# I Direct product of intervals

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1. Sep 23, 2016

### shinobi20

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?

2. Sep 23, 2016

### Math_QED

Do you mean something like:

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

3. Sep 23, 2016

### shinobi20

For examples, yes.

4. Sep 23, 2016

### Math_QED

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.

5. Sep 23, 2016

### shinobi20

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

6. Sep 23, 2016

### micromass

Staff Emeritus
It's neither open nor closed.

7. Sep 23, 2016

### shinobi20

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

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.

8. Sep 23, 2016

### micromass

Staff Emeritus
That domain is indeed not open.

9. Sep 23, 2016

### shinobi20

Yes, just as I thought, so do you have any idea how should the infinite cylinder be covered by a single chart?

10. Sep 23, 2016

### micromass

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

11. Sep 23, 2016

### shinobi20

So can I cover it with
$φ^{-1}: (0,1) × S^1 → ℝ × S^1$
$φ^{-1}: annulus → infinite~ cylinder$

Last edited: Sep 23, 2016
12. Sep 23, 2016

### micromass

Staff Emeritus
$(0,1)\times S^1$ is not an annulus.

13. Sep 23, 2016

### shinobi20

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.

14. Sep 23, 2016

### FactChecker

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.

15. Sep 24, 2016

### shinobi20

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

16. Sep 24, 2016

### FactChecker

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.

17. Sep 24, 2016

### shinobi20

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.

18. Sep 24, 2016

### FactChecker

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.

19. Sep 24, 2016

### shinobi20

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.