# How can we guarantee that $R_{rs}$ has no points in common with the set $\sum_j a_s b_j$ here?

I am reading Zarankiewicz's paper, On a problem of P. Turan concerning graphs, to try to learn more about Zarankiewicz's conjecture, out of my own interest. I have been confused about an assertion he makes about a line segment having no points in common with a set, and my question is how I can understand the possibility of such a line segment not having these points in common. The below 2 paragraphs are Zarankiewicz's exposition of the terms he will use in the portion which confuses me:

( $$\alpha$$ ) in the Euclidean plane two sets of points, $$A$$ and $$B$$, are given, $$A$$ consisting of $$p$$ points $$a_1, a_2, a_3, \ldots, a_p$$, and $$B$$ consisting of $$q$$ points $$b_1, b_2, b_3, \ldots ; b_q$$, ( $$p$$ and $$q$$ are natural numbers); (ß) for each pair of points $$a_i, b_j$$, where $$i=1,2,3, \ldots, p, j=1,2,3, \ldots, q$$, there exists a simple are lying in the plane and having the points $$a_i, b_j$$ as its end points; $$(\gamma)$$ the arcs lie in such a way that no three arcs have an interior point (i. e. a point that is not an end point) in common;

A simple arc having $$x$$ and $$y$$ as its end points will be denoted by $$x y$$. The sum of all the arcs $$a_i b_j$$, where $$i=1,2,3, \ldots, p$$ and $$j=1,2,3, \ldots, q$$, will be called the graph $$G(p, q)$$; so that we can write $$G(p, q)=\sum_{i, j} a_i b_j$$

The last sentence of the 2nd paragraph below is the one which arouses my confusion:

Let us draw a circle $$H_i$$ with the centre $$\boldsymbol{a_i}$$, and a radius so small that: The circles $$H_i$$ have no common points with the set $$\sum_{j, k} a_j b_1+b_1 a_k+a_j b_2+b_2 a_k+a_j b_3+b_3 a_k$$ where $$j$$ and $$k$$ take all the values $$1,2,3$$ except the value $$i$$, the circles $$H_i$$, for $$i=1,2,3$$ are disjoint from one another - which is of course possible.

Let $$e_{r s}$$ be the first point of the simple arc $$b_r a_s$$, going from $$b_r$$ to $$a_s$$, which lies on the circle $$H_s$$; then $$b_r e_{r s}$$ is a simple arc which has only one point in common with the circle $$H_s$$, namely its end point $$e_{r s}$$. Let $$R_{r s}$$ be a segment of the radius of the circle $$H_s$$, with $$e_{r s}$$ as one end point, while the other end point $$\neq a_s$$; let that segment be so small that it has no points in common with the set $$\sum_j a_s b_j$$, where $$j$$ takes all the values $$1,2,3$$ except the value $$s$$.

$$e_{rs}$$, as pointed out in the comments, trivially intersects the arc $$b_r a_s$$, and if the arc $$b_r a_s$$ is straight, then it seems that this segment $$R_{rs}$$ (which is also straight as it is a portion of the radius) will have all of its points intersect $$b_r a_s$$. Below is my attempt to draw a picture of this scenario, with straight arcs and $$R_{21}$$ having points in common with $$a_1b_2$$. So how can we guarantee that this segment $$R_{rs}$$ exists? Some assumptions I made about this text were that $$+$$ between sets of points refers to the union of those sets (elsewhere multiplication is clearly used for the intersection), and also that the center of the circle $$H_i$$ should be $$a_i$$, not $$a$$, which I have rendered as such above (originally it was written merely $$a$$ with a space where the subscript should go, so I assumed the $$i$$ got washed out).

• I'm guessing it's a typo and they meant "except the value $r$" instead of "except the value $s$". That way $R_{rs}$ intersects the arc $b_r a_s$ (at $e_{rs}$), but no other arcs out of $a_s$. May 15 at 7:21