Is the one point compactification of the set $\{xy = 0\}\in \mathbb R^2$ minus any point connected?

Question

Call the subset of $\mathbb R^2$ given by $xy = 0$ (ie. the coordinate axis) $X$. Let the one point compactification of $X$ be given by $X^*$. In my mind, I have each axis headed to the compactification point.

I want to prove that if we remove any point from $X^*$, $X^*$ remains connected.

I can see how this is done if the point at infinity is removed but I cannot prove this for the general case. The difficulty stems from a lack of intuition about what $X^*$ "looks like".

Additionally, is $X^*$ minus a point path connected? If so, how can I explicitly find a path from $(x,0)$ to the compactification point?

Answer 1

It's probably easiest to see this if we project onto the Riemann sphere; there, your set becomes two orthogonal great circles, meeting at the origin and the point at infinity, which are antipodal on the sphere. Clearly, if we remove a point not at one of these poles, the figure remains path connected, and if we remove either pole, the figure is still path connected, because one can use the other pole to get between the great circles.

Answer 2

The space $X^*$ is homeomorphic to a circle inscribed in an ellipse. Small neighborhoods of the point at infinity (call it $\infty$) look like neighborhoods of the origin (call it $O$), $\times$-shaped. Removing either of those points doesn't disconnect the space, as the remaining part of the space is path connected through the other point. Removing any other point doesn't disconnect the space either, as it is path connected through either of those points.

In fact, there are two homeomorphism classes of points in $X^*$: the set $A=\{ O,\infty\}$, and the set $B=X^*\setminus A $. Given any pair of points from such a class, there is a homeomorphism of $X^*$ with itself which interchanges these points, and there is no such homeomorphism interchanging two points from different classes.

Answer 3

Rather than do everything rigorously, I’ll try to give you some intuition about the space.

Let $C$ be a great circle containing the north and south poles, and let $E$ be the equator; then $X^*$ looks like $C\cup E$. $C$ and $E$ intersect in two points; one of them corresponds to the origin in $\Bbb R^2$, and the other corresponds to the point at infinity; call these points $p$ and $q$. It should be clear that if you remove either $p$ or $q$, get a space homeomorphic to the original $X$. What happens if you remove one of the other points?

In that case three of the arcs connecting $p$ and $q$ remain intact, and the fourth is broken. If you flatten everything into a plane, you can arrange it to look like this:

The ‘spikes’ pointing up and down are open at the ends: the missing point is the one removed from $X^*$. In other words, the vertical line is essentially just an open interval.