Compact Sets of Moore Plane

[ForMyThunder]

Is there any easy way to find all the compact sets of 1) the slitted plane and 2) the Moore plane?

1) defined as the topology generated by a base consisting of z\cup A where A is a disc about z with finitely many lines deleted. I believe the compact sets in this topology coincide with the usual real line. How do I prove this?

2) (rough sketch) I know that a closed, bounded set disjoint from the x-axis is compact. I guess if a set touched the x-axis in an infinite number of points, it would not be compact. So the compact sets consisting of a closed, bounded set disjoint from the x-axis along with finitely many points on the x-axis is compact?

 

[fresh_42]

1) I do not think so. A closed section of disc should be compact and is no real line.
2) The number of points on the $x-$ axis shouldn't play a role here.

 

[math771]

Hi there!

For 1), you are correct that the compact sets in this topology are the same as the usual real line. To prove this, we can use the fact that a subset of a compact set is compact and that a closed subset of a compact set is compact. So, let K be a compact set in the slitted plane topology. We know that K is a subset of the usual real line, which is compact. Therefore, K is compact as well. Similarly, if K is a closed subset of the usual real line, it is also closed in the slitted plane topology and therefore compact. This shows that the compact sets in the slitted plane topology are the same as the compact sets in the usual real line.

For 2), your intuition is correct. A closed, bounded set disjoint from the x-axis is compact because it is a subset of the usual real line, which is compact. And a set that touches the x-axis in an infinite number of points is not compact because it is not closed. So, the compact sets in the Moore plane would be the closed, bounded sets disjoint from the x-axis along with finitely many points on the x-axis. This can also be proved using the same reasoning as in 1).

Hope this helps!