# If infinitely many equidistant points on a circle are connected in every possible way, will there be a line that crosses the center of the circle?

I've been thinking about the fact that if an odd number of points are placed on a circle equidistantly (same arc length between points) and these points are connected in every possible way, there will not be any line that cross the center of the circle, no matter how many points there are if this number of points is odd.

Only if there are an even number of points there can be lines connecting points that cross the center of the circle.

But what about when there are infinite points? One can think that if we divide a circle into infinitely equidistant points and connect these points in every possible way, all the internal points of the circle will be crossed by some line, but this is not true given that only an even number of points are able of generate a line that crosses the center of the circle.

My question is: is "infinity" even or odd? If someone were to divide the circle forever and connect the points in every possible way, in the end would the center be crossed or not?

EDIT: In cartesian coordinates, points on a unitary circle are given by: $$P =\left(cos\left(\theta\right),sin\left(\theta\right)\right)$$ Since the circle has been divided into N pieces, the angle between each point is: $$\theta = \frac{2\pi}{N}$$ If you say some point A has angle $$\phi = \frac{n2\pi}{N}$$ where n can be any number from zero to N-1, and point B has angle $$\gamma = \frac{\left(n+k\right)2\pi}{N}$$ where k can be any number from one to N-1, you end up with the condition: $$k = \left(2p+\frac{1}{2}\right)N$$ For the line between the points A and B to be equal to the diameter of the circle. if k and N are integers, then it means N needs to be even. That's why this question popped into my head. Here p is an integer that can vary from zero to infinity.

Infinity is neither even or odd, and there is no end to infinity. Furthermore you cannot meaningfully define an infinity of points that are equidistant on the circle since the separation between adjacent points would be zero.

However, you can determine – given infinitely many points on the circle – whether or not the line formed by any two hits the centre (i.e. the line is a diameter). And the answer can vary:

• Suppose uncountably many points form an interval over only a quarter of the circle. Clearly no two points define a diameter.
• Suppose countably many points are located so that the distance betwen two adjacent points is repeatedly halved: the first arc covers half of the circle, the second a quarter, the third an eighth and so on. The line corresponding to the points of the first arc clearly defines a diameter.
• If we rely on the limiting behaviour of a finite number of $$a_i$$ equally-spaced points where $$\{a_i\}$$ is an infinite monotone increasing sequence of integers satisfying $$a_i\mid a_{i+1}$$, then there are two points defining a diameter iff at least one of the $$a_i$$ is even. Halving the arcs will produce a diameter regardless of initial point count, but dividing arcs into $$3$$ starting from a heptagon will not.
• Use a choice function to select one end of each diameter in the circle. Then no two points form a diameter. Feb 24 at 3:34
• I started this way, assuming only two points and seeing what was the condition for the line between them to be the size of the diameter of the circle. I've edited the question to add what I've done so far. The condition ended up showing that there should be an even number of points for there to be a line the size of the diameter, but then I thought about the problem if this number were infinite. Feb 24 at 3:47
• @ParclyTaxel Yes, it depends on how the circle is divided, but this is what I find intriguing, how the result varies depending on the way you choose to divide the circle. I thought about this when finding areas using integrals: can the way the limit is taken to form the integral change the result like happens here? Feb 24 at 3:59
• @ViniciusAraujoRitzmann If a function is almost everywhere continuous, then the Riemann integral is unique. But take for example the indicator function of the irrational numbers: It can have any integral in [0,1] depending on your choice of points. That's among the reasons for the usefulness of the Lebesgue integral: It is unique for all measurable functions. (And otherwise just not defined) Feb 24 at 12:52

You would need to specify an infinite set to look at, and make some argument that the set is "evenly spaced" (perhaps by relating it to finite sets, or based on some symmetry argument.)

For example, you could have points at angles $$\{\theta = \tfrac {2a \pi}{2^k}| a,k\in \mathbb N \wedge a<2^k\}$$, which are all the fractions with power of two denominators.

The equally spaced thing seems reasonable given symmetry, and how this "looks like" a limit over regular polygons. And, you'd get the center.

If you did the same thing with powers of 3, you wouldn't.

• I think I did something like that when I was thinking about it, I edited the question to add my line of reasoning so far and it started by assuming the angle between two points are the same $\theta = \frac{2\pi}{N}$ where N is the number of points. Feb 24 at 3:42

In the limit there will be a line that crosses the center.

Take the minimal distance a line will have to the center point.

For even numbers of points, this is 0. For odd numbers of points, this decreases as the number of points goes up. The limit of this series is 0, as lines get arbitrarily close to the center the more points there are (I'm reasonably sure that can be proven relatively easily, but I don't want to).

So overall the series limit is 0, meaning in the limit the line crosses the middle.

Note we can't actually have infinite points if we want them to be equidistant, so the best we can do is the limit.