Boolean Algebra Embeddings: Defining and Understanding the Role of Monomorphisms

AI Thread Summary
An embedding from a boolean algebra B to a boolean algebra B' is defined as an injective map f that preserves the supremum and complement operations. The discussion raises a question about whether the condition for infimum preservation, f(inf{x,y}) = inf'{f(x),f(y)}, is necessary for the definition of an embedding. It is noted that this condition may be derived from the other properties due to de Morgan's laws. The conversation emphasizes the relationship between embeddings and monomorphisms in the context of boolean algebras. Understanding these definitions is crucial for a deeper grasp of the structural properties of boolean algebras.
quasar987
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According to my notes, the definition of an embedding from a boolean algebra B in a boolean algebra B' is an injective map f:B-->B' such that for all x,y in B, f(sup{x,y}) = sup'{f(x),f(y)} and f(Cx)=C'(f(x)), where sup is the supremum in B and sup' is the complement in B', and where C is the complement in B and C' the complement in B'.

But I read on wiki that generally, an embedding is supposed to be a monomorphism. Aren't we missing the condition f(inf{x,y}) = inf'{f(x),f(y)}?
 
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quasar987 said:
But I read on wiki that generally, an embedding is supposed to be a monomorphism. Aren't we missing the condition f(inf{x,y}) = inf'{f(x),f(y)}?
Doesn't it follow from the other ones?
 
Right, because of the de Morgan laws !
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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