Compact subset S of R^n, 'a' in R^n fixed. Show that....

[Jolb]

I'm struggling with a problem:

"Suppose S ⊂ ℝⁿis compact and a∈ℝⁿis fixed. Show that there is a point of S closest to a."

The reason I'm having trouble with this is because I immediately thought of a counterexample:

Set n=2
Let S={[x,y]: x² + y² = 1} (this is compact)
Let a = [0,0].

Then there is no closest point to a on S, right?

I'm puzzled.

 

[pka]

Re: Compact subset problem

I'm struggling with a problem:

"Suppose S ⊂ ℝⁿis compact and a∈ℝⁿis fixed. Show that there is a point of S closest to a."

The reason I'm having trouble with this is because I immediately thought of a counterexample:
Set n=2
Let S={[x,y]: x² + y² = 1} (this is compact)
Let a = [0,0].

Then there is no closest point to a on S, right?

I'm puzzled.

You need to check the exact wording of the question.
It is possible that you miss quoted the statement.

However, yours is not a counterexample because each point of S is a point closest to (0,0).

If a is a point then the statement is correct. If \(\displaystyle a \notin S \Rightarrow \,D(S;a) > 0\), the distance of a from S is positive. Use that to get a sequence of points in S that has a limit point closest to a.