# Show that a subset of a linearly ordered set might not have a least element.

I am quite confused about the definition of least element. I know this definition of least element though from Schramm's Real Analysis

If S is a subset of an ordered set, a least element of S is an element x , if there is one, such that (i) x belongs to S and (ii) if y belongs to S and y is comparable to x , then x ≤ y

Also does the least element need to belong to the subset of S or to the subset of S. Please clarify with an example.

• there is no subset of $S$ involved, $S$ is the subset. And the least element has to be an element of $S$. – Thomas Oct 16 '18 at 16:43
• Ok so the well ordering principle says that A linearly ordered set is said to be well-ordered if every nonempty subset of it has a least element. and the well ordering of natural numbers is an axiom. Can you please give an example of a linearly ordered set which doesn't have a least element in its subset always ? – Pratik Patnaik Oct 16 '18 at 16:45
• @Thomas Basically I need an example of a linearly ordered set which is not well ordered. Also what are the advantages of assuming that Natural numbers are well-ordered ? – Pratik Patnaik Oct 16 '18 at 16:49
• look at $\mathbb{Z}$ or $\mathbb{Q}$ or $\mathbb{R}$ – Thomas Oct 16 '18 at 16:54
• How can Z be not well ordered?? – Pratik Patnaik Oct 16 '18 at 16:57

Let $$S=\{z\in \mathbb{Z}| z \text{ is even}\}$$. Then $$S$$ has no least element. If $$z_0\in S$$ were a least element, then $$z_0-2\in S$$ is smaller, so you arrive at a contradiction.