# Defining “minimum” using logic and set-theoretic operations.

Express the notion of a minimum of a set of number (where numbers are defined via sets). That is, define a relation Min(S,x) using logic and set-theoretic operations such that it is true whenever x is the minimum element in S.

• I understand the definition of numbers in terms of sets. I did that in part a of this same question.
• I understand what a function or relation is

But I have no idea what this is asking, nor how to answer it.

Edit: My definition of a number is as follows:

0 = {}
$n+1 = n\cup \{n\}$

So :

• 1 = {{}}
• 2 = {{}, {{}}}
• 3 = {{}, {{}}, {{}, {{}}}}
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Can you include your answer for part (a) of this question? – Benjamin Dickman Nov 4 '12 at 23:00
Try breaking it down into pieces: (1) How would you write '$x$ is a smaller number than $y$' as a logical expression involving sets? (2) How would you write '$x$ is the smallest number in the set $S$' as a logical expression involving the 'less-than' relation? Once you have those, all you need to do is piece them together. – Steven Stadnicki Nov 4 '12 at 23:01
HINT: Have you noticed that by this definition $3=\{0,1,2\}$? – Brian M. Scott Nov 4 '12 at 23:09
Is this an answer? $\forall y\in S(x\in y)$ – agent154 Nov 4 '12 at 23:46
It’s almost part of one: it says that $x$ is less than each number in $S$. Unfortunately, this implies that $x\notin S$, which is not what you want. $\operatorname{Min}(x,S)\leftrightarrow x\in S\land \dots~$, where the missing part has to say in effect that for all $y\in S$, $x\le y$. – Brian M. Scott Nov 4 '12 at 23:50

Hint:

You can notice either of the following observations:

1. $n<m\iff n\in m$; and
2. $n\leq m\iff n\subseteq m$.

Therefore if $S$ is a set of natural numbers, its minimum is the element which is either the member of everyone else; or a subset of everyone (including itself, of course).

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I'd like to thank Brian M. Scott above in the comments for helping me understand the problem and coax me to an answer.

As I understand it, the logic is as follows:

For a number to be the minimum number in a set of numbers, it must

1. Be a member of the set, and
2. Be a member of every other member of that set

For the first part, we have $$Min(S,x)\iff x\in S$$

For the second part, we have

$$Min(S,x)\iff \forall y\in S\{x\subseteq y\}$$

I've chosen $x\subseteq y$ instead of $x\in y\vee x=y$ because I like having less text, but both are equivalent. The reason why $x\subseteq y$ works is because x can legitimately be a subset of itself, but can only also be a subset of any natural number greater than it. So it'll return false only if there is a natural number less than x in the set S.

Putting 1 and 2 together now, we have:

$$Min(S,x)\iff x\in S\wedge \forall y\in S\{x\subseteq y\}$$

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