# Product space of open set

Definition open set Let $(X,Y)$ be topological space. A subset $U \subset X$ is called an open set of $X$ if $U \in Y$

Product topology Let $(X,Y)$ be topological space. The product topology on $X \times Y$ is the topology generated by the basis $\mathcal B$.

$\mathcal B=$ the collection of all sets $U \times V$where $U$ is open in $X$ and $V$ is open in $Y$

Let $X,Y$ be two topological spaces

If $C \subset X$ and $D \subset Y$ are open in $X$ and $Y$, respectively, is $C \times D$ an open set of the product space $X \times Y$ ?

• The answer is yes, but this relies on information you haven't provided. You need to look in your book for the definition of the product topology on $X \times Y$. – user49640 Apr 19 '17 at 3:29
• @user49640 i have improve my question – user273952 Apr 19 '17 at 4:30
• Okay, what is the topology "generated" by a basis $B$. That will give you your answer. – user49640 Apr 19 '17 at 4:32
• Let $\mathcal B$ be a basis of $X$. Let $Y$= the collection of all $U \subset X$ such that for each $x \in U$, there exist an element $B_x \in \mathcal B$ with $x \in B_x$ and $B_x \subset U$. Then $Y$ is the topology on $X$, it is called the topology generated by the basis $\mathcal B$. How to apply this to my question ? – user273952 Apr 19 '17 at 4:38
• The set $C \times D$ belongs to the basis $B$ that generates the product topology, by definition. You just need to check that it belongs to that topology. Can you prove that every element of a basis belongs to the topology generated by that basis? You might have a theorem somewhere that makes this fact even more obvious, so that you don't need to prove it from the definition of a basis. – user49640 Apr 19 '17 at 4:41

Yes, given a collection of topological space $\{X_i\}$ the product topology on $\prod_{i}{X_i}$ is defined by arbitrary unions and finite intersections of sets of the form $\prod_{i}{U_i}$ where $U_i \subset X_i$ and is open $\forall i$.