Product topology

**horowitz** · February 28th 2009, 09:24 PM

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.

**aliceinwonderland** · March 1st 2009, 05:04 AM

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

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

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

The converse is similar to the above.

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