Can $((a,b), \lor, \land)$ be a boolean algebra?

Question

Can I construct Boolean algebra for an open interval $(a, b)$? If I cannot, then why? (I know about $[0, a]$ interval. It constructs with max and min)

Answer 1

It's not really possible to get a good answer to your question without a more specific definition. Here's the specious answer: There exists a function $f:(a,b) \to [0,1]$ so that the range of $f$ is all of $[0,1]$ and no two members of $(a,b)$ are sent to the same point. Fix some Boolean algebra on $[0,1]$, and construct a Boolean algebra on $(a,b)$ as follows: $a \wedge b = f(a) \wedge f(b)$, $a \vee b = f(a) \vee f(b)$. The maximum element will be $f^{-1}(1)$, and the minimum element will be $f^{-1}(0)$.

This answer is probably not very satisfying for you. What I think you wanted to ask was something along the lines of "is there a natural way to construct a Boolean algebra on $(a,b)$" or "is there a way to construct a Boolean algebra on $(a,b)$ so that the operations are familiar functions". But without a formal definition of "natural" or "familiar", there is no good answer to either of those questions.