Definition of compactness in the EXTENDED complex plane?

[AxiomOfChoice]

Definition of "compactness" in the EXTENDED complex plane?

How does one define a compact set in the extended complex plane $\mathbb C^* = \mathbb C \cup \{ \infty \}$? "Closed and bounded" doesn't really make sense anymore, as I'm assuming it's permissible for a compact set to contain the point at infinity...right? I guess the "finite subcover" definition still holds, as always, but this doesn't seem very useful. Are there other, more helpful, equivalent characterizations for compact subsets of $\mathbb C^*$?

 

[HallsofIvy]

The general definition of "compact" is "Given any open cover for the set, there exist a finite subcover". That is, a set, A, is compact if and only if whenever $\{U_\alpha\}$ is a collection of open sets such that every point of A is in at least one of those sets, there exist a finite collection of those same sets that have the same property.

Another, equivalent, definition is that a set, A, is compact if and only if every infinite sequence of points in A has a subsequence that converges to a point in A.

Using either of those you can prove that a compact set is closed and, in a metric space where "bounded" is defined, a bounded set. You can prove, in $R^n$ that any "closed and bounded" set satisfies those definitions.

 

[JSuarez]

As $\mathbb C^* = \mathbb C \cup \{ \infty \}$ is a compact space (in fact it's obtained from $\mathbb C$ by a general process called one-point, or Hausdorff, compactification), all it's closed subset will be compact. What are the closed subsets of $\mathbb C^*$ (don't forget you can identify it with 2-sphere)?

 

[zhentil]

If you take the topology to be the one-point compactification, then it's a definition.

If you don't, then you have to tell us what your topology is. The way the question is phrased is basically "What should the compact sets be?", which of course isn't well-defined.

 

[AxiomOfChoice]

If you take the topology to be the one-point compactification, then it's a definition.

If you don't, then you have to tell us what your topology is. The way the question is phrased is basically "What should the compact sets be?", which of course isn't well-defined.

Interesting! Thanks! That kind of makes sense, since closed subsets of compact spaces are necessarily compact.

As to the question you've posed about characterizing closed subsets of $\mathbb C^*$, though...is there something more intuitive or convenient than just "a set that contains all its limit points" or "a set whose complement is open?"

 

[Landau]

The general definition of "compact" is "Given any open cover for the set, there exist a finite subcover". That is, a set, A, is compact if and only if whenever $\{U_\alpha\}$ is a collection of open sets such that every point of A is in at least one of those sets, there exist a finite collection of those same sets that have the same property.

That's what AxiomOfChoice already mentioned: "I guess the "finite subcover" definition still holds, as always, but this doesn't seem very useful."

Another, equivalent, definition is that a set, A, is compact if and only if every infinite sequence of points in A has a subsequence that converges to a point in A.

No, these are not equivalent. You need the topological space to be first countable, then compact and sequentially compact (= every sequence has a converging subsequence) are equivalent. There are topological spaces which are compact but not seq. compact.

Using either of those you can prove that a compact set is closed

No, this is also not true. You need Hausdorffness for that.

You can prove, in \(\displaystyle R^n\) that any "closed and bounded" set satisfies those definitions.

That's also what TS already mentioned: ""Closed and bounded" doesn't really make sense anymore."

@AxiomOfChoice: also see here (pdf) for some explantion about the Riemann sphere and it's topology.

 

[AxiomOfChoice]

That's what AxiomOfChoice already mentioned: "I guess the "finite subcover" definition still holds, as always, but this doesn't seem very useful."

No, these are not equivalent. You need the topological space to be first countable, then compact and sequentially compact (= every sequence has a converging subsequence) are equivalent. There are topological spaces which are compact but not seq. compact.
No, this is also not true. You need Hausdorffness for that.
That's also what TS already mentioned: ""Closed and bounded" doesn't really make sense anymore."

@AxiomOfChoice: also see here (pdf) for some explantion about the Riemann sphere and it's topology.

Landau, thanks for this post. Some of the things HallsofIvy said confused me, and you cleared them up. Also, thanks for the PDF; I'm looking over it now.