Why the space X=(0,1) is (not sequentially) compact?

In summary: No, the Hopf–Rinow theorem does not exist for a Riemannian submanifold of a Pseudo-Riemannian manifold.
  • #1
santiagorf
2
0
My problem is that the space [itex] X= (0,1)[/itex] is not sequentially compact and compact at the same time.

It is not sequentially compact:
If we define the sequence [itex](\frac{1}{n}) [/itex] we can show that it is not sequentially compact as the sequence converges to 0, but [itex] 0 \notin X[/itex].

It is compact:
On the other hand, for X to be compact we need
1) bounded: The space X is bounded as any ball with center [itex] x \in X [/itex] and radius 2 will X.
2) closed: Is closed as its complement is the empty set (which is open)

Thus, the set [itex] X [/itex] is compact, which is a contradiction as X is not sequentially compact.

Where is my mistake when I show that X is compact?
 
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  • #2
Closed and bounded does not imply compact in general.
 
  • #3
You are using the Heine-Borel theorem which only holds for subsets of Euclidean space.
In that case the interval isn't closed either.

Edit; Ninja'd
Edit 2; Subsets of Euclidean space as far as I know, there might be other special cases.
 
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  • #4
I see, this works for subsets of [itex] R^n [/itex]
 
  • #5
For a subset of a metric space, closed should mean that every Cauchy sequence converges to a point in the space. But a Cauchy sequence in the open unit interval that converges to 0 or 1 will not coverage to a point in the open interval. So the open unit interval is not closed in ##R^1##. If every Cauchy sequence in a metric space converges then the metric space is said to be complete.

Here is a metric space that is closed and bounded but not compact. On a closed disk in the plane, let the distance between two points be the Euclidean distance if the two points lie on the same radial line through the center of the disk; and the sum if their distances to the center of the disk if they lie on different radial lines.
 
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  • #7
A follow-up question that's related to my research, does a similar theorem exist for a Riemannian submanifold of a Pseudo-Riemannian manifold?
 
  • #8
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1. Why is the space X=(0,1) not sequentially compact?

The space X=(0,1) is not sequentially compact because it does not satisfy the definition of sequential compactness. In order for a space to be sequentially compact, every sequence in the space must have a convergent subsequence. However, in the space X=(0,1), there are sequences that do not have convergent subsequences, such as 1/n. Therefore, the space is not sequentially compact.

2. What is the difference between sequential compactness and compactness?

The main difference between sequential compactness and compactness is that compactness is a stronger notion. A space is compact if every open cover has a finite subcover. On the other hand, a space is sequentially compact if every sequence has a convergent subsequence. A sequentially compact space is always compact, but the converse is not always true.

3. Can a non-compact space be sequentially compact?

Yes, a non-compact space can be sequentially compact. For example, the interval (0,1) is not compact, but it is sequentially compact. This is because in the space (0,1), every sequence has a convergent subsequence. However, (0,1) is not compact because it is not closed and bounded.

4. What are the consequences of a space being sequentially compact?

If a space is sequentially compact, it has some important properties. First, it is limit point compact, meaning that every infinite set has a limit point. Second, it is countably compact, meaning that every countable open cover has a finite subcover. Finally, it is also compact, as sequential compactness implies compactness.

5. How can we prove that the space X=(0,1) is not sequentially compact?

To prove that the space X=(0,1) is not sequentially compact, we can use the fact that a space is not sequentially compact if there exists a sequence with no convergent subsequences. In X=(0,1), we can construct the sequence 1/n, which has no convergent subsequences. Therefore, X=(0,1) is not sequentially compact.

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