Well-ordered sets

Definition and examples

1. Well-orderet set

We say that the totally ordered set $(X,\leq)$ is well-ordered if every nonempty subset has a minimum element.

2. Examples

• The set $\mathbb{N}$ with the usual order is well-ordered.
• The set $\mathbb{R}$ with the usual order is not well-ordered.

Properties

1. Properties of well-ordered sets

• If $X$ is well-ordered and $Y\subseteq X$ has the induced order, then $Y$ is well-ordered.
• If $X,Y$ are well-ordered sets, then $X\times Y$ with dictionary order is well-ordered.

Zermelo’s theorem

1. Zermelo

If $A$ is any set, there is a total order $\leq$ on $A$ that is a well-order.

2. There is a set that is uncountable and well-ordered.

Sections

1. Section

If $S$ is an ordered set and $\alpha\in S$, we denote by $S_{\alpha}$ the set: $% $ $S_{\alpha}$ is called the section of $S$ by $\alpha$.

The example

The theorem

1. There is a set that is uncountable and well-ordered, but every section of it is countable.

2. Proof

• Let $X$ be an uncountable, well-ordered set. Then the set $S=\{1,2\}\times X$ is also uncountable and well-ordered, and has sections that are uncountable (for example, those of elements of the form $(2,x)$.)
• The set $\{\alpha\in S\mid \text{$S_{\alpha}$is uncountable}\}$ is nonempty, let $\Omega$ its smallest element.
• Then $S_{\Omega}$ is the required set.

A corollary

1. If $A\subseteq S_{\Omega}$ is countable, then $A$ has an upper bound in $S_{\Omega}$.

2. Proof

• Since for each $a\in A$, we have that $S_{a}$ is countable, then $B=\cup_{a\in A}S_{a}$ is also countable.
• Since $B\subseteq S_{\Omega}$, and $S_{\Omega}$ is uncountable, we can choose $x\in S_{\Omega}-B$.
• If there was $a\in A$ with $% $, then $x$ would be in $B$, which is a contradiction.
• Hence $x$ is an upper bound of $A$.\qed

A nonmetrizable ordered space

1. A nonmetrizable ordered space

The set $S_{\Omega}\cup\{\Omega\}$ is not metrizable

2. Proof

• We have that $\Omega$ is a limit point of $S_{\Omega}$, since any basic element containing $\Omega$ has the form $(a,\Omega]$ and must intersect $S_{\Omega}$. Otherwise we would have $S_{\Omega}=S_{a}\cup\{a\}$, but $S_{a}\cup\{a\}$ is countable.
• However, there is no sequence in $S_{\Omega}$ that converges to $\Omega$, since if $(x_{n})$ is a sequence in $S_{\Omega}$, there is an upper bound $b\in S_{\Omega}$ for all the terms in the sequence. Then $(b,\Omega]$ contains no point of the sequence.\qed
3. We will denote with $\overline{S_{\Omega}}$ the set $S_{\Omega}\cup\{\Omega\}$.