Question on a particular collection of sets

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Let {X} be a set. Let {\mathcal{G}} be a non-empty collection of subsets of {X} such that {\mathcal{G}} is closed under finite intersections. Assume that there exists a sequence {X_h \in \mathcal{G}} such that {X = \cup_h X_h}. Let {\mathcal{M}} be the smallest collection of susbsets of {X} containing {\mathcal{G}} such that the following are true:If {E_h \in \mathcal{M}} {\forall h \in \mathbb{N}} and
{E_h} {\uparrow} {E} then {E \in \mathcal{M}}

If {E}, {F}, {E \cup F \in \mathcal {M}} then {E \cap F \in \mathcal {M}}

If {E \in \mathcal {M}} then {E^c \in \mathcal{M}}

Does {X \in \mathcal{M}} ?
 
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You'll only need to show that if A,B\in \mathcal{G}, then A\cup B\in\mathcal{M}.

For this, use the law of DeMorgan: A\cup B=(A^c\cap B^c)^c.
 
Micromass: I am not sure if it works. Can you explain in more detail?
 
What are you unsure of? What did you already try?
 
If A, B \in \mathcal {G} then A \cap B \in \mathcal{G} as \mathcal{G} is closed under finite intersections and therefore A \cup B \in \mathcal {M} using
the DeMorgan's laws and property II of \mathcal {M} as you have pointed out.

I could not prove that if A, B, C, D, E\in \mathcal {G} then A \cup B \cup C \cup D \cup E \in \mathcal {M}. In other words, the union of some sets (which belong to \mathcal {G}) belongs to \mathcal {M} only when the number of sets is less than five.

So, \cup_{h=0}^n X_h does not belong to \mathcal {M} and I could not
use property one (of the three properties of \mathcal {M} listed originally) to prove that X indeed belongs to \mathcal {M}

I might have made a fatal error somewhere though.
 
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Alright, I see your problem. Have you tried finding counterexamples?

For example,

\mathcal{G}=\{\{0\}\cup\{n\}~\vert~n\in \mathbb{N}\}...
 

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