Compactness of V in Minkowski Space: A Question

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Discussion Overview

The discussion revolves around the question of whether the closed forward light cone in Minkowski space, defined as V = { x ∈ M | x² ≥ 0, x⁰ ≥ 0 }, is a compact set. Participants explore the implications of compactness in the context of Minkowski space and the criteria for compactness.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants express an initial intuition that the light cone might be compact but are uncertain about how to prove it.
  • One participant suggests that a criterion for non-compactness is boundedness, stating that if a set is not bounded, it cannot be compact.
  • A later reply corrects an earlier claim, stating that the light cone is not bounded and therefore not compact.
  • Another participant notes that the Heine-Borel theorem, which applies to subsets of Rⁿ, does not apply to Minkowski space, raising questions about the generalizability of compactness properties.
  • There is a suggestion that Minkowski space could be equipped with a Euclidean topology, which might allow for similar topological properties as R⁴, but uncertainty remains regarding compactness.
  • One participant clarifies that while all compact sets are bounded in any metric space, the specific set in question is not bounded, leading to the conclusion that it cannot be compact.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the compactness of the set, with multiple competing views and uncertainties regarding the application of compactness criteria in Minkowski space.

Contextual Notes

Limitations include the dependence on definitions of compactness and the applicability of the Heine-Borel theorem to Minkowski space, which remains unresolved.

parton
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Consider the closed forward light cone

V = \left \lbrace x \in M \mid x^{2} \geq 0, x^{0} \geq 0 \right \rbrace
and M denotes Minkowski space.

My question is whether V is a compact set or not. If it is a compact set, how do I show it?

Intuitively I would say it is compact, but I don't know how to proof it.

I hope someone can help me.
 
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parton said:
Consider the closed forward light cone

V = \left \lbrace x \in M \mid x^{2} \geq 0, x^{0} \geq 0 \right \rbrace
and M denotes Minkowski space.

My question is whether V is a compact set or not. If it is a compact set, how do I show it?

Intuitively I would say it is compact, but I don't know how to proof it.

I hope someone can help me.

Why does your intuition tell you that a light cone is compact?
 
A simple criterion for deciding if a set is not compact is boundedness: if a set is not bounded, then it's not compact!
 
I'm sorry, I did a mistake and the problem was no cone, but a certain subset of a cone.

Of course the cone defined above is not bounded and therefore not compact. Thanks.
 
That is only true for subsets of Rn. Are you working with subsets of Rn here?
 
HallsofIvy said:
That is only true for subsets of Rn. Are you working with subsets of Rn here?

hmmm, no, I am working with subsets of Minkowski space M. So the Heine Borel theorem doesn't apply to M. On the other side we could equip M with an Euclidean topology. And this space would be homeomorphic to R^4, so the basic topologic properties should be the same in both spaces. But I am not sure whether this holds for compactness, too.

But to be sure, maybe one should use just the general definition of compactness to prove the compactness of a subset of M, i.e. if each of its open covers has a finite subcover. But I think it is rather difficult to prove it in this way, but I don't know another possibility.
 
Sorry, I misread. It is only in Rn or variations that "closed and bounded" sets must be compact. It is true in any metric space that all compact sets are bounded. Since this set is not bounded it cannot be compact.
 

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