## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorJeela
• CommentTimeJun 29th 2023
• (edited Jun 29th 2023)

Hi everyone,

I am new to the forum so pardon the possible selection of the wrong category under which to discsuss this topic. I have been studying Freyd’s algebraic theory of the reals in the past week. I have problems understanding his linear representation theorem that every scale can be embedded in a product of linear scales. To wit, I have two problems:

1) I thought that scale was linearly ordered under the relation of partial order defined by $a \multimap b = \top$. So the embedding should be trivial.

2) In the proof, he shows that the meet $y \vee x$ is strictly smaller than $\top$ in a scale that is an SDI (an algebra is an SDI “if whenever it is embedded into a product of algebras one of the coordinate maps is itself an embedding”). This I understand. But he also says that from this follows the disjonction property that $x \multimap y \vee y \multimap x = \top$ in an SDI scale. Now, I can’t see how this follows.

Ok I understand now. I got confused with the remark about internalization. The equation of linearity $x \multimap y \vee y \multimap x = \top$ is the internalization and the sum is the operation in a lattice not a disjonction. I also see how the sum $x \vee y$ being in the SDI scale, whenever $x$ and $y$ are, imply the disjonction property and hence linearity as an order.