It is true that the intersection between two $n$-sphere in $\mathbb{R}^n$ is a $(n-1)$-sphere if is not empty or a single point? I have tried to prove it but my only idea is to work with equations and apparently is not a good idea.

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    $\begingroup$ I think you mean $n-1$ spheres in $\Bbb R^n$ is an $n-2$ sphere. $\endgroup$ – Adam Hughes Nov 25 '14 at 22:02
  • $\begingroup$ consider a special case first, might be easier, when one sphere is centered at the origin,and the other at a point on the $x$-axis $\endgroup$ – Mirko Nov 25 '14 at 22:02

We pick one center at the origin for sphere $S_0$ (radius $a$) and the other at the point $(1,0,0,...,0)$ (radius $b$) for the sphere $S_1,$ so that the two equations are respectively $$x_1^2+x_2^2+\cdots + x_n^2=a^2,\\ (x_1-1)^2+x_2^2+ \cdots +x_n^2=b^2.$$ Subtracting these gives $2x_1-1=a^2-b^2,$ so that any point in the intersection $S_1 \cap S_2$ must lie in the hyperplane $x_1=c,$ where $c=(a^2+b^2+1)/2.$ In this hyperplane, one views its "origin" as the point $P=(c,x_2, \cdots x_n).$ Then the intersection, lying in $x_1=c,$ is a sphere centered at $P$ in that plane, and with equation $$x_2^2+x_3^2+\cdots + x_n^2=r^2,$$ where $r^2$ may be found from either equation for $S_1$ or for $S_2.$ Hopefully these match, namely $r^2=a^2-c^2=b^2-(c-1)^2.$ These do indeed match (which one would expect by symmetry) and the expression for $r^2$ may be put in the form $$r^2=\frac14[(a+b)^2-1][1-(a-b)^2].$$ We are viewing $a,b>0$ and if the first factor here is negative it corresponds to $|a+b|<1$ so the two radii don't add up to enough for the spheres to meet. (They are outside each other.) In cases I've tried, if the first factor is positive and the second is negative, it corresponds to the case where one sphere is completely inside the other.

For generally located centers $C_1,C_2$ I think one can rescale so the distance between $C_1$ and $C_2$ is one, and apply the preceding, at least to get the center and radius. This means using $C_1+t(C_2-C_1),$ in vector notation, where $t$ is obtained from the previous case (the formula for that $c$ and its radius should be obtained using the rescaled versions of the initial radii for the spheres centered at $C_1$ and $C_2.$ And in this general case the resulting sphere lies in a hyperplane perpendicular to the line joining the two centers, so may not be so easy to visualize.


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