Curious identity involving symmetric difference

Question

While studying the properties of measurable null sets, I found the following identity: $\bigcup_i B_k\triangle B_i=\bigcup_i B_i - \bigcap_i B_i $ Or in other words, the value of the expression is independent of the choice of $k$.

I can prove this using a case separation in first principles, but it doesn't give me much intuition on why it is true.

So I'm asking if the above can be deduced just using basic set algebra? Edit: I mean is there a sequence of rewritings that reveal the symmetry of the statement?

Answer 1

Let $x \in \bigcup_{i} B_i$. If $x$ is contained in all the $B_i$ than $x$ is not contained in $\bigcup_i B_k \triangle B_i$ and also not contained in $\bigcup_i B_i - \bigcap_i B_i$. Now suppose $x$ is contained in some, but not all, of the $B_i$. Then $x$ is contained in $\bigcup_i B_i - \bigcap_i B_i$. Furthermore, if $x \in B_k$ there is a $i \neq k$ such that $x \not\in B_i$ and if $x \not\in B_k$ there is a $i \neq k$ such that $x \in B_i$. That is, there exists $i \neq k$ with $x \in B_i \triangle B_k$ and $x$ is contained in $\bigcup_i B_k \triangle B_i$. Since this holds for all $x$, the conclusion follows.

Answer 2

Using slightly different notation, you ask why, for any $\;k\;$, $$ \tag{0} \langle \cup i :: B_k \triangle B_i \rangle \;=\; \langle \cup i :: B_i \rangle - \langle \cap i :: B_i \rangle $$

Now I'm not sure what you mean by "proving using just basic set algebra", but in any case, I find this kind of equalities are easier proved on the element/logic level. So here is such a proof.

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By set extensionality, the definitions of $\;\cup,-,\cap\;$, and definition $$ x \in X \triangle Y \;\equiv\; x \in X \not\equiv x \in Y $$ we have that $(0)$ is equivalent to the fact that for every $\;x\;$, $$ \langle \exists i :: x \in B_k \;\not\equiv\; x \in B_i \rangle \;\equiv\; \langle \exists i :: x \in B_i \rangle \land \lnot\langle \forall i :: x \in B_i \rangle $$ Abbreviating $\;R(j) \;\equiv\; x \in B_j\;$ to hide irrelevant detail, and applying DeMorgan to the rightmost part for symmetry, this is equivalent to $$ \tag{1} \langle \exists i :: R(k) \;\not\equiv\; R(i) \rangle \;\equiv\; \langle \exists i :: R(i) \rangle \land \langle \exists i :: \lnot R(i) \rangle $$ Now the proof is very simple by a case split on $\;R(k)\;$: if $\;R(k)\;$ then $(1)$ reduces to $$ \langle \exists i :: \lnot R(i) \rangle \;\equiv\; \text{true} \land \langle \exists i :: \lnot R(i) \rangle $$ and if $\;\lnot R(k)\;$ then $(1)$ reduces to $$ \langle \exists i :: R(i) \rangle \;\equiv\; \langle \exists i :: R(i) \rangle \land \text{true} $$ both of which are trivially true. This proves $(1)$, and therefore $(0)$.

Answer 3

Since you tagged boolean algebra, let's try the following.

First let $x,x_i$ be elements of a boolean algebra, then $$\begin{aligned} \prod^n_{i=1}(x+x_i) &= x^n+(x_1+\cdots+x_n)x^{n-1}+(x_1x_2+\cdots)x^{n-2}+\cdots + (x_1\cdots x_n)\\ &=x(1+(x_1+\cdots x_n)+(x_1x_2+\cdots)+\cdots+(x_1\cdots x_n))+(1+x)\prod x_i\\ &=x\prod(1+x_i) +(1+x) \prod x_i\\ \end{aligned} $$ Add $1$ to both sides and replace $x=1+b_k$, $x_i=b_i$, we have $$1+\prod(1+(b_k+b_i))=1+(1+b_k)\prod(1+b_i)+b_k\prod b_i$$ Let $b_i$ be the characteristic function of $B_i$, we have $$\left(\bigcap(B_k\triangle B_i)^c\right)^c = \left(\bigcap B_i^c\right)^c\triangle\bigcap B_i$$ or $$\bigcup B_k\triangle B_i = \bigcup B_i\triangle \bigcap B_i =\bigcup B_i-\bigcap B_i $$