I'm solving some problem about coloring a circle with two colors and I've run into an issue (no, this is not an XY problem).
Lets say I inscribe a regular polyogon with $2^n$ sides into the unit circle. I can prove that, on this polygon, if you take any two opposite (antipodal) points they must have different colors.
As $n$ goes to infinity, it seems like every point on the unit circle is hit, so the conclusion is that the above rule holds for any pair of antipodal points on the entire unit circle.
Is this logic valid? On one hand, every point is hit at infinity. On the other hand, some points are never hit in a finite number of steps, so we could maybe conclude that they are never a part of any polygon and therefore do not folow the rule.
Of course, the rule in my problem is just an example. The question stands for any such rule.
Thanks.
What is true about all points of the form $(\cos\frac{2\pi p}{q}, \sin\frac{2\pi p}{q})$ where $0\le p\lt q$ and the denominator $q$ is a power of $2$, obviously does not carry to all the points on the circle (those of the form $(\cos(2\pi\alpha), \sin(2\pi\alpha))$ for $\alpha\in[0,1)$). One such property is, obviously "$\alpha$ is rational", or "$\alpha$ is rational with a denominator that is a power of $2$".
Thus, any argument that you want to carry over from the smaller set of points to a larger set must include justification why this jump is allowed. Often the arguments here come from mathematical analysis, by invoking the notion of continuity and the limits, and then using the theorems of analysis in your proof. I am not sure if this is applicable here.
In your particular case, lacking any additional information, you can paint all points of the form $(\cos\frac{2\pi p}{q}, \sin\frac{2\pi p}{q})$ with $0\le p\lt q$, $q$ a power of $2$ as "red" if $\frac{p}{q}\in[0, 1/2)$ and "blue" if $\frac{p}{q}\in[1/2, 1)$, but paint all the other points on the circle "blue". This would be an obvious counter-example. However, the words "lacking any additional information" are a key. In your problem, you may know more about your conditions, and those may preclude this example, and perhaps allow for a proof of the type I was talking about above.