How to prove the uniqueness of complement in the algebra of sets ( without using set theory but exclusively the laws of boolean algebra )?

Question

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In abstract algebra, the uniqueness of inverse in a group is proved using more basic laws ( laws of groups themselves, not extraneous laws). Could the same thing be done regarding the algebra of sets?

How to prove, using only the laws of the algebra of sets, that a set has only one complement?

Am I wrong in assuming that complement uniqueness can be proved inside the system of the algebra of sets?

Is it rather a postulate?

Answer 1

Let $A$ be a set in a universal set $U$, and let $A_1$ be a complement for $A$. If $A_2$ is another complement for $A$, we have:

$A_1=U \cap A_1= (A\cup A_2) \cap A_1= (A\cap A_1) \cup (A_2\cap A_1)=A_2\cap A_1$.

Similarly, $A_2=U\cap A_2= (A\cup A_1)\cap A_2=(A\cap A_2)\cup (A_1\cap A_2)=A_1\cap A_2$.

Hence, $A_1=A_2\cap A_1= A_1 \cap A_2= A_2$.

This approach is very similar to showing uniqueness in abstract algebra.

Answer 2

If $$\tag1A\cap B=\emptyset\text{ and }A\cup B=U,$$ then $$x\in B\iff x\in U\land x\notin A $$ hence $B$ is uniquely determind by $U$ and $A$.

Answer 3

Let it be that $B_1$ and $B_2$ are complements of set $A$ in universal set $X$.

Then for $i=1,2$: $$B_i\cup A=X\text{ and }B_i\cap A=\varnothing$$

If $x\in B_1$ then $x\notin A$ since $B_1\cap A=\varnothing$.

Next to that $x\in X=B_2\cup A$ so we conclude that $x\in B_2$.

Proved is now that $B_1\subseteq B_2$ and similarly it can be proved that $B_2\subseteq B_1$.

Our final result is then: $$B_1=B_2$$So complements are unique.

Answer 4

Below a proof that can be found at WikiProof https://proofwiki.org/wiki/Complement_in_Boolean_Algebra_is_Unique