Show that any element of an ordinal is an initial segment of that ordinal

Question

• Show that any element of an ordinal is an initial segment of that ordinal

My proof trying. Let $\alpha$ be an ordinal. Let $\beta\in\alpha$. Then, by the definition, $\beta\subseteq\alpha$. We want to show that $\beta$ is an initial segment of $\alpha$, that is, we need to show that $\beta$ is

$$\left\{ x\in\alpha: x<\beta\right\}$$.

Recall of initial segment:http://mathworld.wolfram.com/InitialSegment.html

So, I couldn't continue. Can you check my proof-trying? Can you help? Can you give a hint?

Answer 1

Hint: Since "$<$" is the only symbol that is yet to be phrased in terms of elementary properties of sets, perhaps you could inspect what "$<$" means for elements of an ordinal.

Answer 2

What you want to show is that if $b\in a\in On$ then the initial segment $\{y\in a:y<b\}$ (of $a$) is equal to $b $ .

Now $<$ and $\in$ mean the same thing for members of $a$ (by definition of $<$ for $a\in On$).

We have $x\in b\implies x\in b\in a\implies x\in a$ because $a$ is a transitive set. So $$x\in b\implies (x\in a \land b\in a \land x\in b)\implies$$ $$\implies (x\in a \land b\in a \land x<b)\implies$$ $$\implies (x\in a\land x<b)\implies $$ $$\implies x\in \{y\in a:y<b\}.$$

$$\text {And we have }\quad x\in \{y\in a:y<b\}\implies (x<b\land x\in a\land b\in a) \implies x\in b $$ because if $x$ and $b$ both belong to $a\in On$ then $x<b$ means $x\in b.$