R2 is NOT Homeomorphic to R3

# R2 is NOT Homeomorphic to R3

Recall that $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^2$. If otherwise, then $\mathbb{R}$ minus a point would be homeomorphic to $\mathbb{R}^2$ minus a point. But $\mathbb{R}$ minus a point is disconnected and $\mathbb{R}^2$ minus a point is connected, and a disconnected space cannot be homeomorphic to a connected space.

With the information we have gained about fundamental groups we can now prove that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$.

Theorem 1: $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$. |

**Proof:**Suppose otherwise, i.e., suppose $\mathbb{R}^2$ is homeomorphic to $\mathbb{R}^3$. Then $\mathbb{R}^2 \setminus \{ (x, y) \}$ is homeomorphic to $\mathbb{R}^3 \setminus \{ (x, y, z) \}$. But homeomorphic spaces have the same fundamental groups, and so:

\begin{align} \quad \pi_1(\mathbb{R}^2 \setminus \{ (x, y) \}, a) \cong \pi_1(\mathbb{R}^3 \setminus \{ (x, y, z) \}) \end{align}

- But $\pi_1(\mathbb{R}^2 \setminus \{ (x, y) \}, a) \cong \mathbb{Z}$ and $\pi_1(\mathbb{R}^3 \setminus \{ (x, y, z) \})$ is the trivial group. So we have come to a contradiction.

- Hence the assumption that $\mathbb{R}^2$ and $\mathbb{R}^3$ are homeomorphic is false. $\blacksquare$