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If $x_1, x_2, x_3, \ldots$ are ordinals such that $x_1\ge x_2\ge x_3\ge \ldots$, then there exists a positive integer $n$ such that $x_n=x_{n+1}=x_{n+2}=\ldots$.
I think that this statement is true, but I do not know if there is a possible way to find such integer $n$.

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up vote 3 down vote accepted

This is true because $\in$ is a well-founded relation, and $\alpha<\beta$ if and only if $\alpha\in\beta$. Therefore there is no infinite sequence which is strictly decreasing.

Note that the $n$ could be any $n\in\omega$. For example, $n,n-1,\ldots,2,1,0,0,0,\ldots$. This would be a sequence which only stabilizes after $n$ steps.

To see that this is true we can either use the fact that $x_i$ are all members of $x_0 + 1$, which is a well-ordered set. Therefore every non-empty set has a minimal element, in particular the set $\{x_i\mid i\in\omega\}$. Let $x$ be the minimum of that set, then $x=x_n$ for some $n$, and since we assumed $x_n\geq x_{n+1}$ we have that $x_k=x$ for all $k\geq n$.

(This claim may have other proofs depending on your definitions of ordinals, and the approach that you wish to take.)

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