My Math Forum  

Go Back   My Math Forum > College Math Forum > Topology

Topology Topology Math Forum


Reply
 
LinkBack Thread Tools Display Modes
March 7th, 2017, 10:09 AM   #1
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

Relatively Open and Closed Sets

Relatively Open and Closed Sets
I periodically come back to this to try and clear it up in my mind. This time I was going to propose it as a question and in the process answered it myself.

Usiing Notation of Rudin Chapt 2:

Let X be a metric space. That's too vaque and ambiquous. Let X = (R^n).
RELATIVE TO X:
1) X open because every point of X has a neighborhood entirely in X.
2) X closed because every limit point of X is in X.

Let Y be a closed or open subset of X.
RELATIVE TO Y:
3) Y open because every point of Y has a neighborhood entirely in Y.
4) Y closed because every limit point of Y is in Y.

If Y is closed relative to X and p is a border point relative to X, then p has a neighborhood entireiy in Y relative to Y, so Y relative to Y is open.

If Y is open relative to X, and p is a border point relative to X, p doesn't exist relative to Y. With respect to Y, points not in Y don't exist.

Example.

Ley Y be interior of a circle with or without its border (closed or open in X). Draw a solid line down the middle. With respect to Y, the line and area to right of line are closed, and the area to left of line is open, regardless of whether or not Y is open or closed with respect to X.

So when Rudin says let X be a metric space, it doesn't have to be specified as closed or open or neither.
zylo is offline  
 
March 7th, 2017, 10:19 AM   #2
Senior Member
 
Joined: Aug 2012

Posts: 1,414
Thanks: 342

Quote:
Originally Posted by zylo View Post
Let X be a metric space. That's too vaque and ambiquous. Let X = (R^n).
RELATIVE TO X:
1) X open because every point of X has a neighborhood entirely in X.
2) X closed because every limit point of X is in X.
I am not happy with (1) and (2). If an open set is defined by (1), then a closed set is defined as the complement of an open set. Then (2) is later shown to be a property of closed sets. So there's a slight logic mismatch between (1) and (2). Minor point but in the context of your question it would be better to be very precise about what is an open set and what is a closed set.

I'm not sure what question you are asking. But being open is a relative property.

Example: In the real numbers $\mathbb R$, the set $\mathbb R$ is open, because every point is interior.

But as a subset of the complex numbers $\mathbb C$, $\mathbb R$ is NOT open, because every open ball around any point of $\mathbb R$ contains non-real complex numbers.

Right? Right. So whether a particular set is open or not is a function of what topological space it's considered part of in the context of a particular discussion.

Now here is a puzzler for you. The metric space $\mathbb R$ is complete. Every Cauchy sequence converges.

Now using the tangent/arctangent we have a homeomorphism (bijection continuous in both directions) between $\mathbb R$ and the open unit interval $(0,1)$. But $(0,1)$ is NOT complete, since the Cauchy sequence $(\frac{1}{n})$ does not converge. What does this tell us?

Last edited by Maschke; March 7th, 2017 at 10:23 AM.
Maschke is offline  
March 7th, 2017, 10:56 AM   #3
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

Quote:
Originally Posted by Maschke View Post
Now here is a puzzler for you. The metric space $\mathbb R$ is complete. Every Cauchy sequence converges.

Now using the tangent/arctangent we have a homeomorphism (bijection continuous in both directions) between $\mathbb R$ and the open unit interval $(0,1)$. But $(0,1)$ is NOT complete, since the Cauchy sequence $(\frac{1}{n})$ does not converge. What does this tell us?
Nothing.
zylo is offline  
March 7th, 2017, 12:04 PM   #4
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

Quote:
Originally Posted by Maschke View Post
Now here is a puzzler for you. The metric space $\mathbb R$ is complete. Every Cauchy sequence converges.

Now using the tangent/arctangent we have a homeomorphism (bijection continuous in both directions) between $\mathbb R$ and the open unit interval $(0,1)$. But $(0,1)$ is NOT complete, since the Cauchy sequence $(\frac{1}{n})$ does not converge. What does this tell us?
Sorry, took me a while to figure out what you were talking about.
Questions of completeness are irrelevant in open and unbounded sets.

Irrelevant to this thread- why I asked it in new thread:
Map (0,1) to R

I assume you meant to discredit me in order to discredit OP, which means it is correct, or you wouldn't have bothered.
zylo is offline  
March 7th, 2017, 12:10 PM   #5
Senior Member
 
Joined: Aug 2012

Posts: 1,414
Thanks: 342

Quote:
Originally Posted by zylo View Post
I assume you meant to discredit me
I've never been anything other than polite and helpful to you. Do you feel otherwise? Can you link any example to the contrary?
Maschke is offline  
March 7th, 2017, 12:22 PM   #6
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

Quote:
Originally Posted by Maschke View Post
I've never been anything other than polite and helpful to you. Do you feel otherwise? Can you link any example to the contrary?
What is the relevance of your puzzler question in Post #2 to OP?
zylo is offline  
March 7th, 2017, 12:25 PM   #7
Senior Member
 
Joined: Aug 2012

Posts: 1,414
Thanks: 342

Quote:
Originally Posted by zylo View Post
What is the relevance of your puzzler question in Post #2 to OP?
I say again. In the year or so that you and I have participated in common threads on this site, have I ever been less than polite and helpful to you? Can you link an example of same?

The purpose of the question was to make you think. Evidently you aren't interested in doing that. If you're looking for me to be rude to you I'm happy to oblige.
Maschke is offline  
March 8th, 2017, 04:14 AM   #8
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

Quote:
Originally Posted by Maschke View Post
Now here is a puzzler for you. The metric space $\mathbb R$ is complete. Every Cauchy sequence converges.

Now using the tangent/arctangent we have a homeomorphism (bijection continuous in both directions) between $\mathbb R$ and the open unit interval $(0,1)$. But $(0,1)$ is NOT complete, since the Cauchy sequence $(\frac{1}{n})$ does not converge. What does this tell us?
I have given the answer in posts 5 and 10 of:
Map (0,1) to R
Apparently you still don't get it.

There is a sequence in R whose inverse map converges to 0, which is not in (0,1), which is not surprising since 0 doesn't map to R. So the inverse map of every convergent sequence in R is convergent.

You should give your sources:
http://math.stackexchange.com/questi...-complete-sets

By the way, the OP was about relatively open and closed sets. If you have any other unrelated questions, please post them in your own thread.

Last edited by zylo; March 8th, 2017 at 04:41 AM.
zylo is offline  
March 11th, 2017, 12:14 AM   #9
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

Quote:
Originally Posted by Maschke View Post
Now here is a puzzler for you. The metric space $\mathbb R$ is complete. Every Cauchy sequence converges.

Now using the tangent/arctangent we have a homeomorphism (bijection continuous in both directions) between $\mathbb R$ and the open unit interval $(0,1)$. But $(0,1)$ is NOT complete, since the Cauchy sequence $(\frac{1}{n})$ does not converge. What does this tell us?

Let f be 1-1 (bijective) and continuous between (0,1) and R.

1/n converges to a limit (which is not in (0,1)).

f(1/n) converges to a limit (which is in R because R is complete). ->

1/n converges to a limit in (0,1). Contradiction.

f is not bijective.
zylo is offline  
March 17th, 2017, 07:38 AM   #10
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

Quote:
Originally Posted by Maschke View Post
Now here is a puzzler for you. The metric space $\mathbb R$ is complete. Every Cauchy sequence converges.

Now using the tangent/arctangent we have a homeomorphism (bijection continuous in both directions) between $\mathbb R$ and the open unit interval $(0,1)$. But $(0,1)$ is NOT complete, since the Cauchy sequence $(\frac{1}{n})$ does not converge. What does this tell us?
This is not a puzzler.

If a bounded subset of R were complete it would be closed and f would be bounded and so couldn't map to R. It has to be incomplete.
zylo is offline  
Reply

  My Math Forum > College Math Forum > Topology

Tags
closed, open, sets



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
I need examples about open, closed and neither open nor closed sets yuceelly Topology 3 March 8th, 2016 10:45 AM
Open / Closed Sets veronicak5678 Real Analysis 1 May 1st, 2012 12:22 PM
neither open nor closed sets mizunoami Real Analysis 3 November 30th, 2011 07:06 PM
Closed and Open Sets in R^d x_saved_kt Real Analysis 1 September 5th, 2011 01:24 PM
Closed/Open sets farzyness Real Analysis 1 February 18th, 2009 04:58 PM





Copyright © 2017 My Math Forum. All rights reserved.