I think I may have this but I think it's best I check as I don't have the solution.

Let $A$ be a set.

$A^0 = \dfrac{A^1}{A^1}$

$A$ / $B$ is everything in $A$ that isn't in $B$

Therefore $A$ / $A$ is everything in $A$ that isn't in $A$ which is an empty set.

Therefore $A^0 = A$ / $A = \{\emptyset\}$

Is this correct?

Thank you.

  • $\begingroup$ What is your definition of an exponent of a set? $\endgroup$
    – Arthur
    Jul 31 '14 at 8:09
  • 3
    $\begingroup$ No. $A^0 \ne A \setminus A$ (And $A \setminus A =\emptyset \ne \{\emptyset\}$. $\endgroup$
    – martini
    Jul 31 '14 at 8:09
  • $\begingroup$ For 'everything in $A$ that isn't in $B$' notations $A\backslash B$ and $A-B$ are practicized. Notation $A/B$ is used for something else: quotients e.g. in the theory of groups. $\endgroup$
    – drhab
    Jul 31 '14 at 8:31
  • $\begingroup$ You are mixing up exponent rules for nubmers, the division sign, and the setminus sign. With this understanding, would you have $A^1$ equal $A^2\setminus A^1$? $\endgroup$ Jul 31 '14 at 8:58

You can think of $A^n$ as the set of functions from a set of $n$ members to $A$.

Thus, $A^0$ is the set of functions from a set of $0$ members, i.e., the empty set, to $A$. Since there is only one such function – the empty function – $A^0=\{\emptyset\}$.

(Here, we use the common convention of identifying a function with its graph; hence the empty function equals the empty set.)

(To make things clearer, the $n$-member set in the first sentence could be $\{1,2,\ldots,n\}$. Or, if you are into ordinals and basic set theory, $\{0,1,\ldots,n-1\}$.)


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