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. $\blacksquare$
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