A discrete subset of a metric space is open and closed

[pob1212]

Hi,

If $X \LARGE$ is a metric space and $E \subset X$ is a discrete set then is $E \LARGE$ open or closed or both?

Here's my understanding:

$E \LARGE$ is closed relative to $X \LARGE$.
proof: If $p \subset E$ then by definition $p \LARGE$ is an isolated point of $E \LARGE$, which implies that $p \LARGE$ is not an interior point of $E\LARGE$. Since a subset of a metric space is open if every point of that subset is an interior point, it follows that $E \large$ is closed.

Now that I write this proof, I believe the problem is that a subset is open IF every point is an interior point, not IF AND ONLY IF. There are other ways for the subset to be open other than all points being interior. Is this my error?

The web tells me a discrete set is open and closed. Proofs I've seen use collections of neighborhoods with radius s.t. the only point contained in the nbhd is the discrete point at its center and since any collection of open sets is open then the discrete set is open.
The Problem: when proving closure by taking complements, wouldn't we need ANY collection of closed sets to be closed when if fact this is only true if the collection is finite?

Thanks for the help,
pob

 

[algebrat]

$X \LARGE$ is a metric space and $E \subset X$ is a discrete set

proof: If $p \subset E$

I think you mean p \in E, not p \subset E.

then by definition $p \LARGE$ is an isolated point of $E \LARGE$, which implies that $p \LARGE$ is not an interior point of $E\LARGE$.

Sounds good.

Since a subset of a metric space is open if every point of that subset is an interior point, it follows that $E \large$ is closed.

Not sure how you got that. I think you mean E is not open. That does not imply that E is closed.

Now that I write this proof, I believe the problem is that a subset is open IF every point is an interior point, not IF AND ONLY IF. There are other ways for the subset to be open other than all points being interior. Is this my error?

This misunderstanding I've heard of before. In defintions, like that of open, the "if" really means "if and only if". But defintions will continue to use just "if". So a set is called open if (and only if) all points are interior.

The web tells me a discrete set is open and closed. Proofs I've seen use collections of neighborhoods with radius s.t. the only point contained in the nbhd is the discrete point at its center and since any collection of open sets is open then the discrete set is open.

You may not have described the problem fully. E is certainly open relative to itself, but not necessarily relative to X.

The Problem: when proving closure by taking complements, wouldn't we need ANY collection of closed sets to be closed when if fact this is only true if the collection is finite?

I'm not even sure what collection of closed sets you're talking about. Please try to be more clear in your questions, it makes it easier to reply to.

Also, be careful about your use of the word collection. I think you need to explicitly specify if you intedn union, or intersection.

For closed sets,

an arbitrary intersection of closed sets is closed.

a finite union of closed sets is closed.

an arbitrary union of closed sets is not generally closed (some are).

 

[pob1212]

I think you mean p \in E, not p \subset E.

yes, my mistake, $p \in E$

Not sure how you got that. I think you mean E is not open. That does not imply that E is closed.

Here's how: every point in E is an isolated point => E has no limit points => E contains all of its limit points => E closed. (As far as I can tell, this is true for a discrete set containing finitely many or infinitely many points)
I was wrong to say not open implies closed, which I know realize is false. I should have gone straight to the definition of closed as I have now done. Is this now correct?

I apologize for not being exact. Let me just state my question: If $\small X$ is a metric space and $\small E$ is a discrete subset of $\small X$, is $\small E$ open, closed, or both? $\small E$ can be either finite or infinite.

Thanks for your help

 

[pob1212]

And I mean is E open/closed/both with respect to X

 

[algebrat]

Here's how: every point in E is an isolated point => E has no limit points => E contains all of its limit points => E closed. (As far as I can tell, this is true for a discrete set containing finitely many or infinitely many points)
I was wrong to say not open implies closed, which I know realize is false. I should have gone straight to the definition of closed as I have now done. Is this now correct?

I think you've decided E is closed. I agree. I can't confirm your proof however, as I can't stand trying to remember or look up all those derived set, limit points, adherent point defintions.

I apologize for not being exact. Let me just state my question: If $\small X$ is a metric space and $\small E$ is a discrete subset of $\small X$, is $\small E$ open, closed, or both? $\small E$ can be either finite or infinite.

Thanks for your help

You've got a great attitude, don't worry about being too exact on a forum, I'm being a crank. I'm imprecise on these things quite often.

My intuition tells me E is not necessarily open, have you decided, can you picture it?

And you're very welcome.

 

[pob1212]

My intuition tells me E is not necessarily open, have you decided, can you picture it?

Rudin states "E is open if every point of E is an interior point of E", so if this "if" is in fact IF AND ONLY IF, then a discrete set cannot be open b/c all points in E are isolated.

 

[algebrat]

Rudin states "E is open if every point of E is an interior point of E", so if this "if" is in fact IF AND ONLY IF, then a discrete set cannot be open b/c all points in E are isolated.

So you are claiming that isolated points cannot be interior? How would you prove this?

 

[pob1212]

Claim: isolated points cannot be interior points of any discrete subset E of a metric space

proof: by definition, all interior points are limit points. If p is an isolated point of E, we can find a nbhd N(p) such that p is the only point in N(p). Therefore p cannot be a limit point of E and hence cannot be an interior point of E.

 

[algebrat]

by definition, all interior points are limit points.

How would you prove this?

 

[pob1212]

I'm actually wrong. The statement - all interior points are limit points - is not true in general, namely when E is a discrete set it is not true! If E = [0,1] then yes every interior point is a limit point.
proof: if p is an interior point then there exists a neighborhood N s.t. N is a subset of E. Consider all neighborhoods which are subsets of N. Each one of these neighborhoods contains a point q not equal to p s.t. q is in E, because for every radius r>0, we can find a q in E s.t. d(p,q)<r for all p in E. It is obvious that any neighborhood with radius greater than N satisfies the requirement of a limit point of E. Therefore every interior point of E is a limit point of E.
(Any improvements are welcome)

What I did not realize was that by definition every subset (singleton) of a discrete set is an open set. However, I don't know why this is the case.

 

[algebrat]

The statement - all interior points are limit points - is not true in general, namely when E is a discrete set it is not true!

You are wrong, find the mistake.

If E = [0,1] then yes every interior point is a limit point.

Hmm. You need to slow down and specify what X is.

proof: if p is an interior point then there exists a neighborhood N s.t. N is a subset of E. Consider all neighborhoods which are subsets of N. Each one of these neighborhoods contains a point q not equal to p s.t. q is in E, because for every radius r>0, we can find a q in E s.t. d(p,q)<r for all p in E. It is obvious that any neighborhood with radius greater than N satisfies the requirement of a limit point of E. Therefore every interior point of E is a limit point of E.

You seemed to have two different claims before you started your proof, so I must confess I am not going to take the time to read this proof, since I don't know what your claim is.

What I did not realize was that by definition every subset (singleton) of a discrete set is an open set.

This is not true in general.

Are you taking a class. You may want to find better feedback through other resources, I personally do not want to continute to check every one of your sentences. You should be checking more of your own logic before I would continue to comment. You might attempt to carefully write a claim, write it's proof, then step away, come back to it a little later, and check each definition, and each step in your logic.

When you carefully state your claim (for instance to your self), including writing it down, then develop your proof, you may find, if you're really careful, that you are having difficulty proving it. With the investigation you have just developed, you may be able to modify your claim, to a better "guess". Lather rinse repeat.

 

[pob1212]

I appreciate the feedback. Yeah I'm taking a class and unfortunately I'm having to rush through the material

 

[algebrat]

I appreciate the feedback. Yeah I'm taking a class and unfortunately I'm having to rush through the material

No sweat, best of luck. Yeah, having to rush through math is not fun, it's sort of like the squirrel who saves nuts for the winter is well off, school is much more of a pain when we're rushing.

But I bet you're enjoying it, and don't let my impulsive crankiness stop you from thinking out loud on here.