Proper Subsets of Ordinals ... ... Searcoid, Theorem 1.4.4 .

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Main Question or Discussion Point

I am reading Micheal Searcoid's book: "Elements of Abstract Analysis" ... ...

I am currently focused on understanding Chapter 1: Sets ... and in particular Section 1.4 Ordinals ...

I need some help in fully understanding Theorem 1.4.4 ...

Theorem 1.4.4 reads as follows:

In the above proof by Searcoid we read the following:

"... ... Now, for each $\gamma \in \beta$ , we have $\gamma \in \alpha$ by 1.4.2, and the minimality with respect to $\in$ of $\beta$ in $\alpha \text{\\} x$ ensures that $\gamma \in x$. ... ...

Ca someone please show formally and rigorously that the minimality with respect to $\in$ of $\beta$ in $\alpha \text{\\} x$ ensures that $\gamma \in x$. ... ...

*** EDIT ***

Is the argument simply that since $\beta$ is the least element of $\alpha \text{\\} x$ ... then if $\gamma \in \beta$ ... then $\gamma$ cannot belong to $\alpha \text{\\} x$ .... otherwise $\gamma$ would be the least element ... and so $\gamma$ must belong to $x$ ...

*** *** ***

Help will be appreciated ...

Peter

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It may help Physics Forum readers of the above post to have access to the start of Searcoid's section on the ordinals (including Theorem 1.4.2 ... ) ... so I am providing the same ... as follows:

It may also help Physics Forum readers to have access to Searcoid's definition of a well order ... so I am providing the text of Searcoid's Definition 1.3.10 ... as follows:

Hope that helps ...

Peter

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andrewkirk
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Is the argument simply that since $\beta$ is the least element of $\alpha \text{\\} x$ ... then if $\gamma \in \beta$ ... then $\gamma$ cannot belong to $\alpha \text{\\} x$ .... otherwise $\gamma$ would be the least element ... and so $\gamma$ must belong to $x$ ...
Almost. Just replace '$\gamma$ would be the least element' by '$\gamma$ would be less than $\beta$, so $\beta$ could not be the least element'

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Thanks Andrew ...

Appreciate your help ...

Peter