countably

If the product of the set {0,1} with itself countably-many times was countable, there would be countably-many real numbers, which we know is not true.

And proof by induction demonstrates that a countably infinite number of subjects can be generated by this method.

Go breadth-first and you'll never get past the first level because the root has (countably) infinitely many children.

The problem is that there are countably infinite squares on the infinite plane so the measure of the infinite plane would add up to 0 instead of 1 by the property of the measure of countably infinite union of measurable sets.

But it does imply that the set of limit points of all countably infinite series of rationals is uncountable.

A has countably many elements and they all have a finite size: A = {{0}, {1/2},...

In conventional mathematics you can also only write down countably many sequences, but the space of all sequences is still uncountable.

A ball in R^3 is composed of a countably infinite number of zero-dimensional points.

The references make it clear that what they're dealing with is the computable reals - only a countably infinite subset of R.

However, this doesn't work for the product of A with itself countably-many times.

So I don't think the idea of dividing a math-monitor into countably many pixels, each with zero area, works.

There are multiple sequences that correspond to any one point, and in the countably infinite construction, there are points that have no sequence.

Edit: I'm taking countable to mean "countably infinite" since this conversation already assumes we're talking about infinite sets of pixels.

B is the result of "running the process to infinity." B is obviously countable since it's the union of countably many finite sets.

I understand under "constructive reals" the computable reals, and there are only countably many of them.

I'm pretty sure you mean "uncountably infinite" instead of "countably infinite".