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I'm trying to solve a simple problem of set theory. I found it in Kuratowski's book (Corollary $8$,chapter $1$), but I'd like to solve it in an slightly different way. This is the statement and a first attempt, until the point where I got stuck.

Let $(A,\trianglelefteq)$ be a well ordered set. Then there is no isomorphism between the set and any of its initial segments.

Let us consider a well ordered set $(A,\trianglelefteq)$ and, by contradiction, suppose that does exist an isomorphism $$f:(A,\trianglelefteq)\rightarrow (pred(A,a,\trianglelefteq )),$$ where $(pred(A,a,\trianglelefteq )$ is the initial segment generated by $a$. Since the set is well ordered, each of its non-empty subset has a least element (w.r. to the ordering). Then consider $B=\{y\in A | f(y)\neq y\}$, that is non-empty (e.g. $f(a)\neq a$, because we are supposing that the isomorphism is not the identity).

I wish to derive a contradiction working only on the least element of $B$, say $m=\min_{\trianglelefteq}B$,

That is, the unique possibility is $m=f(m)$ (it is not possible that $m\trianglelefteq f(m)$ or $f(m)\trianglelefteq m$). Now I have some idea but there's something wrong, and up to now I haven't gotten the exact and rigorous solution. I've found some other questions concerning this topic, but I would like to solve it just in this way, using the $m$ defined above. Thank you very much.

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Hint: We will derive a contradiction to $m$ being the $\trianglelefteq$-minimum element of $B$. As $f(m) \neq m$, then either $f(m) \triangleleft m$ or $m \triangleleft f(m)$. In the former case compare $f(f(m))$ and $f(m)$, and in the latter case consider the unique $x \in A$ such that $f(x) = m$.

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  • $\begingroup$ Thanks, that's wath I was searching for. $\endgroup$ – the_elder Mar 4 '13 at 17:40

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