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I think your argument implies that the rationals are uncountable, but they aren't, so your argument must be wrong somewhere- I'm too rusty at this to form a nic

I think your argument implies that the rationals are uncountable, but they aren't, so your argument must be wrong somewhere- I'm too rusty at this to form a nice argument quickly though.

pdonis 12y ago
> I think your argument implies that the rationals are uncountable No, it doesn't. But it does imply that the set of limit points of all countably infinite series of rationals is uncountable. And that's true: that set of limit points is the set of real numbers, which is uncountable.
- 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 ago
    The 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.

I think your argument implies that the rationals are uncountable, but they aren't, so your argument must be wrong somewhere- I'm too rusty at this to form a nice argument quickly though.

pdonis 12y ago

> I think your argument implies that the rationals are uncountable No, it doesn't. But it does imply that the set of limit points of all countably infinite series of rationals is uncountable. And that's true: that set of limit points is the set of real numbers, which is uncountable.

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 ago
  The 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.