# Homework Help: Metric Spaces Q

1. Oct 7, 2009

### latentcorpse

Given a metric space $(X,d)$, an element $a \in X$ and a real number $r>0$, let

$A:= \{ x \in X | d(a,x) < r \}, C:= \{ x \in X | d(a,x) \leq r \}$

i need to show $\bar{A} \subseteq C$.

The definition of the closure of $A \subseteq X$ is

$\bar{A} = \cap_{C \subseteq X closed, A \subseteq C} C \subseteq X$

*i wanted that writing under the intersection sign but can't do it can someone help me with that LaTeX code?*

anyway, i'm at a complete loss as to waht to take the intersection of....

2. Oct 7, 2009

### aPhilosopher

I would show that a closed set containing A must contain its boundary points.

3. Oct 7, 2009

### Landau

Don't you have a few characterizations of the closure? The definition you gave is not always practical to work with. The idea is that points in the closure of A can be approximated arbitrarily well by points in A, which almost immediately solves the question. So it would be convenient to be able to work with limit points, or converging sequences.

4. Oct 7, 2009

### latentcorpse

yes we have covered material to do with both limit points and converging sequences.

from my understanding the defn of closure is that it's compliment is open. open meaning that it contains an open ball...is that correct?

anyway how do i go about setting up something with limit points?

5. Oct 7, 2009

### snipez90

The definition of a closed set is that its complement is open. Of course the closure of a set is a closed set.

Pick a point y in the closure of A. Then y is either in A or y is a limit point of A. If y is in A, then y is clearly in C so we are done. If y is a limit point of A, every neighborhood of y contains a point a in A with a =/= y. But a is also in C... Can you finish the proof from here?

*EDIT* Of course there are many ways to introduce the concept of closure. One way may use certain definitions to prove other theorems, and a second way could use the theorems in the first way as definitions. This approach may not be available to you depending on what you may use.

6. Oct 8, 2009

### g_edgar

Hey, guys!
Why not help him use his definition to prove his problem? Rather than suggesting other ways to do it? Sometimes suggesting something else is good, when the person with the question has no good idea. But in this case the definition he quotes is PERFECT for doing the proof he wants.

7. Oct 8, 2009

### aPhilosopher

Have you been able to do this? This is almost the definition of a closed set. All that you have to do after this is show that C contains all the limit points of A and you're done. You might want to do that step with a contradiction. What if a limit point of A wasn't in A or C?

8. Oct 8, 2009

### aPhilosopher

Sorry, I was off on my last post. You have to show that C contains all limit points of A and that every point of C is a limit point of A.

Also, instead of contradiction for the case that a limit point p of A isn't in C, try to directly exhibit a closed set containing A that doesn't contain p.

The case of a member of C being a limit point A is easy if you think about the definition of a limit point.

9. Oct 8, 2009

### Landau

No, while it is true that $$\overline{A}=C$$, the exercise only asks to show$$\overline{A}\subseteq C$$.

10. Oct 10, 2009

### latentcorpse

im confused. everyone's saying different things lol!

can someone try and lead me through thsi step by step because im really lost.

11. Oct 10, 2009

### latentcorpse

i dont know how to write thsi mathematically but if we took A as open balls and C as closed balls then for different r values we get the intersection to be 0 and that is definitely contained within or equal to C. or am i completely off track?

12. Oct 10, 2009

### Landau

Yes, you are off track. The number r is fixed, you cannot vary it.

In post #5, snipez90 wrote a beginning of a proof. Can you finish it, and if not, where are you stuck?

13. Oct 10, 2009

### aPhilosopher

It seems as if you are confusing 0 with the empty set. I'm not at all clear about your proof though. It seems as if A is an open ball by assumption and C is a closed ball. This it what you are trying to prove.

What is right in your idea is that you only have to worry about the r value of a point.

You can do this by showing that if p is distance r' > r from the origin, then there is closed subset which contains A, but does not contain p. Then p does not lie in the intersection that defines the closure of A. Then the closure can only contain points p with r' <= r.

As a hint, notice that p has arbitrarily small open neighborhoods. Note that the complement of an open neighborhood is closed.

Last edited: Oct 10, 2009
14. Oct 10, 2009

### latentcorpse

is snipez90's proof ok? i thought that after what was said in post 6 that it was using material i don't have?

aPhilosopher: i'm afraid i need a bigger hint. either that or i dont understand how ur last hint helps.

15. Oct 10, 2009

### Landau

It's not a proof, but a beginning of a proof.
Well, you said you covered sequences and limit points, so you should be able to finish it.

16. Oct 10, 2009

### aPhilosopher

well, if p is r' > r distance from the origin then write r' = r + e. Do you think you can fit an open set around p that doesn't intersect A so that its complement contains it? And wouldn't its complement be a closed set? And wouldn't that closed set not contain p? Then wouldn't that imply that p isn't in the closure of A?

17. Oct 10, 2009

### latentcorpse

what is the exact definition of a limit point?

18. Oct 10, 2009

### aPhilosopher

A limit point of a set S is a point such that any of its neighborhoods contain infinitely many points of S. They are the points that sequences of S can converge towards.

I don't think that we're going to need that though. By your definition of the closure of A, you only need to exhibit a closed set for each p outside of C that contains A and not p.

19. Oct 10, 2009

### latentcorpse

how can it be outside C and contain A if A is in C? would it need to be a union of some sort?

20. Oct 10, 2009

### Landau

p is outside C, not the closed set.