# Direct product of intervals

• I
The 2-D plane is usually constructed as "ℝxℝ" and ℝ is both open and closed. My question is, what is the direct product of a half open and an open interval? Is it also open or half open?

## Answers and Replies

member 587159
Do you mean something like:

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

Do you mean something like:

##(a,b] \times (a,b)## ?
For examples, yes.

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.

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
Yes, that is why I'm asking, what do you think is it?

It's neither open nor closed.

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 domain of the chart" but [0,2π) is not open, so the domain would be [0,2π) x (-∞,∞) not open based on what you said.

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 domain 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.

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

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

Cover it with a chart whose domain is an open annulus.
So can I cover it with
##φ^{-1}: (0,1) × S^1 → ℝ × S^1##
##φ^{-1}: annulus → infinite~ cylinder##

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

##(0,1)\times S^1## is not an annulus.
You mean, there should be a hole,
##φ^{-1}: (a,b) × S^1 → ℝ × S^1## where 0<a<r<b; r is the radius.

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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 the domain of the chart" but [0,2π) is not open, so the domain would be [0,2π) x (-∞,∞) not open based on what you said.
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.

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.
So the domain ##[0,2π) × (-∞,∞)## cannot cover the infinite cylinder since it is also closed?

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So the domain ##[0,2π) × (-∞,∞)## cannot cover the infinite cylinder since it is also closed?
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.

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.
I'm considering ##[0,2π)## as a circle with the usual metric, so ##[0,2π) × (-∞,∞)## does 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.

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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.
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.

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.
Ok, so in this case ##[0,2π) × (-∞,∞)## satisfies the condition and does cover the infinite cylinder with
##U ⊂ [0,2π) × (-∞,∞)##
##φ(U) = (θ,e^z)~## where ##~e^z~## represents the radius.