Let X=RxR. For each (a,b)∈X Let D_(a,b) ={(x,y)∈X :x=a}∪{(x,y)∈X : x^2+y^2=a^2+b^2} Is {D_(a,b):(a,b)∈X} a partition of X ?

I will assume R means the set of real numbers.

No, it is not a partition. The D_(a,b) are not disjoint. The sets of a partition must be disjoint. For example D_(0,0) ∩ D_(1,0) = {(0,1), (0,-1)} ≠ Ø

Actually it is not true that (0, 0) is in every D_(a, b); this would have been true had x^2+y^2=a^2+b^2 been replaced by x^2+y^2<=a^2+b^2.

Note that D_(a,b) is the union of the vertical line through (a,b) and the circle (not including interior) through (a,b) that is centered at the origin.

Still, D_(a,b) is not a partition of X, the coordinate plane, because for example the point (3,0) is on both D_(3,1) and D_(3,2) which are distinct (but not disjoint) sets.

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