# Product topology

• Feb 28th 2009, 09:24 PM
horowitz
Product topology
Let X and Y be the topological spaces, ans assume that A⊂X and B⊂Y. Then the topology on AxB as a subspace of the product XxY is the same as the product topology on AxB, where A has the subspace topology inherited from X and B has the subspace topology inherited from Y.
• Mar 1st 2009, 05:04 AM
aliceinwonderland
Quote:

Originally Posted by horowitz
Let X and Y be the topological spaces, ans assume that A⊂X and B⊂Y. Then the topology on AxB as a subspace of the product XxY is the same as the product topology on AxB, where A has the subspace topology inherited from X and B has the subspace topology inherited from Y.

Let $\displaystyle (X,T_1)$ and $\displaystyle (Y, T_2)$ be topological spaces. A basis element of the product topology on $\displaystyle A \times B$ has the form $\displaystyle V_1 \times V_2$, where $\displaystyle V_1, V_2$ be open sets in the subspace topology on $\displaystyle (A,S_1)$, and $\displaystyle (B,S_2)$, respectively.

Since $\displaystyle S_1=\{A \cap U \text{ } | \text{ }U \in T_1\}$ and $\displaystyle S_2=\{B \cap V \text{ } | \text{ } V \in T_2\}$, we have $\displaystyle V_1 = A \cap U_1$ and $\displaystyle V_2 = B \cap U_2$, where $\displaystyle U_1, U_2$ be open sets in $\displaystyle (X, T_1)$ and $\displaystyle (Y, T_2)$.

Since $\displaystyle V_1 \times V_2 = ((A \cap U_1) \times (B \cap U_2))$, we have a basis element of a topology on $\displaystyle A \times B$ as a subspace of the product $\displaystyle X \times Y$, which is $\displaystyle ((A \cap U_1) \times (B \cap U_2))=((A \times B) \cap (U_1 \times U_2))$, where $\displaystyle U_1, U_2$ be open sets in $\displaystyle (X, T_1)$ and $\displaystyle (Y, T_2)$.

The converse is similar to the above.