Prove that $n$ circles in general position (no three circles share the same point, and every two circles intersect each other), divide a plane in $n^2 -n +2$ areas.

Proof for circles in a plane: $P(n)=n^2-n+2 \implies P(1)=2$

Since one circles divides a plane in two parts, the statemet is true for$\;n_0=1.$

By the induction assumption: $P(k)=k^2-k+2.$

We have to prove: $\;P(k+1)=(k+1)^2-(k+1)+2=k^2+k+2$

If we add the $(k+1)th$ circle to the set of $k$ circles, then those $k$ circles intersect the circle (added to the set) in $2k$ points.

Those $2k$ points form $2k$ consecutive arcs, and every new arc divides the former part (arc) of the circle, in two parts. Therefore, by adding the $(k+1)th$ circle to the set of $k$ intersecting circles, we got $2k$ more areas. $$\implies P(k+1)=P(k)+2k\;\implies P(k+1)=(k^2-k+2)+2k=k^2+k+2.$$ $$\; Q.E.D$$

But what happens in a $3$-$D$ space when the intersection of two spheres is a circle?

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    $\begingroup$ Do the same thing. You are essentially using Euler's formula $V-E+F = 2$ in 2 dimensions. What is the 3-D analogue? $\endgroup$ – Calvin Lin Nov 10 '19 at 11:27
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    $\begingroup$ Is it $V-E+F+C=0$? So the dimension of cross intersection is always less than or equal to the dimension of space? Strictly less, unless one solid is inside another, right? Which isn't the case here? Please, correct me. I understand sphere in 3-D is analogue to circle in 2-D. $\endgroup$ – Praskovya2.718281828 Nov 10 '19 at 12:02

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