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The problem is to show that if A is closed in R^n, and x is outside of A, then there is a point y in A such that d(y,x) = d(x,A).

My method of solution involves letting b = d(x,A) = inf{d(x,z) : z in A} and considering a closed ball of radius b + delta centered on x, and then constructing a sequence of points {z_n} in [itex]\overline{B}_{b+\delta}(x)\cap A[/itex] such that d(z_n,x)-->b (possible because b is an infimum). Because the ball is compact, there is a convergeant subsequence to z_n, but because each z_n is in A and A is closed, that limit is in A. I then show that the distance btw this limit and x is b by using the triangle inequality.

However, I had solved this problem at work in my head 3 weeks ago and I remember this was not the answer I had found. My solution from 3 weeks ago was swifter somehow, I remember. Anyone got any idea what it could have been?

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# Homework Help: Other ways to do this problem? (R^n topology)

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