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In Introduction to Set Theory by J. Donald Monk, he defines ordinal as follows.

Definition (1): $A$ is an ordinal iff $A$ is $\in$-transitive and each member of $A$ is $\in$-transitive. $A$ is $\in$-transitive iff for all $x$ and $y$ , $x \in y \in A \implies x \in A$.

And using definition (1), I have a problem of showing $0$ is an ordinal.

My solution: Let $x \in y \in 0$, then show that $x \in 0$, but in the theorem shown before this, there cannot exist an $x \in 0$ for any $x$, for if $x \in 0$, then $x$ is a set and $x \neq x$, which is absurd. Hence there cannot be an $x$ such that $x\in 0$ .But if this is the case, we cannot shown the above theorem by definition. Any ideas?

thanks for your help.

Edit: Initially I have 2 question to ask. But later on, I have found the answer to my first question. That is why I deleted it and the question post above is my second question.

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    Broad hint: what does the notion 'for each $x$ in $A$' mean when $A$ is the null set? Are you familiar with the notion of vacuous truth?2011-02-25
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    Ok. i have look it up on google wikipedia. now what is the connection of this with vacuous truth?2011-02-25

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Seoral: A statement of the form "for all $x\in A$ blah" literally means "for all $x$, if it is the case that $x\in A$, then blah".

Now: As you mentioned, $y\in 0$ is false for any $y$. We need to show that if $x\in y\in 0$ then $x\in 0$. More precisely, this means: For all $x$ and $y$, if $y\in 0$ and $x\in y$, then $x\in 0$. Since $y\in 0$ is false, we need to show that for all $x$ and $y$, (false and $x\in y$) implies $x\in 0$. Recall that "false and blah" is false, so this is just: For all $x$ and $y$, false implies $x\in 0$.

But false implies anything! (More precisely, statements $p\to q$ are true whenever $p$ is false. They are also true when $q$ is true, but that doesn't matter here.) So, we have "for all $x$ and $y$, true", which is the same as "true". I.e., you have proved what you needed.

One usually doesn't express situations as the one above this way; instead, people talk of statements being "vacuously true", as pointed out in a comment above.

Let's review: 0 is transitive, because any element of 0 is a subset. (Precisely, because there are no elements of 0.)

Similarly, any element of 0 is transitive. Again, because there are no elements of 0.

This is somewhat strange the first time one encounters it, since it is also the case that any element of 0 fails to be transitive (and that any element of 0 is blue, etc).

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    So whole of the argument depend on statements $p \implies q$ are true whenever $p$ is false. since $y\in 0$ is false and $x \in y \implies x\in 0$, hence by vacuous implication it is true and hence $0 \in Ord$?2011-02-25
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    It seems you don't believe it, can you tell us why? @Seoral2011-02-25
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    @awllower: Not that i don't believe it. this is the first time i encounter it and it is just hard to accept such claims. i mean this statement: $p \implies q$ are true whenever p are false. To prove the statement above will require some homework to do. But anyway, i have respect for what is being established so far in mathematics. so as you can see above i have tick that as an answer.2011-02-26
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    @Seoral: If you have access to Munkres "Topology" book, you may want to read the beginning, where he explains how in mathematics common expressions such as "a implies b" have a precise meaning that is not always the one used in common language (this is discussed in other place,s but I like how Munkres explains it). This is actually a serious obstacle for some students, as it is easy not to realize early on that we use some expressions in a technical way rather than in the way they are used in regular conversation.2011-02-26
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    @Seoral: I promise you that if you can walk through the fire, then I will give you three thousand billions, since you cannot walk through the flame, I don't break the word. Is this what you meant?2011-02-27
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    @Andres: thanks for you suggestion. i will try to take a look at it. @awllower:i have no idea what you are talking about...are you talking about example of the vacuous truth here?2011-02-28