountable subset X of R*R, there is a line that partitions X into two uncountable halves (plus some points on the line).

I think that conclusion is correct, but your argument is a bit lacking. The way I would argue for it is, for U an uncountable subset of RxR:

For any real number r, we can partition U into the sets Tr, Ur and Vr whose X co-ordinates are < r, = r, and > r respectively. If Tr and Vr are both uncountable for any r, then the line x=r does the job and we're done.

Otherwise, for any real number r, at least one of Tr and Vr is countable (for the avoidance of doubt, I am using 'countable' to mean either finite or countably infinite - i.e. it doesn't imply infinite, and I'll more specifically say 'countably infinite' when needed).

Suppose there is no r such that Vr is countable. Then Tr is countable for any r, and in particular every one of the countably infinite sequence of sets ..., T(-2), T(-1), T0, T1, T2, T3, ... is countable. Therefore the union of all the sets in that sequence is countable - but U is that union (every point in U is in Ti for all values of i > its X co-ordinate). That contradicts U being uncountable, and therefore we can conclude that there is an r such that Vr is countable.

A symmetrical argument shows that there is an r such that Tr is countable. So there exist real numbers r1 and r2 such that Tr1 and Vr2 are both countable. Those numbers must be such that r1 <= r2, because if r1 > r2, then every point of U has X co-ordinate < r1, > r2 or both, making U equal to the union of Tr1 and Vr2 and so again contradicting its uncountability.

Consider the set S of r such that Tr is countable. It is non-empty, as it contains r1, and it has the upper bound r2 (since if s >= r, Ts is a superset of Tr and so also uncountable, so s is not in S), so it has a least upper bound l. This has the properties that:

* if r > l, Tr is uncountable (since if Tr were countable, r would be in S and so l would not be an upper bound of S);

* if r < l, Tr is countable (since if Tr were uncountable, then for any s > r, Ts would be a superset of Tr and also uncountable, so r would be a smaller upper bound of S).

Symmetrically, the set of r such that Vr is countable is non-empty (containing r2) and has the lower bound r1, so has a greatest lower bound m that has the properties if r < m, Vr is uncountable, and if r > m, Vr is countable.

If l > m, then r = (l+2m)/3 and s = (2l+m)/3 (the 1/3rd and 2/3rd points between them) have the properties that m < r < s < l, so Ts and Vr are both countable. But the union of Ts and Vr contains all points of U whose X co-ordinate is < s, > r or both, which is the whole of U, which again contradicts U's uncountability.

If l < m, then r = (l+m)/2 (the halfway point between them) has the properties that l < r < m, so Tr and Vr are both uncountable. That means that the line x=r has the property that there are uncountably many points of U on either side of it, so we're done in that case.

The only possibility that we're left with is that l = m. In that case, we can use an argument along your lines: l-1, l-1/2, l-1/4, l-1/8, ... are all less than l, so T(l-1), T(l-1/2), T(l-1/4), T(l-1/8), ... are all countable, and so the union of all those sets is countable, being a countable union of countable sets. But that union is Tl (any X co-ordinate that is less than l is also less than l-1/2^n for sufficiently large n), so that implies Tl is countable. Symmetrically, Vl = Vm is countable.

But we know that U is uncountable and is the disjoint union of Tl, Ul and Vl. Since Tl and Vl are countable, Ul must be uncountable. And that is as far as we can get by considering lines of constant X co-ordinate, because if we consider the example of a U that consists of a vertical line segment plus a countable number of points not on that line segment, it's easy to see that

every line x=c (with c a constant) has the property that there are either a countable number of points to its left or to its right (or both in the case that it has the same X co-ordinate as the line segment). So for that example U, there is no line of constant X co-ordinate (i.e. vertical line) with an uncountable number of points of U on either side of it (and possibly some on it).

We have however concluded that given that U is uncountable, there is such a vertical line unless there exists an l such that Tl is countable, Ul is uncountable and Vl is countable. Now restrict our attention to Ul: this is an uncountable set of points all of whose X co-ordinates are all equal to l. Apply the same argument as above, apart from swapping the roles of the X and Y co-ordinates, and we can conclude that there is a line of constant Y co-ordinate (i.e. a horizontal line) such that the sets below it and above it are both uncountable, unless there is a unique such line y=c such that Ul consists of an uncountable number of points with Y co-ordinates = c, a countable number of points with Y co-ordinates < c and a countable number of points with Y co-ordinates > c. But that's impossible, because the set of points in Ul with Y co-ordinate = c consist at most of the single point (x,y) = (l,c).

So there is a horizontal line such that Ul contains an uncountable number of points above it and an uncountable number of points below it. Since Ul is a subset of U, U too has an uncountable number of points above it and an uncountable number of points below it - and so in all cases, there is either a horizontal or a vertical line (or both, of course) such that U contains an uncountable number of points on each side of it (plus possibly some on it).

Whew!

Gengulphus