Continuum crosses over plane

Question

The task is to check whether it is possible to fit continuum number of geometric figures called crosses (which is basically square diagonals (on the picture there are 2 of them)) in a way that they have no intersections at all. Cross sizes are not necessary equal, it may vary from cross to cross.

P.S: I was thinking about drawing some imaginary circle around each cross joint point, then selecting there set of 4 points with rational coordinates, each one for section inbeetween cross lines (in a upper-left "corner", upper-right "corner", lower-left and lower right "corners"). Than we have that unique set of 4 points with rational coordinates, but don't know what to do next with them. Maybe there is a way to prove, that number of that sets of 4 rational coordinates is lower that continuum or somethingg similar to that..

Answer 1

Let's make the task more difficult and try to place Y-shapes (where the three arms are allowed to be arbitrary curves, just that the don't intersect except at the common central node).

For each such shape, we can find a circle around its node such that the three ends are exterior (red circle in the image). By making it sslightly smaller and moving it slightly we find a circle with rational centre and rational radius that also has the node inside and the three ends outside (green circle). Find the first times each arm intersects this circle (green dots). In the three arcs determined this way, pick one point each at a rational angle (blue dots). This assigns $6$ rational numbers with each Y. As $\Bbb Q^6$ is countable, some Y's must be assigned the same green circle and blue dots.

The following image highlights what happens in the rational (=green) circle when another Y shape happens to pick the same rational points (dotted lines and dark green nodes). In the example picture, we see that the second (dotted) Y intersects the first Y - but is this necessarily the case?

In each of the three arcs determined by the rational (=blue) points there must be exactly one end of each of the Y shapes (=one light green and one dark green node). There are only two possible orders of the ends on such an arc, hence we find two arcs where the ends appear in the same order. (In the example, this is the case for the upper arc and the lower-left arc: For both, we have first dark green, then light green when going counter-clockwise). Ends of the same Y are connected within the green disk along the legs of their Y. If we use this for the light green dots of our two arcs and complete this to a closed Jordan curve with a nice arc outside our green disk, then this parts the plane into two regions (using the Jordan curve theorem): The interior (shown shaded below) and the interior of this Jordan curve. By the choice of our arcs, the dark green node on one arc is in the interior and the other in the exterior region. Therefore any curve connecting these two must intersect our Jordan curve. Again, there is a connection along the legs of the (dotted) Y shape that is inside the green disk. Hence the intersection with the Jordan curve is in fact an intersection with the first Y shape, as was to be shown.