Ordinal addition

Conversation started Dec 27, 2011 at 8:24.

Matt

Dec 27, 2011 08:24

@AsafKaragila It might get closed as not a real question but: why is ordinal addition defined in terms of well-ordered sets and not in terms of ordinals? You might say that then it's more general and therefore we have an addition on arbitrary well-ordered sets but then why is it called ordinal addition?

Martin Sleziak

Sorry, going afk.

Matt

See you later, Martin!

Martin Sleziak

I'll be back soon...

Srivatsan

@MartinSleziak What other projections are there? // Yes, I'm talking about orthogonal projection.

Matt

@MartinSleziak You should edit away the "soon" to sound cooler.

Asaf Karagila

Dec 27, 2011 08:25

@Matt Ordinals correspond to order types. So you want the addition to be have like the addition of the order types.

Martin Sleziak

@Srivatsan oh, ok then; if you look at wiki page - basically all maps fulfilling $P^2=P$ are called projections (which of course has geometrical interpretation...)

Matt

@AsafKaragila But order types are ordinals I thought.

robjohn

@MartinSleziak yes, say that $Px-x$ is perpendicular to $Px$

Asaf Karagila

It is called ordinal addition because the sum of well ordered sets is well ordered, and thus can be represented by a unique ordinal.

@Matt No... $(\mathbb Q,\le)$ or $(\{-n\mid n\in\omega\},\le)$ are ordered sets with order types which are not ordinals.

Matt

I thought the order type of the rationals was $\omega$ and so is the order type of the negative numbers. I'm confusing order type with cardinality I think.

Srivatsan

Dec 27, 2011 08:27

@MartinSleziak Yes, but that abstract picture is introduced some time afterwards. [The author also has defined only $P^*$ for a matrix, not an operator.] I'm sure this problem can (and intended to) be done with matrix-fu.

robjohn

Then $x^*P^*Px=x^*P^2x$

If that is true for all $x$, then $P^*=P$

Srivatsan

We should show that, yes.

@robjohn Is that clear?

Matt

Thank you, Asaf!

Asaf Karagila

@Matt Yeah. The order type of the rationals is the rationals themselves, and the negative numbers have order type called $\omega^*$.

The star represents an inverse order of an ordinal.

So the order type of $\mathbb Z$ is $\omega^*+\omega$.