so I need to show that:

$$\bigcap_{i=1}^n(A \times B_i) = A \times \left(\bigcap_{i=1}^nB _i \right)$$ for any $n \geq 1$ and for any sets $A$ and $B_i$

I tried expanding the left side: $$(A \times B_1) \cap (A \times B_2) \cap \cdots \cap (A \times B_n)$$

and then expanding the right side:

$$A \times (B_1 \cap B_2 \cap \cdots \cap B_n)$$ but as far as I know there aren't any rules regarding commutativity and distribution with respect to the Cartesian product.


  • 3
    $\begingroup$ When stuck, just try using the definitions. How do we show that two sets are equal? When is an element in the set $\bigcap_{i = 1}^n (A \times B_i)$? Can you show that it is also in $A \times \bigcap_{i = 1}^n B_i$? What about the converse? $\endgroup$ Apr 15 '21 at 14:19

To prove two sets are equal, there are two things you have to show:

$(1)$ that all members of the first set are in the second set and

$(2)$ that all members of the second set are in the first set

Let's prove $(1)$ first. Take an arbitrary $(a, b) \in \bigcap_{i=1}^n(A \times B_i)$. Clearly because $(a, b) \in \bigcap_{i=1}^n(A \times B_i)$, we must have $(a,b) \in A \times B_i$ for all $i \in [0,n]$. This means $a \in A$ and $b \in B_i$ for all $i \in [1,n]$, and therefore $b \in \bigcap_{i=1}^n B_i$. So, $(a,b)$ must be in $A \times \left(\bigcap_{i=1}^n(B_i)\right)$.

Now try to see if you can prove $(2)$ similarly.

  • $\begingroup$ Thanks for your help! I am not sure if I can prove $(2)$ as well as you can. I'm not particularly good with proofs. Here's what I did: I took an arbitrary $(a,b)$ $\in$ $A \times (\bigcap_{i=1}^n(B_i))$. From there we can conclude that $b \in \bigcap_{i=1}^n B_i$, and subsequently $b \in B_i$ for all $i \in [0,n]$. I don't know how to continue. $\endgroup$
    – john doe
    Apr 15 '21 at 14:46
  • 1
    $\begingroup$ If $a$ is in $A$ and $b$ is in $B_i$, is $(a, b)$ in $A \times B_i$? If so, then for what $i$ is $(a,b)$ in $A \times B_i$? $\endgroup$ Apr 15 '21 at 14:48
  • 1
    $\begingroup$ (also that is one minor detail I left out: $(a, b) \in \bigcap_{i=1}^n (A \times B_i)$ implies $a \in A$) $\endgroup$ Apr 15 '21 at 14:49
  • $\begingroup$ That is also a part of your $(1)$ proof I didn't understand. Yes, $(a,b) \in A \times B$, but for what values of $i$? I'm not sure does it start from $1$ or $0$! Thank you! $\endgroup$
    – john doe
    Apr 15 '21 at 14:54
  • $\begingroup$ $i$ starts from $1$, but that's not really important. If we have that $b$ is in all of the $B_i$ for $i \in [1, n]$, and for each $i$ we know that $a \in A$ and $b \in B_i$ yields $(a, b) \in A \times B_i$, then can we conclude that $(a, b)$ is in all of the $A \times B_i$ for $i \in [1,n]$? $\endgroup$ Apr 15 '21 at 14:57

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