# How can I factorise this union of intersections of sets?

How can I rewrite $$F = F_1 \cup \bigcup_{i=2}^{n \geq2} \left( F_i \cap \bigcap_{j=1}^{i-1} S_j \right)$$ without explicitly writing $$F_1$$?

Assume \begin{align} F_i \cap S_i &= \emptyset \\ \bigcup_{i=1}^n F_i \cup S_i &\subset \mathbb{R}^m. \end{align}

For example, if $$n=3$$, then $$F = (F_1) \cup (F_2 \cap S_1) \cup (F_3 \cap S_1 \cap S_2).$$

• What space are you working on? are there relations between $F_i$ and $S_i$? – macton Jan 22 at 10:09
• @macton, I'm working on $\mathbb{R}^m$. One relation is that $F_i \cap S_i = \emptyset$. Another is $\bigcup_{i \in n} F_i \cup S_i \subset \mathbb{R}^m$, i.e. it doesn't constitute of partition of $\mathbb{R^m}$. – cisprague Jan 22 at 10:21
• You need to get rid of the $\ge 2$ in the upper limit on the union: that’s an external condition on $n$ that has no place in the union notation. – Brian M. Scott Jan 22 at 20:13
• @BrianM.Scott, how should it be expressed then? – cisprague Jan 23 at 21:44
• @ChristopherIliffeSprague: In a separate statement, something like this: Let $S_0=X$; then for $n\ge 2$ we have $$F=\bigcup_{i=1}^n\left(F_i\cap\bigcap_{j=0}^{i-1}S_j\right)\,.$$ – Brian M. Scott Jan 23 at 21:47

If you are working in a whole space called $$X$$, let's say, then define $$S_0 = X$$ then clearly $$F_1 = F_1 \bigcap S_0$$ and $$F_i \cap \bigcap_{j=1}^{i-1} S_j =F_i \cap \bigcap_{j=1}^{i-1} S_j \bigcap S_0 = F_i \cap \bigcap_{j=0}^{i-1} S_j$$ so we can rewrite it as $$F = \bigcup_{i=1}^{n \geq2} \left( F_i \cap \bigcap_{j=0}^{i-1} S_j \right)$$ well it's a little bit cheating by introducing new variable but it works.
• Thank you. This is out of the scope, but, should $S_0 = X$ be changed to $S_0 := X$, or perhaps $S_0 \triangleq X$? – cisprague Jan 26 at 15:48
• @ChristopherIliffeSprague It's a personal preference. For me I've explicitly said "define" before that line, so it should be not that confusing. I use $:=$ only when I want to emphasis that definition. – macton Jan 26 at 15:52