# Proof of non-existence of a certain metric on $S^2$

Proposition: There is no metric $d: S^2 \times S^2 \to \mathbb{R}$ compatible with the usual topology such that $S^2 - \ast$ is isometric to the Euclidean plane.

My proof: [a gap] it is sufficient to prove that there is no metric $d$ on $\overline{\mathbb{C}}$ such that its restriction on $\mathbb{C}$ is the usual metric on $\mathbb{C}$. Consider a convergent sequence $n \mapsto \frac{1}{n}$. An autohomeomorphism $z \mapsto \frac{1}{z}$ maps it to the sequence $n \mapsto n$, which is not Cauchy in the usual metric on $\mathbb{C}$, and is thus divergent. This is a contradiction.

Is this proof correct? It feels fuzzy in a few places, especially in the first sentence :)

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I’d express things as follows: Set $U:=S^ 2\backslash \{ * \}$. The function $d$, being bounded on $S^2\times S^ 2$, is also bounded on $U\times U$. – Pierre-Yves Gaillard Aug 3 '11 at 9:48

Finiteness. If there were a metric on $S^2$ such that $S^2 \setminus \{ * \}$ and $\mathbb{R}^2$ are isometric, the distance between any point in $S^2 \setminus \{ * \}$ and $*$ would be infinite, by continuity of the metric. But the distance between any two points is always finite in a metric space.
Compactness. A subset of a complete metric space is totally bounded if and only if its closure is compact. $S^2$ is compact, so $S^2 \setminus \{ * \}$ is totally bounded; but $\mathbb{R}^2$ is not totally bounded under the usual metric.
It might also be possible to make an argument based on Gaussian curvature, but it's not clear to me that an isometry between $\mathbb{R}^2$ and $S^2 \setminus \{ * \}$ necessarily induces a Riemannian structure on all of $S^2$.
I don't see how the completeness argument works. Completeness is a metric property and you can only assume the usual topology on $S^2 \setminus \{\ast\}$, so it seems like you're assuming your conclusion if you use that argument. – JSchlather Aug 3 '11 at 11:31