Initial segment of a well-order

Let $(W, <)$ be a well-order (Well-order and well-ordered set).

A subset $I \subseteq W$ is an initial segment of $W$ when it is downward closed: if $x \in I$ and $y < x$ then $y \in I$. It is a proper initial segment when $I \ne W$.

For $a \in W$ write

$W_{<a} := \{x \in W : x < a\}, \qquad W_{\le a} := \{x \in W : x \le a\} = W_{<a} \cup \{a\},$

and call $W_{<a}$ the initial segment determined by $a$.

Every initial segment $I$, carrying the order inherited from $W$, is itself a well-order: the inherited order is total, and a nonempty subset of $I$ is a nonempty subset of $W$, so it has a least element, which lies in $I$.

• $\emptyset$ and $W$ are initial segments of $W$; each $W_{<a}$ is a proper initial segment, since $a \notin W_{<a}$ by irreflexivity; and each $W_{\le a}$ is an initial segment.
• Every proper initial segment is $W_{<a}$ for exactly one $a$. Let $I \subsetneq W$ be an initial segment and put $a = \min(W \setminus I)$, which exists because $W \setminus I$ is a nonempty subset of the well-order $W$ (Well-order and well-ordered set). If $x < a$ then $x \notin W \setminus I$ by minimality of $a$, so $x \in I$; hence $W_{<a} \subseteq I$. Conversely let $x \in I$. Then $x \ne a$, because $a \notin I$, and $a < x$ is impossible, because downward closure would then put $a \in I$; so $x < a$ by trichotomy, and $I \subseteq W_{<a}$. Therefore $I = W_{<a}$. For uniqueness, suppose $W_{<a} = W_{<b}$ with $a \ne b$, say $a < b$; then $a \in W_{<b} = W_{<a}$, that is $a < a$, which is impossible.
• Nesting. If $b < a$ then $(W_{<a})_{<b} = W_{<b}$, so an initial segment of an initial segment of $W$ is an initial segment of $W$. This is used whenever two well-orders are compared.
• The initial segments of $W$ are therefore exactly the sets $W_{<a}$ for $a \in W$, together with $W$ itself, and inclusion orders them in the same shape as $W$ with one extra element added on top.
• The corresponding notion for a general poset would be a downward closed set, or "lower set". Nothing on this page needs it outside the well-ordered case, so the definition is stated only there, where the second clause above makes the family of initial segments completely explicit.