A relation \(R\) on a set \(A\) is said to be a partial order if it is reflexive, antisymmetric, and transitive. A partial order is said to be a total order if for any \(x,y \in A\) either \(x\,R\,y\) or \(y\,R\,x\text{.}\)
Note, some authors require partial orders to be irreflexive instead of reflexive. All that this requires is to delete the diagonal from the relation. Basically everything that can be proven about partial orders in our formulation can be proven in the other formulation, and vice versa. Instead, we we call a relation that is irreflexive, symmetric and transitive a strict partial order.
Let \(\prec\) be a partial order on a set \(A\text{.}\) An element \(x \in A\) is said to be minimal relative to the ordering \(\prec\) if for every \(y \in A\text{,}\) \(y \not\prec x\text{.}\) In other words, there are no elements smaller than \(x\text{.}\) Maximal elements are defined analogously.
An element \(x \in A\) is said to be a minimum if for all \(y \in A\text{,}\) \(x \prec y\text{.}\) Maximum elements are defined analogously.
Prove that if \(A\) has a minimum element relative to the partial order \(\prec\text{,}\) then it is unique (similarly for maximum elements).
Consider the less than partial order \(\lt\) on the real numbers \(\mathbb{R}\text{.}\) Does it have maximum or minimum elements? What about maximal or minimal elements?
Consider a universe \(U\) and the subset relation on the subsets of \(U\text{.}\) Does \(U\) have maximum and minimum elements?
Consider the divides partial order on \(\mathbb{N} \setminus \{1\}\text{,}\) i.e., then integers greater than or equal to \(2\text{.}\) Describe the minimal elements of this partial order and explain your reasoning.
Suppose \(\prec\) is a total order on \(A\text{.}\) There is an induced ordering \(\prec_n\) on \(A^n\) call the lexicographical ordering or dictionary order.
It works like this, given \((x_1,\ldots,x_n), (y_1, \ldots, y_n) \in A^n\text{,}\) if \(x_j = y_j\) for all \(j = 1,\ldots, n\) then \((x_1, \ldots, x_n ) \prec_n (y_1,\ldots, y_n)\text{.}\) Otherwise, let \(k\) be the first index for which \(x_k \not= y_k\text{.}\) If \(x_k \prec y_k\text{,}\) then \((x_1,\ldots, x_n) \prec_n (y_1, \ldots, y_n)\text{,}\) and otherwise \((y_1,\ldots, y_n) \prec_n (x_1, \ldots, x_n)\text{.}\)
Prove that \(\prec_n\) is a total order on \(A^n\text{.}\)
A total order \(\prec\) on \(A\) is said to be a well order if every nonempty subset of \(A\) has a minimum element with respect to \(A\text{.}\)
On \(\mathbb{N}\text{,}\) we know that \(\lt\) is a well-order. This is precisely the content of the Well-Ordering Principle.
Is \(\lt\) a well ordering on \(\mathbb{R}, \mathbb{Q}\) or \(\mathbb{Z}\text{?}\)
Do you think \(\mathbb{R}, \mathbb{Q}, \mathbb{Z}\) have well-orders of some form or another?
Every set can be well-ordered.
This theorem seems absolutely ridiculous. What does a well-order of \(\mathbb{R}\) look like? The answer is, in short, we don't know. It is actually impossible to write down a formula for one. However, the proof of this theorem and hence the existence of such a well-order is a consequence of the Axiom of Choice, which was originally the most highly contested axiom of early set theory, but not it is almost ubiquitously accepted. Why? Because it makes lots of nice theorems true, like the previous ones (it also makes some very annoying theorems true, but that is a discussion for another day). We will state the Axiom of Choice after we discussion functions.
Let \(\prec\) be a partial order on \(A\) and let \(a \in A\) be a minimum. Then \(a\) is the unique minimal element.
We first show \(a\) is minimal. Notice that for any \(b \in A\text{,}\) we have \(a \prec b\) since \(a\) is a minimum. If \(a \not= b\text{,}\) then by antisymmetry \(b \not\prec a\text{,}\) so \(a\) is minimal.
Now suppose that \(c\) is minimal. Since \(a\) is a minimum, \(a \prec c\text{.}\) By minimality, this can only happen if \(a = c\text{.}\)
There is an analogue of Theorem 3.3.14 for maximum and maximal elements.