# Math Help - Closed subsets

1. ## Closed subsets

Let AcX, A is not the empty set.

Prove if every b in X there exists an a in A such that d(b,A)=d(b,a) then A is a closed subset of X.

So for A to be a closed subset, that means that it has to include all of its limit points. So should I assume that b is also in A?

I was thinking that I could say that A has a sequence {an} that converges to a, but how does that relate to b? with d(b,A)=d(b,a)? I am not sure how to put the two together.

2. Originally Posted by EricaMae
Let AcX, A is not the empty set.
Prove if every b in X there exists an a in A such that d(b,A)=d(b,a) then A is a closed subset of X.
See if you can show that if c is a limit point of A then d(c,A)=0.

3. So I'm assuming c is in A and is a limit point of A. Therefore, there is a ball B(c,r) that contains a point, a, not equal to c?

I don't understand what d(b,A) means? How can I measure the distance from a point to a whole set? Would that mean the distance of b to the boundry of A? If so, I don't get why d(b,A)=d(b,a)

The distance d(c,A)=0 would mean that c is on the boundry of A, does that mean that it is an accumulation point? If so, then d(c,A)=0, d(c,a)= r, d(b,a)>r, d(b,c)=d(b,A)?

I'm not sure where this takes me?

4. So you problem is really about definitions.
$D\left( {A;b} \right) = \inf \left\{ {d(a,b):a \in A} \right\}$.
Clearly $D\left( {A;b} \right) = \inf \left\{ {d(a,b):a \in A} \right\} \geqslant 0$.
Suppose that b is a limit point of A and $\varepsilon > 0$.
$\left( {\exists a \in A} \right)\left[ {a \in B(b;\varepsilon ) \wedge a \ne b} \right]$.
But that means that $d(a,b) < \varepsilon \Rightarrow \quad D(A;b) < \varepsilon$.
By definition of infimum, that means that $D(A;b) = 0$.

5. The way I have the picture drawn is that there is a metric space X, that contains a space A. I have b in X and not in A, and a in A.

Assuming that I say that a is a limit point because it is in A, and b is not a limit point because it is not in A. How would I show that b is not a limit point?

6. Okay,

So could I say that b is in X and not A. there is a ball B(b,r) that is completely contained in X such that B(b,r) intersect A= the empty set. This directly implies that b is not a limit point.

Assuming that c is a limit point in A, there is a ball B(a,E) and a not equal to c. So d(a,c)<E implies D(A,c)<E so D(A,c)=0?

My problem is that b is not in A. So I'm not sure how to relate d(a,c) and D(A,c) with d(a,b) and D(A,b) because c is in A, and b is not.

7. Originally Posted by EricaMae
The way I have the picture drawn is that there is a metric space X, that contains a space A. I have b in X and not in A, and a in A. Assuming that I say that a is a limit point because it is in A, and b is not a limit point because it is not in A. How would I show that b is not a limit point?
I am sorry to say but I don’t think that you have any idea what this material is all about.
Now, don’t be offended by that remark! It is meant as a very constructive remark.
What you need to do is to go to the instructor and have a one-to-one tutorial with him/her. Any competent lecturer would be glad to help you. I say this having taught this material for more that thirty years.

The crux of this problem is the following theorem.
In any metric space, the closure of any set $A$ is $\overline A = \left\{ {x(A,x) = 0} \right\}" alt="\overline A = \left\{ {x(A,x) = 0} \right\}" />.
A point is in the closure of a set iff its distance to the set is zero.

8. I don't take any offense to that at all. Yeah, I'm very confused about this material. I went to the professor today with questions, and he just made me more confused.

9. Originally Posted by EricaMae
I don't take any offense to that at all. Yeah, I'm very confused about this material. I went to the professor today with questions, and he just made me more confused.
Maybe you can find a graduate student or a fellow classmate who can help you more than the instructor.
Sometimes student to student is better.
I hope you can get help. I know you cannot get it online.

10. I actually think I understand what you were saying now. So thank you very much for trying to help me!