Question on a particular collection of sets

[turiya]

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 $${X \in \mathcal{M}}$$ ?

 

[micromass]

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: $$A\cup B=(A^c\cap B^c)^c$$.

 

[turiya]

Micromass: I am not sure if it works. Can you explain in more detail?

 

[micromass]

What are you unsure of? What did you already try?

 

[turiya]

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.

 

[micromass]

Alright, I see your problem. Have you tried finding counterexamples?

For example,

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