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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 $F^c = X\backslash F \in \mathfrak{F}$?

1. Is $X\hspace{1mm}\in\mathfrak{F}$?

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 $X\subset\hspace{-1.5mm}\bigcup\limits_{D\in\mathfrak{D}}\hspace{-1.5mm} D$.


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

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.

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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 $F$].

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  • $\begingroup$ 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? $\endgroup$ – mathebeginner Jun 18 at 4:54
  • $\begingroup$ @mathebeginner I have done that. $\endgroup$ – Kabo Murphy Jun 18 at 5:02
  • $\begingroup$ That's awesome! I was trying to do something with induction but I like this approach! $\endgroup$ – mathebeginner Jun 18 at 5:07

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