4

The following apparently defines a function, but there is something wrong with it which I am struggling to find:

$$m:\mathcal P(\{1,2,3,\dots,100\})\to\{1,2,3,\dots,100\}$$ $$m(A) =\text{the smallest element of }A$$

Is it the fact that when the domain is the empty set, the codomain would have to be 0, which is not actually inlcuded in the codomain? Am I right to assume that the smallest element of the empty set is 0?

Parcly Taxel
  • 103,344
kafo
  • 93
  • 1
    The empty set has no elements, so it isn't true that the smallest element is $0$, or anything else. The problem is that the function is not defined at $\emptyset$. (You are misusing the terms "domain" and "codomain." These refer to the entire set, not individual elements.) – saulspatz Oct 26 '20 at 12:41
  • Related. The infimum of an empty set is usually $+\infty$. That said, minimums and infimums, although related, are not identical terms. Minimums must be elements of the set in question. – JMoravitz Oct 26 '20 at 12:41
  • @saulspatz So how would I define it at ∅? Would i have to inlcude ∅ in the domain? – kafo Oct 26 '20 at 12:45
  • 2
    @kafo $\emptyset$ is in the domain. The empty set is an element of $\mathcal{P}({1,2,\dots,100})$. The minimum of $\emptyset$ doesn't exist and is undefined. Related, the infimum of $\emptyset$, namely $\infty$, is not an element of the codomain. Those are the problem. You could get around this by rewording $m$ to be infimum instead of minimum and include $+\infty$ in the codomain, you could reword $m$ to say explicitly that $m$ acts on $\emptyset$ in a specific way... for instance that $m(\emptyset)$ is defined to equal $42$... or you could remove $\emptyset$ from the domain. – JMoravitz Oct 26 '20 at 12:48
  • 1
    While you could if you insist include $0$ in the codomain and explicitly define $m$ to be the minimum of a set for non-empty sets in the domain and $0$ for the empty set, this is ill-advised as it loses several of the nice properties of $m$ unless you bend over backwards trying to fix all the problems. For instance, if it weren't for the empty set we would have that $m(A\cup B) = m({m(A),m(B)})$. Similarly, ignoring the empty set we have $m(A\cup B)\leq m(A)$. Letting $m(\emptyset)$ be smaller than the other outputs rather than larger breaks this. – JMoravitz Oct 26 '20 at 12:55

1 Answers1

4

The function is indeed ill-defined, for "the smallest element of $A$, $m(A)$" has no meaning if $A=\varnothing$. It is nonsense to define $m(\varnothing)=0$ without adding $0$ to the codomain set first.

Parcly Taxel
  • 103,344