Can you cover a unit square with a countable union of circles?

no

Hah, exactly the man I wanted to hear from in this thread.

Well, no if they aren't allowed to go outside the unit square.

otherwise, just stick a circle at the centre of radius at least one over root two.

If I'm understanding the problem correctly, you'll never be able to do it with a countable union. With an infinite union you could fill .9 repeating of the square, which we've shown is one. The problem lies in the corners of the square.

But there are two kinds of infinite (that I know of)... countably infinite and uncountably infinite.

This question sucks, cause I don't know what you're asking

Well if you change the coordinates to something other than cartesian ...

Matty: Basically, you have a square with sides length of one. Can you draw enough circles in the square, not overlapping, such that they would fill the entire area of the square?

On second thoughts, i believe it is possible with a countably infinite number of circles

ps, answer is no.

Sounds like a no. I'm not big with math, but it seems no matter how small the circles would be, there would still be space between them.

For every circle you add, you create at least two new sub-problems.

That doesn't make it uncountable.

It'll be the same reason that the rationals are countably infinite. You count the first circle, then the circles you add to fill the spaces left by the first circle, then the circles you add to fill the spaces left by the previous round, and so on.

I haven't worked out a proof, but it LOOKS countable

[QUOTE=IRG SithLord]Matty: Basically, you have a square with sides length of one. Can you draw enough circles in the square, not overlapping, such that they would fill the entire area of the square?[/QUOTE]

The way I look at it, no. Because that means they'd all have to be tanget to eachother, which will leave gaps, no matter the size of them. You'd have to reduce the diameter of the circles to 0, which results in a point, not a length, which will never fill the square. (Since as I remember, points do not have a length)

Now if the circles were allowed to travel outside the box, even partially, then this would be solved with only one circle. Just make a circle that is tanget with each corner of the square.

Isn't a cartesian unit circle a unit square in polar?

No, because they'll be too small for me to count them. I win. :p

Based on the stipulation of countable: No


If we had an infinate number than probably.

Well, countable != finite.

The set of positive integers is infinite and very much countable.