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 March 7th, 2017, 10:09 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90 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.
March 7th, 2017, 10:19 AM   #2
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 Originally Posted by zylo 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.

March 7th, 2017, 10:56 AM   #3
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 Originally Posted by Maschke 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.

March 7th, 2017, 12:04 PM   #4
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 Originally Posted by Maschke 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.

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.

March 7th, 2017, 12:10 PM   #5
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 Originally Posted by zylo 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?

March 7th, 2017, 12:22 PM   #6
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 Originally Posted by Maschke 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?

March 7th, 2017, 12:25 PM   #7
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 Originally Posted by zylo 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.

March 8th, 2017, 04:14 AM   #8
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 Originally Posted by Maschke 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.

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.

March 11th, 2017, 12:14 AM   #9
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Quote:
 Originally Posted by Maschke 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.

March 17th, 2017, 08:38 AM   #10
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Quote:
 Originally Posted by Maschke 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.

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