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I tried finding Structures $(X,+,\cdot)$ where X is a set (either finite or infinite) and $+,\cdot$ are operators, for which the following laws apply:

Let $a,b,c\in X$:

$ a\cdot(b+c)=(a\cdot b)+(a\cdot c)\qquad \text{(Distributivity)}\\ a+(b\cdot c)=(a+ b)\cdot (a+ c)\\ a+b=b+a\qquad \text{(Commutativity)}\\ a\cdot b = b\cdot a\\ a\cdot (b \cdot c)=(a\cdot b)\cdot c\qquad \text{(Associativity)}\\ a+(b+c)=(a+b)+c$

And there should be no $a,b\in X$ for which $\forall c\in X: a*b=a*c$ (meaning I dont want 2 elements acting invariant under the Operators, eliminating trivial answers)

I do not require multiplicative or additive inverse (and identity) elements.

Now my Questions: (1) Is there a classification for these structures? I have not found anything on the internet, except for boolean algebras, for which $X=\{0,1\}$ and $+:=\vee,\; \cdot:=\wedge$ (side note: this is where I got the idea from)

(2) I have found, that for any subset of $\mathbb{R}$, $a+b:=\max\{a,b\}$, $a\cdot b:=\min\{a,b\}$ satisfy these rules. However, I have been unable to prove that these are exactly the operators which fulfill them, nor have I been able to find alternative definitions that do. (In boolean algebra, if $true:=1\in\mathbb{R}$ and $false:=0\in\mathbb{R}$ these Operators fit the common definition of $\vee, \wedge$)

Are there other sets of operators (excluding trivial examples, such as $+=\cdot$) that comply with these rules? If not, how is it proven (Over $X\subset\mathbb{R}$)?

I have been trying this for quite a time, but I seem unable to find a proof. ANy clues/hints are welcome, I am willing to work :)

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Maybe you would be interested in ? – Hurkyl Nov 3 '12 at 11:30
Indeed. The structures I was searching for are called lattices. Thank you. – CBenni Nov 3 '12 at 11:59
up vote 1 down vote accepted

Given any set, consider the set of all its subsets, with the two operations, union and intersection.

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$\mathbb{R}$ is isomorphic to $\mathcal{P}(\mathbb{N})$. Starting here, a definition can be found in $\mathbb{R}$ that is not equivalent to my other definition. Problem solved. Thank you! – CBenni Nov 3 '12 at 12:05

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