The Subspace and Product Topologies
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Table of Contents
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The Subspace and Product Topologies
The Subspace Topology
| Proposition 1: Let $(X, \tau)$ be a topological space and let $Y \subseteq X$ be nonempty. Let $\tau_Y = \{ Y \cap U : U \in \tau \}$. Then $(Y, \tau_Y)$ is a topological space. |
- Proof: There are three things to prove.
- First, since $X \in \tau$ we have that $Y \cap X = Y \in \tau_Y$. Similarly, since $\emptyset \in \tau$ we have that $\emptyset = Y \cap \emptyset \in \tau_Y$. So $\tau_Y$ contains both $Y$ and $\emptyset$.
- Let $\{ Y \cap U_i : i \in I \}$ be an arbitrary collection of sets in $\tau_Y$. Since $\tau$ is a topology on $X$ we have that $\displaystyle{\bigcup_{i \in I} U_i \in \tau}$. So:
\begin{align} \quad \bigcup_{i \in I} [Y \cap U_i] = Y \cap \bigcup_{i \in I} U_i \in \tau_Y \end{align}
- Lastly, let $\{ Y \cap U_i : 1 \leq i \leq n \}$ be a finite collection of sets in $\tau_Y$. Since $\tau$ is a topology on $X$ we have that $\displaystyle{\bigcap_{i=1}^{n} U_i \in \tau}$. So:
\begin{align} \quad \bigcap_{i=1}^{n} [Y \cap U_i] = Y \cap \bigcap_{i=1}^{n} U_i \in \tau_Y \end{align}
- Thus $(Y, \tau_Y)$ is a topological space. $\blacksquare$
| Definition: Let $(X, \tau)$ be a topological space and let $Y \subseteq X$ be nonempty. The Topological Subspace $Y$ of $X$ is defined to be the topological space $(Y, \tau_Y)$ where $\tau_Y = \{ Y \cap U : U \in \tau \}$. |
Proposition 1 above tells us that the topological subspace $(Y, \tau_Y)$ is indeed itself a topological space.
The Product Topology
| Definition: Let $(X_1, \tau_1)$ and $(X_2, \tau_2)$ be two topological spaces. The Product Space is defined to be the Cartesian product $X_1 \times X_2$ with the topology generated by sets of the form $U_1 \times U_2$ where $U_1 \in \tau_1$ and $U_2 \in \tau_2$. |