0
$\begingroup$

Prove: Let $B_1, B_2,\ldots, B_n$ be the bases for topology spaces $(X_1,T_1), (X_2,T_2),\ldots,(X_n,T_n)$, respectively. Then the family

$$\{U_1 \times U_2 \times\ldots\times U_n: U_i\in B_i, i = 1, 2,\ldots,n\}$$

is a basis for the product topology on $X_1 \times X_2 \times\ldots\times X_n$.

Definition. The product topology on $X_1\times X_2\times\ldots\times X_n$ is the topology having the family $$\{O_1\times O_2\times\ldots\times O_n:O_i\in T_i, i=1,\ldots,n\}$$ as a basis.

Proof:

First, I let $A$ be an open set in $X_1 \times X_2 \times\ldots\times X_n$. Suppose a point ($\langle x_1, x_2,\ldots, x_n\rangle$) is in $U$. I know that I need to show that there exist $U_1 \in B_1, U_2\in B_2,\ldots, U_n \in B_n$ such that the point $\langle x_1, x_2,\ldots, x_n\rangle\in U_1 \times U_2 \times\ldots\times U_n$ is the subset of $U$. But I'm not sure how to do it. Could you please help me? Thanks

$\endgroup$
5
  • $\begingroup$ What exactly is your definition of the product topology? It can be defined in several (different but equivalent) ways, and what you have to prove depends on which definition you’re using. $\endgroup$ Commented Dec 1, 2013 at 2:15
  • $\begingroup$ I'm using Topology without tears book of Sidney. The definition is a little bit long to type. $\endgroup$
    – user110969
    Commented Dec 1, 2013 at 2:19
  • $\begingroup$ do you want me to post the definition or can you find it? $\endgroup$
    – user110969
    Commented Dec 1, 2013 at 2:22
  • $\begingroup$ Sidney Morris’s book that’s available online? I’ll add it. $\endgroup$ Commented Dec 1, 2013 at 2:29
  • $\begingroup$ yes, it's online $\endgroup$
    – user110969
    Commented Dec 1, 2013 at 2:30

1 Answer 1

1
$\begingroup$

HINT: Suppose that $p=\langle x_1,\ldots,x_n\rangle\in U$, where $U$ is an open set in $X=X_1\times\ldots\times X_n$. Then by the definition of the product topology there are open sets $V_k\in T_k$ for $k=1,\ldots,n$ such that $p\in V_1\times V_2\times\ldots\times V_n\subseteq U$. This implies that $x_k\in V_k$ for $k=1,\ldots,n$. Now use the fact that $B_k$ is a base for $T_k$ for $k=1,\ldots,n$ to conclude that for each $k=1,\ldots,n$ there is a $U_k\in B_k$ such that ... ?

$\endgroup$
4
  • $\begingroup$ such that p ∈ Uk ⊆ Vk. so p ∈ U1×U2×…×Un ⊆ V1×V2×…×Vn⊆U $\endgroup$
    – user110969
    Commented Dec 1, 2013 at 2:51
  • $\begingroup$ @user110969: You’ve got it. As you can see, the argument is really just a matter of using the definition of base for a topology in two different places. $\endgroup$ Commented Dec 1, 2013 at 2:54
  • $\begingroup$ so I just need to conclude: Therefore, {U1×U2×…×Un:Ui∈Bi,i=1,2,…,n} is a basis for the product topology on X1×X2×…×Xn $\endgroup$
    – user110969
    Commented Dec 1, 2013 at 2:55
  • $\begingroup$ @user110969: That’s right: you’ve shown that whenever $p\in U$, where $U$ is an open set in the product $X$, there is a set of the form $U_1\times\ldots\times U_n$, with $U_k\in B_k$ for $k=1,\ldots,n$, such that $$p\in U_1\times\ldots\times U_n\subseteq U\;,$$ which is exactly what’s needed to show that the family of such sets is a base for the product topology. $\endgroup$ Commented Dec 1, 2013 at 2:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .