# Clarify the number of path compnents...

• Nov 8th 2009, 11:26 AM
bigdoggy
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...?
• Nov 8th 2009, 11:53 AM
Opalg
Quote:

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??
• Nov 8th 2009, 12:00 PM
bigdoggy
Quote:

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 ?
• Nov 8th 2009, 12:15 PM
Opalg
Quote:

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. (Happy)
• Nov 8th 2009, 12:26 PM
bigdoggy
Quote:

Originally Posted by Opalg
Yes. (Happy)

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??(Wondering)