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I'm reading Kuratowski's "Set theory", and here is a question from the Chapter 7.

Let $A$ be a well-ordered set and $X$ its initial segment, i.e. $X$ is a proper subset of $A$, and $\forall x\in X$ if there exists the predecessor $x^-$ of $x$, then $x^-\in X$. Let $r$ be the first element in $A-X$. Let $O(r)=\{a\in A:a < r\}$. Author says that $X=O(r)$, but I can't prove that $X\subset O(r)$. Can you please help me with it?

Andrés E. Caicedo
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Bertrand Haskell
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    The definition of the initial segment seems wrong to me. If $A=\omega+1$ and $X={\omega}$, then clearly $X$ is a proper subset of $A$, and since as limit ordinal $\omega$ has no predecessor, the condition that every existing predecessor is in $X$ is also fulfilled. But $X$ clearly isn't an initial segment of $A$, and furthermore, $A\setminus X=\omega$, and thus $r=0$ and $O(r)={0}\ne X$. – celtschk Dec 01 '19 at 20:02
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    The usual definition of "initial segment" requires that, if $x\in X$ and $y<x$ then $y\in X$. That is, $X$ contains all predecessors of its elements, not just immediate predecessors. (I conjecture that Kuratowski had the usual definition and a translator messed it up.) – Andreas Blass Dec 01 '19 at 20:04
  • @Andreas your conjecture was right! – Bertrand Haskell Dec 01 '19 at 20:30

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With these definitions, this is not true.

With this definition - assuming a predecessor of $x$ to be an element of $x^-\in A$ such that $x=\min\{y\in A\,:\, x^-<y\}$ - it is entirely possible that a set $X$ is "predecessor-closed" without there being any $r\in A$ such that $X=O(r)$. Namely, consider the set $X=\{\omega\}$ in $A=\omega+2$: the condition on predecessors is vacuously true, but $X$ is not in the form $[0,r)$ or $[0,r]$ for any $r\in A$.

Typically, the definition we give of initial segment of an ordered set $(A,\le)$ is a subset $X\subsetneq A$ such that, for all $x\in X$ and for all $y\in A$ such that $y\le x$, $y\in X$. With this definition, the claim is true.