# Inverse of set operations

$$A \cup B \equiv C$$ then what is $$A$$ in terms of $$B,C$$?

I tried to use $$A\cup B \equiv (A-B)\cup(A \cap B)\cup(B-A)$$ to find a similar expression for $$A \cap B$$ but got nowhere.

From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.

Complement is easy as it is it's own inverse $$(A^C)^C=A$$

not sure if inverse of difference operator is unique, for example, one can use $$(A-B)\cup A \equiv A$$ to construct one inverse of difference $$-X$$.

when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?

• What does $\equiv$ mean in this context? Why aren't you saying they are equal? – fleablood Jan 20 at 0:38

Inverse operators for union, intersection, or set difference are impossible in general.

Knowing $$B$$ and $$C$$ is not enough to determine what $$A$$ is. Consider for example $$? \cup \{1,2\} = \{1,2,3\}$$ Then either $$\{1,3\}$$ or $$\{2,3\}$$ would be possible solutions (and there are two more), so there's no operator that given just $$\{1,2\}$$ and $$\{1,2,3\}$$ can tell you "which of them $$A$$ really was".

On the other hand symmetric difference $$A \mathbin{\triangle} B = (A\setminus B)\cup(B\setminus A)$$ has an inverse operation, namely itself: $$(A\mathop\triangle B)\mathop\triangle B = A$$.

If you consider the algebra of subsets of some universe $$U$$ under the operations $$\triangle$$ and $$\cap$$, you get a Boolean ring which satisfies the usual ring properties with $$\triangle$$ as addition and $$\cap$$ as multiplication. The ring's $$0$$ is $$\varnothing$$ and $$1$$ is $$U$$ itself.

This gives an opportunity to use more of the usual algebraic rules on sets. And you can express the remaining set operations in this vocabulary too: $$A^\complement = 1 \mathop\triangle A \qquad\qquad A\cup B = A\mathop\triangle B \mathop\triangle(A\cap B)$$

What you lose by doing things this way is the nice duality between $$\cup$$ and $$\cap$$ and De Morgan's laws for sets.

(The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $$\mathbb Z$$).

• Nice counter example! – fleablood Jan 20 at 1:31
• So if $C=A\triangle B$ then $C\triangle B=(A\triangle B)\setminus B \cup B\setminus(A\triangle B)=[(A\setminus B \cup B\setminus A)\setminus B ]\cup [B\setminus(A\setminus B \cup B\setminus A]=(A\setminus B)\cup B\setminus(B\setminus A) = (A\setminus B)\cup (B\cap A) = A$. .... nifty! – fleablood Jan 20 at 2:04

It's not possible.

Given sets $$C$$ and $$B$$ there four states of being we can describe. $$x$$ in or not in $$C$$ and $$x$$ in or not in $$B$$ and we can only define sets by some combination of those conditions. Those conditions define the following $$4$$ disjoint basic sets and any set we can possibly be described in terms of $$A$$ and $$B$$ will be a union of these sets.

1) $$\{x \not \in C, x\not \in B\} = C^c$$ (as $$B \subset C$$)

2) $$\{x \not \in C, x \in B\} = \emptyset$$ (as $$B \subset C$$)

3) $$\{x \in C, x \not \in B\} = C\setminus B$$.

4) $$\{x\in C, x \in B\} = C \cap B = B$$ (as $$B \subset C$$).

Although $$A$$ is disjoint from 1) and 2) and 3) $$\subset A$$ set 4) will typically contain elements in $$A$$ and as well as elements not in $$A$$.

Hence in general we can not define $$A$$ solely on the conditions of whether they are or are not in $$C$$ or $$B$$.

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FWIW

We can list all possible sets possible to describe they are

The four above:

1) and [1 and 2] $$C^c$$, 2)$$C^c\cap B = \emptyset$$, 3) and [2 and 3] $$C\setminus B$$ and 4) and [2 and 4] $$B$$.

[1 and 3] and [1 and 2 and 3] = $$C^c \cup C\setminus B = B^c$$.

[1 and 4] and [1 and 2 and 4] = $$C^c \cup B$$

[3 and 4] and [2 and 3 and 4] = $$(C\setminus B )\cup B = C$$

[1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.