Recursive definition isomorphism

If $(X, \lt)$ is a well - getting I can show by transfinite recursion over the ordinals that the function $f(x) = \text{ran} f |_{\hat{x}}$ exists (where $\hat{x} = \{ y : y \lt x\}$).

I have actually gotten $f$ in this manner, Let $V$ be the class of ready and also $F:V \to V$ be a class - function, after that there is an one-of-a-kind $G:ON \to V$ where $ON$ is the class of all ordinals such that $F(\alpha) = F(G|_\alpha)$. So I can use this to get a function $f$ such that $f(x) = F(f|_\hat{x})$. Currently I allow $F = \{(x, \text{ran} x) : x \in V\}$ and afterwards I get the function as above.

Currently, this need to be an isomorphism (order preserving bijection) in between $X$ and also the set of real first sectors of $X$, $I_X$ gotten by incorporation.

Nonetheless, when I have $x < y$, after that I see that $\text{ran} f|_\hat{x} \subset \text{ran} f|_\hat{y}$. So $f(x) \leq f(y)$. Why do I have $f(x) \neq f(y)$?

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2019-12-02 02:53:01
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Answers: 2

You require to make use of the reality that a well - getting can not be isomorphic to any one of its first sectors.

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2019-12-03 04:20:02
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Okay, it resembles you are considering your $(X,\lt)$ as an ordinal, as opposed to an approximate set.

I assert that for all $y\in X$, if $x\lt y$, after that $f(x)\neq f(y)$ and also $f(x)\subseteq f(y)$. You have actually currently revealed $f(x)\subseteq f(y)$, so we simply require to show the inequality.

If $y=\emptyset$, the least component of $X$, after that there is absolutely nothing to do and also the case holds.

Think the case holds for all $z\lt y$. Allow $x\lt y$. If $x^+$, the follower of $x$, is additionally much less than $y$, after that $f(x)\subseteq f(x^+)\subseteq f(y)$, and also $f(x)\neq f(x^+)$ by the induction theory, so $f(x)\neq f(y)$.

If $y=x^+$, after that $\hat{y} = \hat{x}\cup\{x\}$. So $f(y) = f(x)\cup\{f(x)\}$. If $f(x)\cup\{f(x)\} = f(x)$, after that $f(x)\in f(x)$, which is difficult given that ordinals are well - started about $\in$. Consequently, $f(y)=f(x)\cup\{f(x)\}\neq f(x)$.

By transfinite induction, the case holds for all $y\in X$.

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2019-12-03 04:18:00
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