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I refer to pg.4, Lemma 12 of this article on ordinal numbers.

It says "If $x,y\in \alpha$, then $x<y, x>y, \text { or }x=y$".

My question: As $\alpha$ is an ordinal, we know $x<\alpha$ and $y<\alpha$. However, how do we know that $x$ and $y$ can be compared in this manner? We know that as $\alpha$ is transitive, $x,y\subset\alpha$. However, that does not necessarily imply $x\subset y$, $y\subset x$ or $y=x$! Moreover, $x,y$ are not real numbers such that we could assume that one of the three relations $x<y,x>y$ or $x=y$ should hold.

Although $x$ and $y$ are ordinal numbers themselves by the fact that $x,y\in\alpha$, we can't assume this now as the lemma I am currently studying will be used to establish this fact.

Thanks in advance!

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This follows directly from the definition of ordinals used in that text:

Definition 8. An ordinal is a transitive set well-ordered under the relation $x < y$ if $x \in y$.

This means that $\in$ is a well-ordering of $\alpha$. By definition, well-ordering is a linear ordering.

To address your comment requiring more clarification for the same proof: $\alpha\in\alpha$ cannot hold for any set (in ZFC) by Axiom of regularity. See also this question and the questions linked there.

Notice that this is used in that proof to arrive to contradiction. (So you are right in claiming that this cannot hold. But since this is a proof by contradiction, we want to get something that does not hold.)

It is perhaps worthwhile noticing, that this proof in fact does not use Axiom of Regularity. But I thought it was worth mentioning, so that you know that $x\notin x$ is true for any set (if you're working in the axiomatic system ZFC). It is not a special property valid only for ordinals.

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Martin Sleziak
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  • This is on a related note: another line in the same proof says $\alpha\in\alpha$, as $\alpha$ is transitive. Definition 7 says that $y\in X$ implies $y\subset X$. How can $\alpha$ be a proper subset of itself? Thanks! –  Jul 17 '13 at 08:22
  • @AyushKhaitan I have expanded my answer to address your comment. – Martin Sleziak Jul 17 '13 at 08:44
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Note that $<$ is merely a binary relation symbol. It is not inherently the order of the real numbers, or some other partial order.

I claim that a well-order is a linear order. Consider Definition 6, given $(A,\prec)$ which is a well-order, let $a,b\in A$ then $\{a,b\}$ is a non-empty subset of $A$ and therefore has a least element, therefore $a$ and $b$ are comparable.

Now it's easy to deduce that ordinals are linear orders by Definition 8 which requires them to be well-ordered by $\in$.

Asaf Karagila
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  • Good point. Many sources include directly in the definition of well ordered set that it is a linearly ordered. But we get an equivalent definition if we require only partial order -- your argument shows that it must be linearly ordered. – Martin Sleziak Jul 17 '13 at 07:04
  • Perhaps you meant Definition 6 (well-ordered set) and not Definition 7 (transitive set)...? – Martin Sleziak Jul 17 '13 at 07:04
  • @Martin: Yes, Definition 6, thank you. Note that often require that a well-order is a linear well-founded set, in which case linearity is required. – Asaf Karagila Jul 17 '13 at 07:11
  • This is on a related note: another line in the same proof says $\alpha\in\alpha$, as $\alpha$ is transitive. Definition 7 says that $y\in X$ implies $y\subset X$. How can $\alpha$ be a proper subset of itself? I'm not sure if $\subset$ means the same thing it does generally , or if it is an arbitrary binary relation. Thanks! –  Jul 17 '13 at 08:28
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    @Ayush: Note that the use of $\alpha\in\alpha$ is for arriving at a contradiction. It is in order to prove that either $\alpha\in\beta$ or $\beta\in\alpha$ or $\alpha=\beta$. – Asaf Karagila Jul 17 '13 at 09:33