First of all, everywhere I see the typical example of $\left(0,1\right)\cup\left(2,3\right)$ and its one-point compactification that is two circles that are tangent in one point.
I see that $\left(0,1\right)$ can be homeomorphic to a circle deleting one point. So I was thinking about the reason.
So, I thought that this map can be combined with the other that comes from $\left(2,3\right)$ (I am not looking for the mathematical definition of the map, just the idea)
And making those circles like connect by this point that is missing, and adding it (that is where I thought the infinity point $\infty$ really participates) can make the one-point compactification.
- Am I thinking it the wrong way?
- What about $\left(0,1\right)\cup\left[2,3\right]$? Which is its one-point compactification and why?
Best Answer
If you have a compact Hausdorff space $C$ and a point $p \in C$ such that $C\setminus \{p\}$ is homeomorphic to some space $X$, then a standard theorem on the essential unicity of the one-point compactification, tells us that the one-point compactification of $X$ is homeomorphic to $C$.
We can apply this to the touching circles $C$ and their common "touching point" $p$ : if we remove it we ar left with two disjoint open arcs, which are each homeomorphic to an open interval.
As to $[2,3] \cup (0,1)$: consider a circle and a disjoint compact segment, like $S^1 \cup ([2,3]\times \{0\})$ in the plane. It's compact and removing a point from the circle again leaves an open interval, disjoint from a compact interval.