Disc. Topological Spaces Homeo. to the Topo. Sum of their Sep.

# Disconnected Topological Spaces Homeomorphic to the Topological Sum of their Separation

Recall from the Connected and Disconnected Topological Spaces page that a topological space $X$ is said to be disconnected if there exists open sets $A, B \subset X$ such that $A, B \neq \emptyset$, $A \cap B = \emptyset$ and:

(1)\begin{align} \quad X = A \cup B \end{align}

Also recall from the Topological Sums of Topological Spaces that if $A$ and $B$ are disjoint topological spaces then the topological sum $A \oplus B$ is the set $A \cup B$ whose topology is given by the basis:

(2)\begin{align} \quad \mathcal B = \{ U \subseteq A \cup B : U \: \mathrm{is \: open \: in \: } A \: \mathrm{or} \: B \} \end{align}

We will now see that if the topological space $X$ is disconnected and has a separation $\{ A, B \}$ then $X$ is homeomorphic to the topological sum $A \oplus B$.

Theorem 1:A topological space $X$ is disconnected and has a separation $\{ A, B \}$ if and only if $X$ is homeomorphic to $A \oplus B$. |

**Proof:**$\Rightarrow$ Suppose that $X$ is a disconnected topological space and that $\{ A, B \}$ is a separation of $X$. To show that $X$ is homeomorphic to $A \oplus B$ we show that the topologies on the two spaces are the same.

- Let $U$ be an open set in $X$. There are three cases to consider.

- If $U \subset A$, then $U \cap A = U$ is open in $A$ and so $U$ is open in $A \oplus B$.

- If $U \subset B$, then $U \cap B = U$ is open in $A \oplus B$.

- Lastly, if $U \not \subset A$ and $U \not \subset B$ then $U \cap A \neq \emptyset$ and $U \cap B \neq \emptyset$. But $A$ and $B$ are open sets in $X$. In particular, $U \cap A$ is open in $A$ and $U \cap B$ is open in $B$. Therefore $U = (U \cap A) \cup (U \cap B)$ and so $U$ is open in $A \oplus B$.

- Now let $U$ be an open set in $A \oplus B$. Then $U$ is a union of open sets from $A$ and/or from $B$. So $U \cap A$ is open in $X$ and $U \cap B$ is open in $X$. Hence $U = (U \cap A) \cup (U \cap B)$ is open in $X$.

- Since the open sets in $X$ are the same as the open sets in $A \oplus B$, we must have that $X$ is homeomorphic to $A \oplus B$.

- $\Leftarrow$ Suppose that $X$ and $A \oplus B$ are homeomorphic. Then there exists a homeomorphism $f : X \to A \oplus B$, i.e., $f$ is bijective, continuous, and open.

- Since $A$ and $B$ are open sets in $A \oplus B$ and since $f$ is continuous we have that $f^{-1}(A)$ and $f^{-1}(B)$ are open sets in $X$. We will show that $\{ f^{-1}(A), f^{-1}(B) \}$ are a separation of $X$.

- We have already established that these two sets are open in $X$. Furthermore, these two sets are nonempty since both $A$ and $B$ are nonempty as topological spaces. We also see that $f^{-1}(A) \cap f^{-1}(B) = \emptyset$. If not, say suppose that $x \in f^{-1}(A) \cap f^{-1}(B)$ then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$ so $f(x) \in A$ and $f(x) \in B$. So $f(x) \in A \cap B$, but $A$ and $B$ are disjoint as a topological sum $A \oplus B$ - a contradiction.

- Lastly we see that $X = f^{-1}(A) \cup f^{-1}(B)$ since $f(X) = A \cup B$ as $f$ is a bijection (and is hence surjective, so the range of $f$, $f(X)$ is equal to the codomain $A \cup B = A \oplus B$.

- So $\{ f^{-1}(A), f^{-1}(B) \}$ is a separation of $X$ which shows that $X$ is disconnected. $\blacksquare$