# Let $\mathfrak{D}$ be the collection of subsets of $X$ of the form $E_1^{\lambda_1}\cap\dots\cap E_n^{\lambda_n}$. Is $X=\bigcup_{D\in\mathfrak{D}}D$?

## Full problem statement:

Let $$E_1,\dots,E_n$$ be distinct but not necessarily disjoint subsets of $$X$$.

Let $$\mathfrak{D}$$ be the disjoint collection of all subsets of $$X$$ of the form $$E_1^{\lambda_1}\cap\dots\cap E_n^{\lambda_n}$$ where $$\lambda_i \in \{0,1\}$$. Note that $$E_i^1 = E_i$$, $$E_i^0 = E_i^c$$.

Let $$\mathfrak{F}$$ be collection of arbitrary unions of members of $$\mathfrak{D}$$.

1. Is $$X\hspace{1mm}\in\mathfrak{F}$$?
2. Let $$F\in\mathfrak{F}$$. Is $$F^c = X\backslash F \in \mathfrak{F}$$?

## Author's solution

Let $$x\in X$$. Since $$\forall i\in\{1,\dots,n\}\hspace{1mm} E_i^1\cup E_i^0 = X$$, $$x\in E_i$$ or $$x\in E_i^c$$.

Then, every $$x$$ is contained for some $$D = E_1^{\lambda_{1}}\cap\dots E_n^{\lambda_{n}} \in\mathfrak{D}$$. So $$X=\bigcup\limits_{D\in\mathfrak{D}} D\in\mathfrak{F}$$.

## My Questions

Q1. Can we just end the proof there? It seems like all we've proven is that $$X\subset\hspace{-1.5mm}\bigcup\limits_{D\in\mathfrak{D}}\hspace{-1.5mm} D$$.

## Author's solution

Let $$F\in \mathfrak{F}$$. Then $$F$$ is a union of member of $$\mathfrak{D}$$.

Since $$\mathfrak{D}$$ is a disjoint collection and $$X=\hspace{-1.5mm}\bigcup\limits_{D\in\mathfrak{D}}\hspace{-1.5mm} D$$, $$F^c=X\backslash F$$ is also a union of members of $$\mathfrak{D}$$. So $$F^c\in\mathfrak{F}$$.

## My Questions

Q1. How does the statements that $$\mathfrak{D}$$ is a disjoint collection and $$X=\hspace{-1.5mm}\bigcup\limits_{D\in\mathfrak{D}}\hspace{-1.5mm} D$$ work together to imply $$F^c=X\backslash F$$ is a union of members of $$\mathfrak{D}$$? I did try some DeMorgan stuff but it all went nowhere :(

I didn't put all my questions at the end because I think having to scroll up to the relevant block and scroll back down to the questions is going to be annoying.

First question: since $$D \subset X$$ for all $$D \in \mathfrak D$$ it is understood that the reverse inclusion holds so equality holds.

Second question. It is advisable to look at some simple examples. If $$F$$ is expressed as union of certain members of $$\mathfrak D$$ then $$F^{c}$$ is precisely the union of the remaining members of $$\mathfrak D$$.

[$$\mathfrak D$$ is a partition of $$X$$: its members are disjoint and their union is $$X$$. Call these sets $$(D_i)_{i\in I}$$. For any subset $$J$$ of $$I$$, let $$F$$ be the union of the sets $$D_i$$ with $$i \in J$$. Let $$G$$ be the union of the sets $$D_i$$ with $$i \in I\setminus J$$. Then $$F\cup G$$ is the union of all the $$D_i$$'s which is $$X$$. Also $$F$$ and $$G$$ are disjoint. These two facts imply that $$G$$ is the complement of $$F$$].

• Sorry I tried but I can't seem to form a proof for your second answer. Can you edit your answer to include a formal proof for the second part? – mathebeginner Jun 18 at 4:54
• @mathebeginner I have done that. – Kabo Murphy Jun 18 at 5:02
• That's awesome! I was trying to do something with induction but I like this approach! – mathebeginner Jun 18 at 5:07