Proving two path connected subsets of $\Bbb{R}^2$ are not homeomorphic when the number of cut-points/pairs are the same.  
Given the two path connected subsets of $\Bbb{R}^2$ depicted in the image attached, where the end of the line segment on the left is included, how do I go about proving they are not homeomorphic?
This question comes from a larger set of questions in which the lack of topological equivalence was proven with simple arguments based on the number of cut-points and pairs. However, as far as I can tell these have the same number of n-type cuts (they have infinite 1 points and 2 points, infinite 1 pairs, 2 pairs, and 3 pairs etc).
They do seem intuitively different to me as the 1-pairs on the left are based on selecting the end point and any point on the circle, where as the choice of points for the pairs is not as restrictive on the right. But I can't figure any argument from this.
 A: In the left hand picture if we have a cutpoint, one of its induced components is contractible, while this is not the case for the cutpoints of the right hand picture (these all lie on the middle segment). This is a topological property that shows they're not homeomorphic.
A: Here are two approaches which use cut-points:


*

*The space on the left has the property that one can remove a non-cut point (namely any point on the left on the circle) such that the remaining space has exactly one non-cut point. For the space on the right this is not true since after removing a non-cut point almost any point on the remaining full circle is a non-cut point.

*The space on the right has the property that one can remove one point (namely the point in the middle of the line segment) and get two components such that each component has infinitely many non-cut points. For the space on the left this is not true as after removing a point and getting two components one of them is a half open intervall which has only one non-cut point.
