# 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.

• @Clayton edit: Surely all points along the line segment bar the end disconnect the sets though? – Goethe Mar 18 at 19:58
• Path-components and connected components are the same here, since the space is locally path-connected. The idea described by @Clayton comes from the fact that homeomorphisms induce bijections on path-components (or connected components, but again, for this space there is no difference). – o.h. Mar 18 at 20:03