# Ordering on pairs

Define $(a,b) < (a',b')$ if $\max(a,b) < \max(a',b')$ or $\max(a,b) = \max(a',b')$ and $b < b'$ or $\max(a,b) = \max(a',b')$ and $b = b'$ and $a < a'$.

Now I want to prove that the order-type of $(\{(b,c) : \max(b,c) = a\}, <)$ is equal to $a + a + 1$, does someone have a hint how to do this? I can't find the bijection. All elements are ordinals.

Thanks.

-
@Arturo: $a,b,c$ are ordinals. –  Jonas Teuwen Dec 12 '10 at 23:23
T: I think you've misstated the order relation: shouldn't it be "$\max(a,b)\lt \max(a',b')$, or $\max(a,b)\mathbf{=}\max(a',b')$ and $b'\lt b$, or $\max(a,b)\mathbf{=}\max(a',b')$ and $b=b'$ and $a\lt a'$"? –  Arturo Magidin Dec 12 '10 at 23:28
@Arturo: Yes, you're right. Sorry, I corrected it. –  Jonas Teuwen Dec 12 '10 at 23:29
You and Arturo seem to have $b'$ and $b$ mixed up. –  JDH Dec 13 '10 at 3:30
@JDH: Oops; quite right. I was just focusing on the issue of maxima. Thanks, Joel. –  Arturo Magidin Dec 13 '10 at 4:26

Hint: The set naturally breaks up into three disjoint parts: $\{(a,a)\}$, the set $\{(b,a)\mid b\lt a\}$; and the set $\{(a,b)\mid b\lt a\}$. Every element of the latter is less than each element of the middle one, and each element of the middle one is stricly smaller than $(a,a)$.
Oh, haha. I staired at this for some while but I was thinking about a $\leq$ in the definition instead of $=$. Thanks! –  Jonas Teuwen Dec 13 '10 at 15:51