# Non-Homeomorphic Topological Spaces Classified by Connectivity

Recall from the Connected/Disconnected Criterion for Non-Homeomorphic Topological Spaces page that if $X$ and $Y$ are topological spaces such that for all $x \in X$, $X \setminus \{ x \}$ is connected (with the subspace topology) and there exists a $y \in Y$ such that $Y \setminus \{ y \}$ is disconnected (with the subspace topology) then $X$ and $Y$ cannot be homeomorphic.

This result is very useful for showing that two topological spaces are not homeomorphic.

For example, consider the topological subspace $X = (0, 1)$ of $\mathbb{R}$ with the usual topology inherited from $\mathbb{R}$ and the topological subspace $Y = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \}$ of $\mathbb{R}$ with the usual topology inherited from $\mathbb{R}^2$.

We claim that $X$ and $Y$ are not homeomorphic.

Notice that if we take any point $x \in X$ then the set $X \setminus \{ x \}$ is disconnected and $\{ (0, x), (x, 1) \}$ is a separation of $X$:

However, if we take any point $\mathbf{y} \in Y$, then the set $Y \setminus \{ \mathbf{y} \}$ is still connected. Visually, this is rather obvious:

As a result, $X$ cannot be homeomorphic to $Y$.