I 2-sphere with any topology can't be homeomorphic to the plane

[cianfa72]

TL;DR Summary

A 2-sphere as set can't be homeomorphic to the euclidean plane with any topology assigned to it

Suppose there was a bijection $\varphi$ between the 2-sphere $M$ and the euclidean plane $\mathbb R^2$.
Then one could endow $M$ with the initial topology from $\mathbb R^2$ through $\varphi$ turning it into an homeomorphism (this topology on $M$ would be different from the subset topology inherited from sitting in $\mathbb R^3$ with its standard topology).

Since there is no bijection $\varphi: M \to \mathbb R^2$ as sets, then the 2-sphere as set cannot be endowed with any topology such that it would result homeomorphic to the euclidean plane, right ?

 

[Orodruin]

What makes you think there is no bijection $\mathbb R^2 \to \mathbb S^2$? Both sets are $\aleph_1$.

 

[cianfa72]

What makes you think there is no bijection $\mathbb R^2 \to \mathbb S^2$? Both sets are $\aleph_1$.

Ah..and how can one define such a bijection? For example stereographic projection from North pole maps bijectively all points on $\mathbb S^2$ to the plane but the North pole itself.

 

[Orodruin]

Start with the stereographic projection $S_p$. You now have a bijection from the sphere except a single point $p$ to $\mathbb R^2$. Now make room for $p$ in $\mathbb S^2 \setminus p$: Let $s = (x_i)_{i \geq 0}$ be any countably infinite sequence of distinct points in $\mathbb S^2$ such that $x_0 = p$ and define
$B: \mathbb S^2 \to \mathbb S^2 \setminus p$ such that
$$
B(x) = \begin{cases} x_{i+1},& \quad x = x_i \in s\\
x, &\quad x\notin s \end{cases}
$$
which is clearly a bijection. The composition $S_p \circ B$ is now a bijection from $\mathbb S^2$ to $\mathbb R^2$.

 

[cianfa72]

The composition $S_p \circ B$ is now a bijection from $\mathbb S^2$ to $\mathbb R^2$.

Very interesting! So by mean of $S_p \circ B$ one can assign $\mathbb S^2$ the initial topology from $\mathbb R^2$. With this topology $\mathbb S^2$ is homeomorphic to the Euclidean plane.

Therefore the claim that the 2-sphere $\mathbb S ^2$ is not homeomorphic to the Euclidean plane, assumes $\mathbb S^2$ endowed with the subspace topology from sitting in Euclidean $\mathbb R^3$.

 

[Orodruin]

Very interesting! So by mean of $S_p \circ B$ one can assign $\mathbb S^2$ the initial topology from $\mathbb R^2$. With this topology $\mathbb S^2$ is homeomorphic to the Euclidean plane.

Therefore the claim that the 2-sphere $\mathbb S ^2$ is not homeomorphic to the Euclidean plane, assumes $\mathbb S^2$ endowed with the subspace topology from sitting in Euclidean $\mathbb R^3$.

I mean, if you don’t want to endow $\mathbb S^2$ with the subspace topology, one might question the point of letting it be that particular subspace at all …