Let be a disjoint family of topological figures in . Can B be uncountable?
I want to say no since, each should contain a point with rational coordinates, but feel that this is wrong somehow.
I haven't thought through the details, but I think this approach should work.
Suppose to start with that each figure T has the same dimensions. For each figure T, let be the point where the upright part of the T meets the crossbar. For two figures and , the distance between and can be arbitrarily small (think of one T being placed upside down on top of another T). But a bit of geometrical reasoning should convince you that if you try to add a third figure , then the distance from to cannot be less than some minimum value r (depending on the dimensions of the Ts). Thus each disk of radius r/2 can contain at most two points It follows that there are at most countably many Ts.
Now suppose that the Ts do not all have to have the same dimensions. Let m(T) be some quantity that measures the size of T (it might for example be the minimum of the lengths of the upright and the crossbar). For each n, the previous argument should show that there are at most countably many Ts with m(T) > 1/n. Take the union over n to see that there are only countably many Ts altogether.
I saw some where in the book, but I cant remeber which book .
It asserts that the plane can be fully fiilled with the letter T. By the baire category, the collection is uncoutable!
sorry, I am not sure whether I remember it accurately .
I hope I haven't confused you !