# Light cone

1. Sep 7, 2009

### parton

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.

2. Sep 7, 2009

### wofsy

Why does your intuition tell you that a light cone is compact?

3. Sep 7, 2009

### quasar987

A simple criterion for deciding if a set is not compact is boundedness: if a set is not bounded, then it's not compact!

4. Sep 7, 2009

### parton

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.

5. Sep 7, 2009

### HallsofIvy

Staff Emeritus
That is only true for subsets of Rn. Are you working with subsets of Rn here?

6. Sep 8, 2009

### parton

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.

7. Sep 8, 2009

### HallsofIvy

Staff Emeritus
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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