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

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Discussion Overview

The discussion centers around the properties of the space X = (0,1) in relation to sequential compactness and compactness. Participants explore the definitions and implications of these concepts within the context of metric spaces and subsets of Euclidean space.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant argues that the space X = (0,1) is not sequentially compact because the sequence (1/n) converges to 0, which is not in X.
  • Another participant states that closed and bounded does not imply compactness in general.
  • A different participant points out that the Heine-Borel theorem, which is referenced, only applies to subsets of Euclidean space and claims that the interval (0,1) is not closed in this context.
  • It is noted that for a subset of a metric space, being closed means that every Cauchy sequence converges to a point within the space, and since Cauchy sequences in (0,1) can converge to 0 or 1, it is not closed in R.
  • One participant provides an example of a closed and bounded metric space that is not compact, specifically describing a closed disk with a unique distance metric.
  • There are follow-up questions regarding the existence of similar theorems for Riemannian submanifolds of Pseudo-Riemannian manifolds.

Areas of Agreement / Disagreement

Participants express disagreement regarding the compactness of the space X = (0,1) and the application of the Heine-Borel theorem. The discussion remains unresolved as multiple perspectives are presented without consensus.

Contextual Notes

Participants highlight limitations in the application of certain theorems and definitions, particularly regarding the closedness of the interval (0,1) in the context of metric spaces and the implications for compactness.

santiagorf
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My problem is that the space X= (0,1) is not sequentially compact and compact at the same time.

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

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 x \in X and radius 2 will X.
2) closed: Is closed as its complement is the empty set (which is open)

Thus, the set X 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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Closed and bounded does not imply compact in general.
 
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.
 
Last edited:
I see, this works for subsets of R^n
 
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.
 
Last edited:
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A follow-up question that's related to my research, does a similar theorem exist for a Riemannian submanifold of a Pseudo-Riemannian manifold?
 

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