Topological Sums of Topological Spaces

We will now look at a rather nice topological space that we can create from a collection of other topological spaces. We define this topological space below.

Definition: Let $\{ (X_i, \tau_i) : i \in I \}$ be any collection of pairwise disjoint topological spaces. The Topological Sum denoted $\displaystyle{\bigoplus_{i \in I} X_i}$ is the topological space with the set $\displaystyle{\bigcup_{i \in I} X_i}$ with the topology $\tau$ given by the basis $\displaystyle{\mathcal B = \left \{ U \subseteq \bigcup_{i \in I} X_i : U \in \tau_i \: \mathrm{for \: some \: } i \in I \right \}}$.

Theorem 1: Let $\{ (X_i, \tau_i) : i \in I \}$ be a collection of pairwise disjoint topological spaces. Then a set $V$ is open in the topological sum $\displaystyle{\bigoplus_{i \in I} X_i}$ if and only if $V \cap X_i$ is open in $\tau_i$ for all $i \in I$.

• Proof: $\Rightarrow$ Suppose that $V$ is open in the topological sum $\displaystyle{\bigoplus_{i \in I} X_i}$. From the definition above, we have that the basis for the topology on $\displaystyle{\bigoplus_{i \in I} X_i}$ is given by:

(1)

\begin{align} \quad \mathcal B = \left \{ U \subseteq \bigcup_{i \in I} X_i : U \in \tau_i \: \mathrm{for \: some \:} i \in I \right \} \end{align}

• So then for $\mathcal B^* \subseteq \mathcal B$ we have that:

(2)

\begin{align} \quad V = \bigcup_{B \in \mathcal B^*} B \end{align}

• Each $B \in \mathcal B^*$ is such that $B \in \tau_i$ for some $i \in I$. So $V \cap X_i$ is either an arbitrary union of open sets from $X_i$ which is open in $X_i$, or $V \cap X_i$ is the empty set (which is open in all of the topological spaces), so $V \cap X_i$ is open for all $i \in I$.

• $\Leftarrow$ Suppose that $V \cap X_i$ is open for each $i \in I$. Then since each of the topological spaces are disjoint we see that:

(3)

\begin{align} \quad V = V \cap \bigcup_{i \in I} X_i = \bigcup_{i \in I} \underbrace{[V \cap X_i]}_{\mathrm{open \: in \: X_i}} \end{align}

• Each of the $V \cap X_i$ are contained in $\mathcal B$, so since $V$ is an arbitrary union of basis elements (which are open) we have that $V$ is open in the topological sum $\displaystyle{\bigoplus_{i \in I} X_i}$. $\blacksquare$