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This question already has an answer here:

''There exists a point in a closed set which is at minimum distance from a point not in the set.''

I have no idea why this is true.

Any help will be appreciated.

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marked as duplicate by Travis, user228113, Ben Sheller, user296602, JKnecht Apr 26 '16 at 0:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It does not have to be unique. Suppose your closed set is $S^1\subset \mathbb R^2$ and let $(0,0)$ be the chosen point not in the set. $\endgroup$ – lulu Apr 25 '16 at 14:53
  • $\begingroup$ It's phrased a bit opaquely. A clearer way to say the same thing: given a closed set $C$ and a point $x$, there is a point $y \in C$ with $d(x,y) \leq d(x,z)$ for all $z \in C$. Note that I did not say $d(x,y)<d(x,z)$, so in particular this $y$ is not necessarily unique. $\endgroup$ – Ian Apr 25 '16 at 14:57
  • $\begingroup$ Post-edit, my earlier comment no longer applies. Existence follows from the fact that continuous functions from compact sets take a minimum. Granted, your closed set need not be compact...but just take the intersection of your set with sufficiently big closed balls around the target point. $\endgroup$ – lulu Apr 25 '16 at 14:58
  • $\begingroup$ @ervx Thanks it was a great help. $\endgroup$ – user333900 Apr 25 '16 at 15:04
  • $\begingroup$ :-) Glad to help $\endgroup$ – ervx Apr 25 '16 at 15:05
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Before I begin, here are my sources:

Definition of a closed set Equivalence of Definitions of Closed Sets

My answer won't be technical, but general. We will prove this by assuming that your statement is wrong and finding a contradiction.

Let's say that there is a point x in set C and a point y in set C - R (the set C - R is the set of points in R that are not in set C.

Let's assume that your statement is wrong. That means that there will exist a point z1 (a point in set C) is less than the distance between point x and point y: (this means that point z1 is closer to point x than point y.)

Remember that point z1 is a point in set C.

Since point z1 is a point in C, this means that there will exist a point z2 (a point in set C) whose distance is less than the distance between point x and point z1: (this means that point z2 is closer to point x than point z1.)

Remember that point z2 is a point in set C.

Since point z2 is a point in C, this means that there will exist a point z3 (a point in set C) whose distance is less than the distance between point x and point z2: (this means that point z3 is closer to point x than point z2.)

Now we are creating a sequence of points: z1, z2, z3, ... zn.

And set C contains the sequence z1, z2, z3, ... zn.

You can prove that this sequence of points converge at point x. So you can prove that point x is the limit of this sequence.

So set C contains the convergent sequence z1, z2, z3, ... zn.

Since set C is closed, it must contain the limit of all convergent series within itself. In other words, set C must contain the limit of the convergent sequence z1, z2, z3, ... zn.

But point x is not inside set C.

We have reached a contradiction.

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