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This is surely a trivial question but I want to be sure I understand correctly what happens.

Given a set $A = \{1\}$, what is $\min A$ and $\max A$?

Is it $\min A = 1$ and $\max A = 1$?

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    $\begingroup$ Definitions are your friends. How do you define $\min A$ and $\max A$? You need to have an order relation $\le$ specified for $A$ for this to make sense. It seems likely you are thinking of $A$ as a subset of the real numbers. $\endgroup$
    – hardmath
    Jun 19, 2016 at 9:01
  • $\begingroup$ $\min(A)$ and $\max(A)$ are values belonging to the set. So... $\endgroup$
    – user65203
    Jun 19, 2016 at 9:30

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Yes, by definition $\min(A)$ gives you the smallest element in the set. Since it is $1$ in this case, you will get $1$. Likewise for $\max(A)$. Of course, $\min(A)$ and $\max(A)$ will not generally be equal but they are equal in this special case.

In a more general scenario a set $B$ with only one element need not satisfy $\min(B)=\max(B)$ since I could, for example, make $B$ a subset of some field without an order relation. In such a set, on which no order relation is defined, you usually also don't know what a definition of $\min(B)$ and $\max(B)$ would be. It might still be possible by utilizing, for example, a norm. I can make a set with one vector as its element and while vectors have no order relation, I could define the maximum and minimum with relation to a vector's norm. In that case the maximum and minimum would coincide again.

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Yes.

However, I'm sure someone can create some pathological cases where this isn't true (like when the set is unordered).

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    $\begingroup$ No, if $A=\{a\}$ is a one-element subset of an ordered set, then it’s always true that $\min A=\max A=a$. $\endgroup$
    – Brian M. Scott
    Jun 19, 2016 at 9:10
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    $\begingroup$ Brian is right, there are counter examples of some non-ordered sets but in subsets of ordered sets like $\mathbb{R}$ or $\mathbb{Z}$ with only one element, you cannot construct a counterexample. $\endgroup$
    – MM8
    Jun 19, 2016 at 9:16
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    $\begingroup$ I suppose you could just say that $a$ isn't an element of an ordered set, but then you can define a trivial order on ${a}$ and you're fine. $\endgroup$
    – Neil
    Jun 19, 2016 at 9:16
  • $\begingroup$ Yea, when I said "pathological," I meant the unordered case as well. $\endgroup$
    – Mustafa Eisa
    Jun 19, 2016 at 9:20

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