5
$\begingroup$

Define an ordinal as any transitive set, well-ordered by $\in$. I want to prove that if $\alpha\ne\beta$ are ordinals and $\alpha\subset\beta$ (where in my book $\subset$ permits equality) then $\alpha\in\beta$. The book gives a proof but there is a step which confuses me.

Proof: Let $\gamma$ be the least element of $\beta-\alpha$. Since $\alpha$ is transitive then $\alpha$ is the initial segment of $\beta$ given by $\gamma$. That is to say $\alpha = \{\xi \in \beta:\xi < \gamma\} = \gamma$. Thus $\alpha\in\beta$.

I don't get two things that are probably kind of the same thing, which are in the last two sentences. How do we know $\gamma=\{\xi\in\beta:\xi<\gamma\}$? I get the intuition is that ordinals are the set of all elements less than themselves but that's not the formal definition, so I'm not seeing how this inference gets made.

All I have about ordinals and initial segments up to this point are: 1) any two well-ordered sets are isomorphic or an initial segment of one is isomorphic to an initial segment of the other, 2) $0=\emptyset$ is an ordinal, and 3) if $\alpha\in\beta$ where $\beta$ an ordinal then $\alpha$ is an ordinal.

The other thing I don't get his how we then infer $\alpha\in\beta$ because again, while the intuition makes sense I'm not even sure how we infer $\alpha\in\beta$ when we don't know the conditions for being in $\beta$. All we know about $\beta$ is that it's an ordinal and $\alpha\subset\beta$. The only way I can imagine inferring something is in $\beta$ is by knowing it's in $\alpha$.

$\endgroup$
4
$\begingroup$

There are two things here.

  1. An ordinal $\xi$ is equal to the set $\{\zeta\mid \zeta<\xi\}$. Exactly because the ordering of ordinals is given by $\in$.

  2. Since $\gamma$ is an element of $\beta$, and $\alpha=\gamma$ it follows that $\alpha$ is an element of $\beta$.

$\endgroup$
3
$\begingroup$

By definition, the relation $<$ on $\beta$ is $\in$. That is, for $x,y\in\beta$, $x<y$ is defined to mean $x\in y$. So we have $$\alpha=\{\xi\in\beta:\xi<\gamma\}=\{\xi\in\beta:\xi\in\gamma\}=\gamma\cap\beta.$$ But since $\beta$ is transitive and $\gamma\in\beta$, $\gamma\subseteq\beta$, so $\gamma\cap\beta=\gamma$.

Since $\alpha=\gamma$ and by definition $\gamma$ is an element of $\beta$ (namely, the least element of $\beta-\alpha$), we conclude that $\alpha\in\beta$.

$\endgroup$
1
$\begingroup$

Let $\gamma=\beta\setminus\alpha$ and $c$ be the least element of $\gamma$. We will prove $\alpha=c$.

For all $t\in \alpha, t\in c$. If not, there exists $t'\in\alpha\subsetneq\beta, t'\notin c$. Since $t',c\in\beta$, which is well-ordered under $\in$, then $c\in t'\subsetneq\alpha$ [Since $t'\in\alpha$, which is an ordinal]. This implies $c\in\alpha$ and $c\in\gamma$, which is a contradiction.

For all $t\in c, t\in \alpha$. If not, there exists $t'\in c, t'\notin \alpha$, hence $t'\in\gamma$. Thus $t'\in c$ and $c,t'\in \gamma$, which contradicts to the fact that $c$ is the least element of $\gamma$.

To sum up, $\alpha=c\in\gamma\subsetneq\beta$. Thus $\alpha\in\beta.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.