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

1. Jul 20, 2007

### quasar987

1. The problem statement, all variables and given/known data
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 $\overline{B}_{b+\delta}(x)\cap A$ 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?

2. Jul 20, 2007

### jostpuur

If A was compact, and if we knew that continuous functions $f:X\to\mathbb{R}$, where X is a compact metric space, always reach their minimum and maximum somewhere, then we could just note that mapping $y\mapsto d(y,x)$ must reach it's minimum somewhere in A.

Since A is not compact, it should be first somehow showed that it is sufficient to study some compact part of it, like set $A\cap \overline{B}(0,R)$ with some large R. And if the theorem about continuous mappings out of compact metric spaces is not known, then its proof is probably equally complicated as your proof. So I'm not sure it this is really any easier. Although going through some more general theorems is usually more elegant, in some sense.

3. Jul 21, 2007

### matt grime

A is not empty. Pick an arbitray point a in A. Now let r be the distance from x to a. The thing you're looking for must be smaller than r, now, is there anyway to consider only points that are at most distance r from x and lying in A, I wonder......

4. Jul 21, 2007

### quasar987

Other than looking at $\overline{B}_{r}(x)\cap A$? (which comes down to my method essentially)

5. Jul 22, 2007

### matt grime

It comes down to the fact that a continous function on a compact subset of R^n has a maximium, quasar - there is no point or need to invoke sequences. It is quite simple - you can assume A is a compact set.

6. Jul 22, 2007

### quasar987

Well, ok, but this is just the method of jostpuurand he was correct that the max-min thm on compact is now known at this time. :tongue:

But thanks for your input matt & jostpuurand. I guess my original solution was just erroneous or was and incomplete version of that of post #1.