Connected/Disconnected Criterion for Non-Homeo. Topo. Spaces.

# Connected/Disconnected Criterion for Non-Homeomorphic Topological Spaces

Consider two topological spaces $X$ and $Y$. Suppose that if we remove any point $x \in X$ then the resulting space $X \setminus \{ x \}$ with the subspace topology is still connected. Furthermore, suppose that there exists a specific point $y \in Y$ such that the resulting space $Y \setminus \{ y \}$ with the subspace topology is disconnected. Then it seems reasonable to assume that $X$ cannot be homeomorphic to $Y$. Fortunately this is true and we prove it in the following theorem.

Theorem 1: Let $X$ and $Y$ be topological spaces. If for every $x \in X$ we have that $X \setminus \{ x \}$ is connected (with the subspace topology), and if there exists a $y \in Y$ such that $Y \setminus \{ y \}$ is disconnected (with the subspace topology) then $X$ is not homeomorphic to $Y$. |

**Proof:**Suppose otherwise, i.e., suppose that $X$ is homeomorphic to $Y$. Then there exists a homeomorphism $f : X \to Y$.

- Let $y \in Y$ be such that $Y \setminus \{ y \}$ is disconnected. Let $\{ A, B \}$ be a separation of $Y \setminus \{ y \}$. Then $A, B \subset Y \setminus \{ y \}$, $A, B \neq \emptyset$, $A \cap B = \emptyset$, and:

\begin{align} \quad Y \setminus \{ y \} = A \cup B \end{align}

- Since $f$ is continuous we have that $f^{-1}(A)$ and $f^{-1}(B)$ are open in $X \setminus \{ f^{-1}(y) \}$. We claim that $\{ f^{-1}(A), f^{-1}(B) \}$ is a separation of $X \setminus \{ f^{-1}(y) \}$.

- We have already established that $f^{-1}(A)$ and $f^{-1}(B)$ are open in $X \setminus \{ f^{-1}(y) \}$. Furthermore, $f^{-1}(A) \neq \emptyset$ and $f^{-1}(B) \neq \emptyset$. We also have that $f^{-1}(A) \cap f^{-1}(B) = \emptyset$ since $A \cap B = \emptyset$. Lastly, we note that since $A \cup B = Y \setminus \{ y \}$ that then:

\begin{align} \quad f^{-1}(A) \cup f^{-1}(B) &= f^{-1}(Y) \setminus \{ f^{-1}(y) \} \\ \quad f^{-1}(A) \cup f^{-1}(B) &= X \setminus \{ f^{-1}(y) \} \end{align}

- Therefore $\{ f^{-1}(A), f^{-1}(B) \}$ is a separation of $X \setminus \{ f^{-1}(y) \}$. But since $f$ is a bijection, $f^{-1}(y) = x$ for some $x \in X$, so $X \setminus \{ x \}$ is disconnected. But this is a contradiction since for all $x \in X$ we have that $X \setminus \{ x \}$ is connected.

- Hence the assumption that $X$ and $Y$ were homeomorphic was false. So if $X$ is such that for all $x \in X$, $X \setminus \{ x \}$ is connected and if there exists a $y \in Y$ such that $Y \setminus \{ y \}$ is disconnected then $X$ is not homeomorphic to $Y$.