# One point compactification of the punctured plane $\Bbb{R}^2\setminus\{(0,0)\}$

I am thinking about the Alexandroff one point compactification of $\Bbb{R}^2\setminus\{(0,0)\}$. Clearly $\Bbb{R}^2\setminus\{(0,0)\}$ is homeomorphic to a sphere which does not has both North and South poles. As I think the compactification can be done two ways.

1. Bring both poles into the sphere towards the center, and then glue them at the center by adding a point.

2. Pull both poles away sphere and then glue them together by adding a point.

(Unfortunately I have no picture or animation to explain these constructions.)

I doubted that whether these two spaces are homeomorphic or not. Can someone explain this? thank you.

• Both spaces are homeomorphic. You're worrying about being inside or outside the sphere, but this is only a problem because you're trying to embed your space in $\mathbb R^3$. – Pedro Tamaroff Nov 10 '16 at 21:33
• Could you please explain it bit more? – Bumblebee Nov 11 '16 at 16:07
• There is a natural choice for a bijection between the two apparently different versions of the one-point compactification that you describe. Try to show that this bijection is a homeomorphism. Seems you won't have much work to do... – Mirko Nov 11 '16 at 17:16
• @Nil You're thinking of constructing your space inside three space: don't! Define it as a quotient space. – Pedro Tamaroff Nov 11 '16 at 22:20

Let $X$ be a non-compact, locally compact Tychonoff space with topology $T_X$. Let $p\ne S$ and let $aX=\{p\}\cup X$ be a compact Hausdorff space such that the subspace topology on $X$ (as a subspace of $aX$) is $T_X.$ We show that the topology on $aX$ is unique up to homeomorphism.

(i). If $p\in U\subset aX$ where $U$ is open in $aX,$ then $aX$ \ $U$ is closed in $aX$, hence $aX$ \ $U$ is a compact subspace of $aX$ and is a subset of $X.$

(ii). If $V\subset X$ and $V$ is a compact subspace of $aX$ then $V$ is closed in $aX,$ so $U=aX$ \ $V$ is open in $aX$ and contains $p.$

(iii). From (i) and (ii), if $p\in U\subset aX$ then $U$ is open in $aX$ iff $aX$ \ $U$ is a compact subspace of $aX.$

Now let $X_1=X\cup \{p1\}$ and $X_2=X\cup \{p_2\}$ be one-point compactifications of $X.$ Let $f(x)=x$ for $x\in X,$ and let $f(p_1)=p_2.$

(iv). If $U_1$ is an open subset of $X_1$ and $U_1\subset X$ then $f(U_1)=U_1$ is open in $X_2$ because every member of $T_X$ (the original topology on $X$ )is open in $X_2.$

(v). If $U_1$ is open in $X_1$ with $p_1\in U_1$ then $X_1$ \ $U_1$ is a compact subspace of $X_1$ by (iii), while $$X_2 \backslash f(U_1)=X_1 \backslash U_1.$$ Important point: $X_1$ \ $U_1 \subset X$, and the subspace topology on $X$ as a subspace of either $X_1$ or $X_2$ is $T_X,$ so the subspace topologies on $X_1$ \ $U_1$ as a subspace of either $X_1$ or $X_2$ are the same topology.

And $X_1$ \ $U_1$ is a compact subspace of $X_1$. So $X_2$ \ $f(U_1)=X_1$ \ $U_1$ is a compact subspace of $X_2$. By (iii), $f(U_1)$ is open in $X_2.$

(vi). Interchanging the subscripts 1, 2 and changing $f$ to $f^{-1}$ in (iv) and (v) , we see that $$f \text { is a homeomorphism.}$$