1
$\begingroup$

Provide either a proof or a counterexample.

I haven't been able to find an example where these two statements are not equivalent, as trying to equate any of A, B or C to the Empty Set doesn't work, and I've also tried setting them to shared variables and separate variables. So I'm inclined to believe that this is true, however, I'm not sure how to definitively prove it. If I say, for example, "Let $x∈A, y∈B, z∈C$," that doesn't cover every possibility, such as $x∈A, x∈B$, etc. If anyone could point out even the first step to creating a proof that definitively covers these options, I'd greatly appreciate it.

$\endgroup$
2
  • $\begingroup$ If $x\in A\times (B\setminus C)$ then there are $a,b$ such that $x=(a,b)$ and $a\in A$, $b\in B\setminus C$ $\endgroup$
    – user160738
    Feb 24, 2017 at 6:50
  • $\begingroup$ Ah, I see, and by the same logic you can say $y = (a,b)$ for the right hand side, regardless of $C$ in both cases. Thank you! If you make this an answer I'll accept it. $\endgroup$
    – Mock
    Feb 24, 2017 at 6:59

1 Answer 1

3
$\begingroup$

If $(x,y)\in A\times(B\setminus C)$ then $x\in A$ and $y\in(B\setminus C)$. Hence, in particular $y\in B$ so $(x,y)\in A\times B$ and $y\not\in A\times C$. Summing-up $(x,y)\in A\times B\setminus A\times C$. We conclude: $A\times(B\setminus C)\subseteq A\times B\setminus A\times C$.

For the opposite inclusion. Assume $(x,y)\in A\times B\setminus A\times C$. Then, of course, $x\in A$ and $y\in B$, but since $(x,y)\not\in A\times C$, then $y\in B\setminus C$. Hence, $(x,y)\in A\times(B\setminus C)$.

Remark that the statement is false if $B=C$ or $B\subset C$. As a counter-example in this case choose $A=\{x\}$ and $B=C=\{y\}$.

Edit: After discussion with @J. Davis Taylor, I get convinced that in fact, there is no counter-example. Plz, look at the discussion below.

$\endgroup$
7
  • $\begingroup$ In your "counterexample", both sets are the empty set, thus equal. $\endgroup$ Feb 24, 2017 at 7:25
  • $\begingroup$ @J.DavidTaylor I consider $A\times\emptyset$ not empty and $\{(x,y)\}\setminus\{(x,y)\}$ really empty. $\endgroup$
    – Maczinga
    Feb 24, 2017 at 7:29
  • 1
    $\begingroup$ Would you name an element of the set $A\times\varnothing$? Where $A=\{x\}$, as in your example. $\endgroup$ Feb 24, 2017 at 7:31
  • $\begingroup$ I would say $(x)$ ie. there is no pair... but maybe it is matter of conventions? $\endgroup$
    – Maczinga
    Feb 24, 2017 at 7:33
  • $\begingroup$ Two constructions of the product I know, Choice functions $\{A,\varnothing\}\rightarrow A\cup\varnothing$ Ordered Pairs, $(a,b)$, where $a\in A,b\in\varnothing$ An alternative Defintion: the product in the category of sets, i.e. set functions $Z\rightarrow A\times\varnothing$ are in natural bijection with pairs of set functions, $Z\rightarrow A$ and $Z\rightarrow\varnothing$. The first construction is empty because the only set function into $\varnothing$ is the empty function, $\varnothing\rightarrow\varnothing$. The second, because $\varnothing$ has no elements, $\endgroup$ Feb 24, 2017 at 7:40

You must log in to answer this question.

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