Does F, the collection of all finite subsets of the real numbers, have the finite intersection property?

I'm not sure how to approach this questions but my intuition tells me that F does not have the finite intersection property. So if one set is {1, 2, 3, 4} and another set is {5, 6, 7, 8} the intersection of these 2 sets would be empty thus F does not have the FIP. Is this logic right?

Also, I am trying to figure out:

Does G, the collection of complements of the subsets in F (the collection of all co-finite subsets of R), have the finite intersection property?

It makes sense to me that the intersection of co-finite subsets would be non-empty because all of the sets would be "bounded" by negative infinity and positive infinity if that makes sense.. but I don't know how to explain or prove that logically.

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    $\begingroup$ Hint: the empty subset is in particular a finite subset of the real numbers. $\endgroup$ – Mees de Vries Nov 7 '16 at 13:19
  • $\begingroup$ @MeesdeVries so F doesn't have the FIP since any set intersect with the empty set is empty? $\endgroup$ – cshooo Nov 7 '16 at 13:22
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    $\begingroup$ Can you define the finite intersection property you have in mind? $\endgroup$ – Mikhail Katz Nov 7 '16 at 13:23
  • $\begingroup$ You're right that $F$ does not have the finite intersection property. Showing that $G$ does have it is more interesting. Hint: You are working with complements and intersections, so recall De Morgan's law. (Your current plan might lead you astray: the collection of all unbounded sets does not have the finite intersection property. Consider the set of integers and the set of non-integers.) $\endgroup$ – Nate Eldredge Nov 7 '16 at 13:24

As smartly pointed out in the comments, it is trivial to show that $F$ does not have the $FIP$.

Now what about $G$? Let us assume that $G$ does not have the $FIP$ and arrive at a contradiction:

If $G$ does not have the $FIP$ then there is a finite $G' \subset G $ for which

$$\bigcap_{g \in G'} g = \emptyset$$

But given that $G$ is composed of the complements of sets in $F$, we get: $\forall g\in G'\ \ \exists f_g\in F:\ g = \mathbb{R} - f_g$

Rewriting the intersection, you get

$$\bigcap_{g \in G'} (\mathbb{R} - f_g) = \emptyset$$

Now if there is no number $x$ that appears in all $g$ then for every number $x \in \mathbb{R}$ there is some $g_x$ such that $x \notin g_x \iff x \notin (\mathbb{R} - f_{g_x}) \iff x \in f_{g_x}$. But since that is true for any $x$, then all the real numbers are in a finite union of finite sets, which is itself finite! So we would be proving that $\mathbb{R}$ is finite. But we know that $\mathbb{R}$ isn't finite so we arrived at a contradition and therefore $G$ has the $FIP$.


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