union of cartesian products problem! $$\bigcup_{i \in I}A _{i}\times \bigcup_{j \in J}B _{j} = \bigcup_{(i\times j) \in I\times J}A_{i} \times B_{j}$$
$$\bigcup_{k\in N} \mathbb{N}\times\{k\} = \mathbb{N}\times\mathbb{N}$$
i'm not yet skilled at proving something like this. how can I prove this??
 A: Prove it by making use of: $$A=B\iff\forall x\left[x\in A\iff x\in B\right]$$
You have:
$$\left(a,b\right)\in\bigcup_{i}A_{i}\times\bigcup_{j}B_{j}\iff a\in\bigcup_{i}A_{i}\wedge b\in\bigcup_{j}B_{j}\iff$$$$\left(a,b\right)\in A_{i}\times B_{j}\text{ for some pair }\left(i,j\right)\in I\times J\iff\left(a,b\right)\in\bigcup_{\left(i,j\right)\in I\times J}(A_{i}\times B_{j})$$
and:
$$\left(r,s\right)\in\bigcup_{k\in\mathbb{N}}(\mathbb{N}\times\left\{ k\right\}) \iff r\in\mathbb{N}\wedge s=k\text{ for some }k\in\mathbb{N}\iff$$$$ r\in\mathbb{N}\wedge s\in\mathbb{N}\iff\left(r,s\right)\in\mathbb{N}\times\mathbb{N}$$
A: Forward inequality. This should get you started.
Let ${x} \in \cup_{i \in I} A_i$
Then, $\{x\} \times \cup_{j \in J} B_j \subset \cup_{i \in I} Ai \times \cup_{j \in J} B_j $
Similarly, $\{y\} \in \cup_{j \in J} B_j \Rightarrow \cup_{i \in I} A_i 
\times \{y\} \subset \cup_{i \in I} Ai \times \cup_{j \in J} B_j $
Together, $x \times y \subset \cup_{i \in I} Ai \times \cup_{j \in J} B_j$
A: What does it mean for two sets to be equal? 
Generally, if you want to show that $A=B$, where $A$ and $B$ are sets, try to prove each of the inclusions $A \subset B$ and $B \subset A$ separately. 
To prove that a set $A$ is a subset of another set $B$, it suffices to show that every element of $A$ is an element of $B$. 
Say that you would like to prove that $$\bigcup_{k\in \mathbb{N}} \mathbb{N} \times \{k\}=\mathbb{N} \times \mathbb{N}$$
Then let $x \in \bigcup_{k\in \mathbb{N}} \mathbb{N} \times \{k\}$, implying that $x \in \mathbb{N} \times \{k\}$ some $k \in \mathbb{N}$, implying that $x=(n,k)$ some$n \in \mathbb{N}$ and $k \in \mathbb{N}$, hence $(n,k) \in \mathbb{N} \times \mathbb{N}$
Conversley let $x \in \mathbb{N} \times \mathbb{N}$, then $x=(n,k)$ some $n \in \mathbb{N}$ and $k \in \mathbb{N}$, implying that $x \in \mathbb{N} \times \{k\}$, and hence $x \in \bigcup_{k\in \mathbb{N}} \mathbb{N} \times \{k\}$.
