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Boundary of the n-dimensional ball

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In R^n with the usual topology, let A be the set of points x=(x_1,x_2,...,x_n) such that x_1^2+x_2^2+...+x_n^2 le 1. Prove that Bdry(A) is the (n-1)-dimensional sphere S^n-1.ie. x in Bdry(A) iff x_1^2+x_2^2+...+x_n^2=1.

Intuitively it is easy.. but I am not sure where to start. Probably looking at bdry(A)=Cl(A) intersect with Cl(C(A))? But how do I connect to the function.

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Solution Summary

We show that the boundary of the n-dimensional ball in the usual topoogy on R^n is the (n-1)-dimensional sphere.

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I would use the fact that the open unit ball in the usual topology is given by the equation
A={x in R^n: dist(x,0)<1} (that's what the inequality x_1^2+...+x_n^2<1 amounts to).
Then, it's easy to show that its closure is the closed ball B={x in R^n: ...

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