Bounded and unbounded sets I am having trouble with the idea that sets can be bounded/unbounded.
I understand how functions and sequences are bounded/unbounded because they are related with numbers. But with sets, they can be anything so like set $A=\{a,b,c,d\}$ but how can this be bounded? There are no numbers in it...
I don't understand.
Could someone give some explicit examples of bounded sets and unbounded sets. Although I think sets like naturals, reals and rationals and complex sets are all unbounded.
 A: Since the comments got a bit sidetracked, let me try to lay it out a little more clearly:


*

*The concepts of "bounded" and "unbounded" are not defined for arbitrary sets. In order to discuss boundedness, you need some additional structure defined on the set.

*The most familiar structure on a set $S$ that defines boundedness is that of an order: a binary relation "$\le$" that satisfies the conditions:


*

*reflexivity: for all $a \in S, a \le a$.

*anti-symmetry: for all $a, b \in S$, if $a \le b$ and $b \le a$, then $a = b$.

*transitivity: for all $a, b, c \in S$, if $a \le b$ and $b \le c$, then $a \le c$.

*totality: for all $a, b \in S$, either $a \le b$ or $b \le a$


*On sets $S$ with an order relation, we define a subset $A$ to be bounded if there exist $a, b \in S$ such that for all $x \in A$, $a \le x \le b$. A set is unbounded if it is not bounded. 

*Note that any subset of a bounded set must also be bounded. For if $A, a, b$ are as in the definition, and $B \subseteq A$, then for all $x \in B$, we have that $x \in A$, and therefore $a \le x \le b$. This also implies that if any set has an unbounded subset, then the set itself is unbounded.

*Clearly $\Bbb R$ has an order. The Archimedean principle states that for any real number $b$, there is a natural number $n$ such that $b < n$. As a consequence, the natural numbers $\Bbb N$ are unbounded. Since $\Bbb N$ is a subset of the integers $\Bbb Z$, the rationals $\Bbb Q$ and the real numbers $\Bbb R$, all three of these sets are also unbounded.

*The order on $\Bbb R$ can be leveraged to define a different structure that also defines a concept of boundedness: a metric - a function from $S \times S$ to the real numbers $\Bbb R$ that satisfies the conditions


*

*positivity: For all $x, y \in S, d(x, y) \ge 0$.

*definiteness: For all $x, y \in S, d(x, y) = 0$ only if $x = y$.

*triangle inequality: For all $x, y, z \in S, d(x, z) \le d(x, y) + d(y, z)$.


*On sets $S$ with a metric, a subset is bounded if there is an $M \in \Bbb R$ such that for all $x, y$ in the subset $d(x, y) \le M$. A subset of $S$ is unbounded if it is not bounded.

*As before, subsets of bounded sets are also bounded, and if any set contains an unbounded subset, then the set itself is unbounded.

*On $\Bbb R$, letting $d(x, y) = | x - y |$ defines a metric. It is not hard to show that a subset of $\Bbb R$ is bounded with respect to this metric if and only if it is bounded with respect to the order. So this sense of boundedness agrees completely with the earlier one. In particular, $\Bbb N, \Bbb Z, \Bbb Q,$ and $\Bbb R$ are all still unbounded.

*On $\Bbb C$, letting $d(w, z) = | w - z | = \sqrt{(w - z)(w - z)^*}$ also defines a metric, extending the metric on $\Bbb R$ to the complex numbers. Since $\Bbb N \subseteq \Bbb C$, and $\Bbb N$ is still unbounded, so is $\Bbb C$.

*Another metric on $\Bbb C$ is $d_1(x, y) = \min\{| x - y|, 1\}$. This metric is topologically equivalent to the earlier one: every limit that converges under $d$ also converges under $d_1$ to the same value, and vice versa. But under this metric $\Bbb C$ is bounded: for all $x, y \in \Bbb C, d_1(x, y) \le 1$. Since $\Bbb C$ is bounded, so are all of its subsets, including $\Bbb R, \Bbb Q, \Bbb Z, \Bbb N$. Thus boundedness is a property of the metric, and even equivalent metrics can give different answers for boundedness.

*There are more general structures that provide a definition for boundedness (as linked by Henno Brandsma). However, it is uncommon to need more than a slight loosening of the two structures mentioned here (such as dropping the totality condition from the order, or positive definiteness from the metric). You are unlikely to meet any examples outside of graduate-level mathematics unless you go looking for them.

A: I'm not sure where you got the idea that all possible sets have to be bounded or unbounded.  Obviously for sets of things that don't have some sense of measure such as animals or letters or colors, it doesn't make sense to talk of boundedness.
But for sets of things that do have some sense of measure it does. 
So some bounded sets:  {all numbers between -$\sqrt{537}$ and $\pi^{25}/17$}; {7, 9, 3, -2},  {all positive rational numbers whose square is less than 2}{1/n| n $\in \mathbb N$}, etc.
Some unbounded sets: {all the real numbers. period}, {all even integers} {all the numbers less than -5 except those between -17 and negative two billion}, {$2^n | n \in \mathbb N$}, etc.
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Now bounded/unbounded sets don't need be numbers per se.  But they do have to be measurable, if not by themselves then against each other.  For example, we can measure the distances between points on a geometric plane.  No point by itself has any measure but you can always find a distance between them.  A set can consist of points.  The set of, say, all the points in a triangle, would be a bounded set as any two points in the set are within a certain bounded distance of each other.  A set of, say, the points on an infinite line would be unbounded.
There is a subtle difference between sets that are bounded because the individual elements all fall within a range by themselves, such as {all the positive numbers less than 3} and sets that are bounded because pairwise the elements are within a certain distance of each other, such as my geometry examples.
If you could find some way to relate ... bears and ice cream flavors... to measurable values.  Then hypothetically, you could talk about a set of bears being bounded.  But you'd need a we of measuring, otherwise it just doesn't make sense.
