In my algebra notes :
Definition of upper bound
$x \in X$ is an upper bound of $Y$ if $y \le x\ \ \forall y \in Y$.
Definition of maximal element
We say $m \in X $ is a maximal element if $m\leqslant x\ \forall \in X\ \implies x=m$
Definition of inductive set
Let $X$ be a set and $\le $ an ordering over $X$
$X$ is said to be inductive if any subset $Y \subset X$ , $Y$ totally ordered has a least an upper bound in $X$
Then they give a couple of examples I don't understand:
Any finite set has maximal elements and it is partially ordered and inductive, for any ordering defined over it.
$\Bbb N, \Bbb Z,\Bbb Q,\Bbb R$, with the usual ordering, are not inductive
Can someone explain or prove why?
In 1) how can I be sure this is valid for any ordering?
In 2) If I considered any finite set in any of these sets, with the usaul ordering, it has to be totally ordered an therefore the greatest one would be an upper bound and hence, inductive. I don't get why they say they are not inductive .
If $\langle A,\le\rangle$ is a non-empty finite partial order, let $X$ be a totally ordered subset of $A$. $X$ is finite, so we can list its elements as $X=\{x_1,x_2,\ldots,x_n\}$ for some $n\in\Bbb Z^+$. Moreover, since $X$ is totally ordered by $\le$, we can index its elements so that $x_1\le x_2\le\ldots\le x_n$. But then $x_n$ is an upper bound for $X$, so $A$ is inductive.
To see that $A$ has maximal elements, let $a_1$ be any element of $A$. If $a_1$ is not maximal, there is some $a_2\in A$ such that $a_1<a_2$. If $a_2$ is not maximal, there is some $a_3\in A$ such that $a_2<a_3$. Keep going; $A$ is finite, so the process must stop after some finite number of steps, and at that point we have a maximal element of $A$. (This argument even shows that if $a_1$ is any element of $A$ whatsoever, $A$ has a maximal element $m$ such that $a_1\le m$.)
$\Bbb N$ is a totally ordered subset of $\Bbb N$, $\Bbb Z$, $\Bbb Q$, and $\Bbb R$ that has no upper bound, so none of these four sets is inductive.