3
$\begingroup$

The problem states:

If there is function $f:A\to B$ and $$\left(f=A \times B \right) \iff \left( A= \varnothing \text{ or } |B| = 1\right)$$

(where $A \times B$ is the cartesian product and $|B|$ is the cardinality of $B$)

So I must demonstrate that equivalence. I've tried to slove it, but i can't wrap my head around what $f=A\times B$ means. Please help me!

$\endgroup$
  • $\begingroup$ $f=A\times B$ means that $f$ is the set of all pairs $(a,b)$ where $a\in A$ and $b\in B$. $\endgroup$ – Xoff Nov 17 '13 at 15:48
2
$\begingroup$

Recall that $A\times B$ is the set of all possible ordered pairs from $A$ and $B$, a function is just a subset of these pairs. For example if $A=\{x,y\}$ then a function whose domain is $A$ will only have two elements $\langle x,f(x)\rangle$ and $\langle y,f(y)\rangle$.

You are asked to show that if $f$ has all the possible ordered pairs, then either $A$ is the empty set or $B$ is a singleton. To do so, first you have the show that if $A$ is the empty set, or $B=\{b\}$, then $A\times B$ is a function; and in the other direction show either that from the assumption $f=A\times B$ you can conclude at least one of the things hold ($A=\varnothing$ or $B=\{b\}$), or work towards contradiction or contrapositive and assume that $A$ is not empty, $B$ is either empty, or have two elements, and conclude that $A\times B$ cannot be a function whose domain is $A$.

$\endgroup$
  • $\begingroup$ Thanks amWhy. I'm just so hungry that my brain works at half the power. $\endgroup$ – Asaf Karagila Nov 17 '13 at 15:55
  • $\begingroup$ You know I only posted because you're a good sport, and I'm half-teasing you :P $\endgroup$ – Namaste Nov 17 '13 at 15:56
  • $\begingroup$ I know... food is going to be ready in 10 minutes, too. $\endgroup$ – Asaf Karagila Nov 17 '13 at 15:57
  • $\begingroup$ Now you're making me hungry! ;-) $\endgroup$ – Namaste Nov 17 '13 at 15:58
  • $\begingroup$ And it's just my laziest pasta possible (olive oil, salt and pepper). You should taste my food when I put some effort into it. $\endgroup$ – Asaf Karagila Nov 17 '13 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.