Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Are there interesting algebraic structures whose cardinality is greater than the continuum? Obviously, you could just build a product group of $\beth_2$ many groups of whole numbers to get to such a structure, but the properties of this group seem to be not very different from those you get when you just put together $\beth_1$ many whole number groups here.

share|improve this question
    
I think this is a duplicate but can't find the other question... it was about groups, maybe? –  Qiaochu Yuan Jun 2 '11 at 13:44
    
Dear Qiaochu Yuan: even if not an exact duplicate, the OP could read the answers to the similar question mathoverflow.net/questions/32370/…. Was you refering to it? –  Giuseppe Tortorella Jun 2 '11 at 18:06
add comment

3 Answers

The MO question Martin is referring has several good examples of algebraic structures larger than the continuum; one that I did not see talked about there is Conway's field of surreal numbers.

The surreal numbers form a proper class, so their cardinality dwarfs the cardinality of any set.

share|improve this answer
add comment

This MO question "Cardinalities larger than the continuum in areas besides set theory" has recieved a couple of interesting answers, also algebraic ones (e.g. Zariski cotangent space of a manifold which is not smooth, automorphism tower length of a group)

share|improve this answer
add comment

The set $\beta\mathbb{N}$ of ultrafilters on $\mathbb{N}$ is not only a topological structure, but also an algebraic structure as well which is partialy compatible with the topological structure of $\beta\mathbb{N}$ (i.e. it is continuous in one variable). If $\mathcal{U},\mathcal{V}$ are ultrafilters on $\mathbb{N}$, then define an ultrafilter $\mathcal{U}+\mathcal{V}$ by letting $R\in\mathcal{U}+\mathcal{V}$ if and only if $\{n|\{m|m+n\in R\}\in\mathcal{U}\}\in\mathcal{V}$. With this operation, $\beta\mathbb{N}$ becomes a well known semigroup of cardinality $2^{2^{\aleph_{0}}}$.

Also, Boolean algebras are interesting structures, but the cardinalities of Boolean algebras are often greater than the continuum. For instance, the weakly compact and strongly compact cardinals can be characterized in terms of Boolean algebras. An inaccessible cardinal $\kappa$ is strongly compact if and only if every Boolean algebra $B$ that satisfies the identity $$\bigwedge_{i\in I}\bigvee_{j\in J_{i}}x_{i,j}=\bigvee_{f}\bigwedge_{i\in I}x_{i,f(i)}$$ (where $|I|<\kappa,|J_{i}|<\kappa$ and $f\in\prod_{i\in I}J_{i}$) has a $\kappa$-complete ultrafilter. There is a similar characterization of weakly compact cardinals since for weakly compact cardinals the above holds only when $|B|\leq\kappa$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.