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Right, yes. I was taking the product of the process of subdividing to not actually include the limit points at all. You are only performing the operation a coun
by roywiggins 12y ago
Right, yes. I was taking the product of the process of subdividing to not actually include the limit points at all. You are only performing the operation a countable number of times, and at each point you only add a finite number of points, so having an uncountable set pop out at the end seems unlikely.
A bonehead way to formalize this:
The first step outputs {0}, the second, {0, 1/2}, the third {0, 1/4, 1/2, 3/4}, etc.
We drop them all into one big set A. A has countably many elements and they all have a finite size:
A = {{0}, {1/2}, ... }
If B is the union of A, it has to have every point that could ever be output by the process. B is the result of "running the process to infinity." B is obviously countable since it's the union of countably many finite sets.
If you can think of a way to define running the process to infinity and -including- the limit points, I'd be interested, but the most natural interpretation to me is that the process produces a set that doesn't include the limit points at all.
- SamReidHughes 12y agoThe points {0, 1/4, 1/2, 3/4, 1}, etc, are the boundaries between pixels, not the pixels themselves.
- pdonis 12y ago> I was taking the product of the process of subdividing to not actually include the limit points at all. That's not correct. The limit point is "the product of the process of subdividing", by definition. > You are only performing the operation a countable number of times, and at each point you only add a finite number of points, so having an uncountable set pop out at the end seems unlikely. By this reasoning, the limit point of every sequence of rational numbers would be a rational number, and the set of all such limit points would be countable. But it isn't; it's the set of reals, which is uncountable.