There are infinite black and white dots on a plane
Answers
Answer:
You don't, because it's false. If all black dots happen to be on the line y=0 and white dots on the line y=π (and the rest of the plane is neither white nor black), there is no such pair.
Now if each point of the plane were either black or white (and there were infinitely many of each type), that would be different. In fact, it is sufficient to have at least one of each color.
Why? Pick any two points A and B that have different colors. Starting at A , we can reach B using a finite number of steps, each of length exactly 1: just go directly towards B until the distance becomes less than 1, and at the end, if we didn't reach B exactly, we make two steps "to the side and back" to reach it. (Formally, if you are currently at C , imagine circles with radius 1 centered at B and C . Pick one of their two intersections, go from C to that intersection and from there to B .)
As the first and the last point on this path have opposite colors, there has to be a pair of consecutive points with opposite colors, q.e.d.
(Alternately, you could prove the new statement by contradiction. Pick any black point. All points in distance 1 from that point have to be black. This is the circle with radius 1. All points in distance 1 from those points have to be black as well. Here we can observe that the set of all points known to be black at this moment is the entire disc of radius 2 centered where we started. Continuing this argument, we can now grow the black disc indefinitely and thus prove that the entire plane has to be black, which is the contradiction we seek. Of course, this is basically the same proof as above, just seen from a different point of view.)
You don't, because it's false. If all black dots happen to be on the line y=0 and white dots on the line y=n (and the rest of the plane is neither white nor black), there is no such pair.
Now if each point of the plane were either black or white (and there were infinitely many of each type), that would be different. In fact, it is sufficient to have at least one of each color.
Why? Pick any two points A and B that have different colors. Starting at A, we can reach B using a finite number of steps, each of length exactly 1: just go directly towards B until the distance becomes less than 1, and at the end, if we didn't reach B exactly, we make two steps "to the side and back" to reach it. (Formally, if you are currently at C, imagine circles with radius 1 centered at B and C. Pick one of their two intersections, go from C to that intersection and from there to B.)
As the first and the last point on this path have opposite colors, there has to be a pair of consecutive points with opposite colors, q.e.d.
(Alternately, you could prove the new statement by contradiction. Pick any black point. All points in distance 1 from that point have to be black. This is the circle with radius 1. All points in distance 1 from those points have to be black as well. Here we can observe that the set of all points known to be black at this moment is the entire disc of radius 2 centered where we started. Continuing this argument, we can now grow the black disc indefinitely and thus prove that the entire plane has to be black, which is the contradiction we seek. Of course, this is basically the same proof as above, just seen from a different point of view.)
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