Question

Draw a Venn diagram for 4 rectangles with all sides vertical or horizontal.

Answer 1

Answer is as follows :

Diagram for given scenario :

Explanation :

Let us Assume, There exists a family R of 4 (congruent) rectangles representing Venn diagram. Define two orders ≤x and ≤y on RR. Namely, for rectangles R,R′∈R put R≤tR′. if t-coordinate of the left side of the rectangle R in not bigger that t-coordinate of the left side of the rectangle R′.

Let R1≤xR2≤xR3≤xR4 be the enumeration of the family R.

Assume first that R1 or R4 is a minimal or a maximal element of (R,≤y). Applying, if needed, to the plane transformations x↦−x and y↦−y, without loss of generality we may suppose that R1 is a minimal element of (R,≤y). If R2≤yR4 then R1∩R4⊂R2 , contradicting the property of Venn diagram.

If R3≤yR4 then R1∩R4⊂R3, contradicting the property of Venn diagram. Thus R4≤yR2 and R4≤yR3. Now if R2≤yR3 then R1∩R3⊂R2, contradicting the property of Venn diagram and if R3≤yR2 then R2∩R4⊂R3, contradicting the property of Venn diagram.

Thus in the order ≤y the elements R1 and R4 are placed between elements R2 and R3. But then R1∩R4⊂R2∪R3, contradicting the property of Venn diagram.

Indeed, assume to contrary that there is an (interior) point R1∩R4∖(R2∪R3) Let ℓ be a horizontal straight line going through the point x. It is easy to see that if ℓ intersects any of rectangles R2 or R3 then it contradicts the choice of R1 or R4 as a leftmost or a rightmost rectangle. Since R2 and R3 intersects, both R2 and R3 lie above ℓ or both R2 and R3 lie below ℓ, a contradiction with that R2 and R3 are topmost and bottommost rectangles.

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