Thread: Clarify the number of path compnents...

I have two circles joined by a straight line, which are path-connected subsets of the plane.
If I remove any point from either circle, how many path components would there be?
Intuitively, I think there could be the same number as a circle, namely 1, but I'm not sure this is correct...?

Originally Posted by bigdoggy

I have two circles joined by a straight line, which are path-connected subsets of the plane.
If I remove any point from either circle, how many path components would there be?
Intuitively, I think there could be the same number as a circle, namely 1, but I'm not sure this is correct...?

In general, yes, but suppose the point that is removed is the point where the straight line meets the circle??

Originally Posted by Opalg

In general, yes, but suppose the point that is removed is the point where the straight line meets the circle??

Hi, thankyou. that would create 2 path components.
But if I removed a point from either circle, I would be left with the subset
o-c .....is this still one path component ?

Originally Posted by bigdoggy

Hi, thankyou. that would create 2 path components.
But if I removed any other point from either circle, I would be left with the subset
o-c .....is this still one path component ?

Yes.

Originally Posted by Opalg

Yes.

Therefore it has infinitely many points which create 1 path component....

So to show this subset is not homeomorphic to o- (where the line s the circle and the end point of the line is included) I need to show that they have different types of cut points. But, o- has infinitely many cut points of type 1(any point on the circle) and infinitely many cut points of type 2(anypoint on the line).
So via this method, how do I show they aren't homeomorphic??