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!

  • $\begingroup$ $f=A\times B$ 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

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 $A$.

  • $\begingroup$ 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

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