Can one call a linear order a lattice? If not

[nomadreid]

Can one call a linear order a lattice? If not...

I have problems putting together the three ideas
(1) the meets and joins of a lattice are unique, hence lattices must have discrete elements
(2) the truth values of a logic are arranged in a lattice
(3) there exist probability logics, whereby each truth value is a probability which can be any real number between 0 and 1, inclusive.
These three do not seem to fit together, so which one is wrong?
Thanks

 

[micromass]

I have problems putting together the three ideas
(1) the meets and joins of a lattice are unique,

OK

hence lattices must have discrete elements

Why does that follow? And what do you mean with discrete elements?

And yes, any linear order is a lattice. So $\mathbb{R}$ is a lattice.

 

[nomadreid]

micromass, thanks for the answer. I must have been brain dead in writing that question, as I was somehow thinking of the meet of a and b being less than rather than less than or equal to a and b.
By discrete elements I meant separate elements (Hausdorff); another symptom of brain death.
Thanks for reviving me. I promise to be a good zombie.
Actually, I was trying to work my way up to a slightly more involved question, that of whether any partial order on S can be extended to a lattice so that no new elements are added and the order stays the same. (The specific one I was thinking of was partially ordering complex numbers with modulus less than or equal to 1, whereby the partial order is defined by: a+bi < c+di iff a²+b² < c²+d² (essentially concentric circles). But I do not think it can be turned into a lattice.)

 

[micromass]

You can always extend the partial order to a linear order. So yes, in particular you can extend your order to a lattice. This can be proven using Zorn's lemma.

In your specific case, all you need to do is define some ordering on the circles. This can be done in many ways (but none of them are very natural). Once you have done that, you will have a linear order on $\mathbb{C}$.

 

[nomadreid]

thanks again, micromass.
True, it would be easy enough to just say z₁ < z₂ iff
[either
||z₁||< ||z₂||
or
(||z₁||= ||z₂|| & angle 1-0-z₁ < 1-0-z₂)].