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- #1

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- #2

member 587159

Do you mean something like:

##(a,b] \times (a,b)## ?

##(a,b] \times (a,b)## ?

- #3

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For examples, yes.Do you mean something like:

##(a,b] \times (a,b)## ?

- #4

member 587159

For examples, yes.

It seems a weird question to me:

##(a,b] \times (a,b) := \{(x,y)| x \in (a,b] \land y \in (a,b)\}## How can this set be closed or open? It's a set containing ordered pairs. This might be related to topology, so I'm not qualified to answer this question.

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Yes, that is why I'm asking, what do you think is it?It seems a weird question to me:

##(a,b] \times (a,b) := \{(x,y)| x \in (a,b] \land y \in (a,b)\}## How can this set be closed or open? It's

- #6

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It's neither open nor closed.

- #7

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It's neither open nor closed.

There is a problem that I found that asks to construct a single chart to cover an infinite cylinder. It is talked about in this thread that the direct product of a half open interval and an open interval somehow yields an open set

https://www.physicsforums.com/threads/infinite-cylinder-covered-by-a-single-chart.879193/

andrewkirk said in post #6 that "What has to be open is the

- #8

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There is a problem that I found that asks to construct a single chart to cover an infinite cylinder. It is talked about in this thread that the direct product of a half open interval and an open interval somehow yields an open set

https://www.physicsforums.com/threads/infinite-cylinder-covered-by-a-single-chart.879193/

andrewkirk said in post #6 that "What has to be open is thedomain of the chart" but [0,2π) is not open, so the domain would be [0,2π) x (-∞,∞) not open based on what you said.

That domain is indeed not open.

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Yes, just as I thought, so do you have any idea how should the infinite cylinder be covered by a single chart?That domain is indeed not open.

- #10

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Cover it with a chart whose domain is an open annulus.

- #11

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So can I cover it withCover it with a chart whose domain is an open annulus.

##φ^{-1}: (0,1) × S^1 → ℝ × S^1##

##φ^{-1}: annulus → infinite~ cylinder##

Last edited:

- #12

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##(0,1)\times S^1## is not an annulus.

- #13

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You mean, there should be a hole,##(0,1)\times S^1## is not an annulus.

##φ^{-1}: (a,b) × S^1 → ℝ × S^1## where 0<a<r<b; r is the radius.

- #14

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If the set [0,2π) is being treated as the points of the circle rather than on R, then 0 = 2π and this is both open and closed. Then [0,2π) x (-∞,∞) is both open and closed. But [0,2π) is misleading notation because it implies the metric of R where 0 and 2π have a positive distance. It would be better to specify something like Ci as a circle and say that Ci x (-∞,∞) is both open and closed.There is a problem that I found that asks to construct a single chart to cover an infinite cylinder. It is talked about in this thread that the direct product of a half open interval and an open interval somehow yields an open set

https://www.physicsforums.com/threads/infinite-cylinder-covered-by-a-single-chart.879193/

andrewkirk said in post #6 that "What has to be open is thedomain of the chart" but [0,2π) is not open, so the domain would be [0,2π) x (-∞,∞) not open based on what you said.

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So the domain ##[0,2π) × (-∞,∞)## cannot cover the infinite cylinder since it is also closed?If the set [0,2π) is being treated as the points of the circle rather than on R, then 0 = 2π and this is both open and closed. Then [0,2π) x (-∞,∞) is both open and closed. But [0,2π) is misleading notation because it implies the metric of R where 0 and 2π have a positive distance. It would be better to specify something like Ci as a circle and say that Ci x (-∞,∞) is both open and closed.

- #16

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The topology of [0,2π) as a subset of R with the usual metric is different from the topology of [0,2π) as a circle where 2π = 0 with the usual metric.So the domain ##[0,2π) × (-∞,∞)## cannot cover the infinite cylinder since it is also closed?

In the first case, [0,2π) × (-∞,∞) is neither open nor closed. It cannot cover the infinite cylinder.

In the second case, [0,2π) × (-∞,∞) is the entire space and is both open and closed. It is an infinite cylinder.

So if you consider [0,2π) × (-∞,∞) as a subset of R x R, the answer is no. It can not cover the infinite cylinder.

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I'm considering ##[0,2π)## as a circle with the usual metric, so ##[0,2π) × (-∞,∞)## does cover the infinite cylinder.The topology of [0,2π) as a subset of R with the usual metric is different from the topology of [0,2π) as a circle where 2π = 0 with the usual metric.

In the first case, [0,2π) × (-∞,∞) is neither open nor closed. It cannot cover the infinite cylinder.

In the second case, [0,2π) × (-∞,∞) is the entire space and is both open and closed. It is an infinite cylinder.

So if you consider [0,2π) × (-∞,∞) as a subset of R x R, the answer is no. It can not cover the infinite cylinder.

##φ: U → φ(U)~## where ##~U ⊂ [0,2π) × (-∞,∞)~## and ##~φ(U) ⊂ ℝ×ℝ##

But I'm wondering about the definition of a chart, the domain should be open but in this case it is both open and closed.

- #18

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You should only worry about the properties that the definition needs. If there are other properties (like closed), that doesn't mean it fails to satisfy the definition.But I'm wondering about the definition of a chart, the domain should be open but in this case it is both open and closed.

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Ok, so in this case ##[0,2π) × (-∞,∞)## satisfies the condition and does cover the infinite cylinder withYou should only worry about the properties that the definition needs. If there are other properties (like closed), that doesn't mean it fails to satisfy the definition.

##U ⊂ [0,2π) × (-∞,∞)##

##φ(U) = (θ,e^z)~## where ##~e^z~## represents the radius.

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