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 $$ F = (F_1) \cup (F_2 \cap S_1) \cup (F_3 \cap S_1 \cap S_2). $$

  • $\begingroup$ What space are you working on? are there relations between $F_i$ and $S_i$? $\endgroup$ – macton Jan 22 at 10:09
  • $\begingroup$ @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 $\mathbb{R^m}$. $\endgroup$ – cisprague Jan 22 at 10:21
  • $\begingroup$ 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. $\endgroup$ – Brian M. Scott Jan 22 at 20:13
  • $\begingroup$ @BrianM.Scott, how should it be expressed then? $\endgroup$ – cisprague Jan 23 at 21:44
  • $\begingroup$ @ChristopherIliffeSprague: In a separate statement, something like this: Let $S_0=X$; then for $n\ge 2$ we have $$F=\bigcup_{i=1}^n\left(F_i\cap\bigcap_{j=0}^{i-1}S_j\right)\,.$$ $\endgroup$ – 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.

  • $\begingroup$ Thank you. This is out of the scope, but, should $S_0 = X$ be changed to $S_0 := X$, or perhaps $S_0 \triangleq X$? $\endgroup$ – cisprague Jan 26 at 15:48
  • $\begingroup$ @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. $\endgroup$ – macton Jan 26 at 15:52

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