Is the proper segment of a TOSET is initial segment?

Question

I can't fimd its answer. I have learnt that for a well ordered set proper segment is the initial segment. But i am unable to find segment of a toset which is proper and not initial.

Answer 1

The term "segment" isn't standard in the literature. Some texts, like Munkres' topology book, use "segment" to mean "set of the form $\{x: x\le a\}$ for some $a$." If we use this definition then the answer to your question is trivially yes.

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However, I've also seen (in talks, not in books or papers) the term "segment" used synonymously with "interval" or "convex set:" that is, a segment of a linear order (= a more common synony for "toset") is a set $A$ such that for all $x<y<z$, if $x,z\in A$ then $y\in A$. For example, according to this definition each of $[0,1]$, $(0,1)$, $[0,1)$, and $(0,1]$ would be segments of $\mathbb{R}$ (with the usual ordering). To avoid confusion I'll use the term "interval" for this notion of segment in what follows.

According to this definition, not even well-orderings have the "interval = initial segment" property; indeed, the only linear order which does have this property is the one-element linear order (this is a good exercise). However, well-orderings do satisfy a weak version of this property: namely, every interval is (order-)isomorphic to an initial segment.

This property in fact characterizes well-orderings:

• Suppose $A$ is a linear order such that every interval of $A$ is isomorphic to an initial segment of $A$, but $A$ is not well-ordered; we want to get a contradiction.

• Since $A$ is not well-ordered it has some subset $B$ with no least element. From $B$ we can produce an interval with no least element: the set $$\tilde{B}=\{x: \exists y\in B(x>y)\}$$ is an interval and has no least element (this is a good exercise).

• Because $\tilde{B}$ is an interval in $A$, it must be isomorphic to an initial segment of $A$. This means that $A$ has no least element, since $\tilde{B}$ has no least element.

• But now consider any interval in $A$ which does have a least element - for example, $\{x\}$ for some $x\in A$. Such an interval cannot be isomorphic to an initial segment of $A$, since every initial segment of $A$ must have no least element (do you see why?). This gives a contradiction.