Proving Discreteness of M: Set Theory Problem

[ehrenfest]

Homework Statement

THIS PROBLEM IS DRIVING ME INSANE! HELP!
Let M be a metric space in which the closure of every open set is open. Prove that M is discrete.

Homework Equations

The Attempt at a Solution

 

[matt grime]

Hmm, so one wishes to show that each one point set is both open and closed. I wonder, what can one do to make a one point set in a metric space?

 

[ehrenfest]

Hmm, so one wishes to show that each one point set is both open and closed. I wonder, what can one do to make a one point set in a metric space?

Take the complement of (the metric space minus that point)? I really do not know where you are going

 

[Dick]

Let S={x} for x a point in the metric space. S is closed. Prove it. The complement S^C is open. Prove it. Your assumptions says S^C is also closed. What does this tell you about d(x,y) for y in S^C? Or even simpler, if S^C is closed, what is (S^C)^C? See? I've left all the hard work to you.

 

[ehrenfest]

Your assumptions says S^C is also closed.

The assumption is that the closure of any open set is open. How did you use the assumption to get that S^C is closed (or anywhere else)?

In order for S^C to be closed S needs to be open. Yes there are other ways to prove something is closed but I do not see how you used any method to show that S^C is closed. Obviously it is open because S is closed. Sorry if you wanted me to prove that S^C is closed I just do not see it :(

 

[Dick]

You are right. I was oversimplifying the problem. Sorry. Let me rethink this. Matt's clue is probably the one to think about.

 

[tronter]

You want to show that each singelton set is clopen in the metric space. First of all write out the definition of a discrete metric space. This should lead you to the answer. So show that the metric space doesn't contain any accumulation points.

 

[ehrenfest]

That is an interesting term for open and closed if that is what you mean by clopen.

Anyway, a metric space X is discrete iff every subset of X is open.

If x is a limit point of X then, every neighborhood of x must contain a point in X - {x}. I am not sure how that is inconsistent with the assumption that the closure of every open set is open (or clopen as you would say!).

 

[matt grime]

I'll say it again. A one point set. Is it open?

 

[HallsofIvy]

To simplify even further: {x} is a one point set. What is its closure?

 

[Dick]

To simplify even further: {x} is a one point set. What is its closure?

I think you simplified too much (I did too). Regarding closures, it only says that the closure of OPEN sets is open. If I knew {x} were open so I could apply this, I wouldn't need to apply it.

 

[Dick]

Ok, try this. Follow tronter's suggestion. Assume x is an accumulation point in the space X. Cleverly select a sequence of points S approaching x and a set of radii r_s for each s in S such that i) x is an accumulation point of the union of all of the open balls B(s,r_s) for s in S (call it U) and ii) x is also an accumulation point of X-closure(U). Then x is in closure(U), U is open, but every neighborhood of x contains points not in closure(U). So closure(U) is not open. Contradiction. So X has no accumulation points.

To start the construction of S etc, imagine that x is a limit point of the union of a sequence of disjoint balls monotonically approaching x. Show you can make such a thing. Choose even numbered balls only to make S.