Let $$\mathbf{B}=\{B_i : i \in I_1\} \subseteq 2^\Omega$$ and suppose that $$\cup(\mathbf{B}) \neq \Omega$$ If $I_2 \subseteq I_1$ then $$\cap\{B_i : i \in I_1\} \subseteq \cap\{B_i : i \in I_2\}$$ therefore the operator $$\cap : 2^{2^\Omega} \mapsto 2^\Omega$$ is, loosely speaking, decreasing. Perhaps, this could be a naif reasoning in favor of $$\cap \varnothing = \Omega$$ but it raises a paradox: given that each $B_i$ "doesn't remember" what set it has been cutted out from, we can also conceivably conjecture $$\cap \varnothing = \cup(\mathbf{B})$$ and, by hypothesis $$\cap \varnothing = \cup(\mathbf{B}) \neq \Omega = \cap \varnothing$$ Obiviously, something is dead wrong. I bumped into this pitfall because I'm studying general topology, and I suspect that it can lead to major misunderstandings.
3 Answers
In SET THEORY by Kunen, he defines $\cap F=\{x:\forall y\in F (x\in y)\}$ only for $F\ne \phi$ and says this "should" make $\cap \phi$ equal to the set of all sets (if applied when $F=\phi$). The convention is that $\cap \phi =\phi$ to avoid this. Then your formula for "monotonicity" only applies to non-empty $B_1,B_2$.
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You should reject the conjecture that $\bigcap\varnothing=\bigcup\mathbf B$.
Working in universe $\Omega$ and $\mathbf B\subseteq\wp(\Omega)$ think of $\bigcap\mathbf B$ defined as:$$\bigcap\mathbf{B}:=\left\{ x\in\Omega:\forall B\in\mathbf{B}\left[x\in B\right]\right\}\subseteq\Omega$$ Then: $$\bigcap\varnothing=\Omega$$
This because for every $x\in\Omega$ no $B\in\varnothing$ exists with $x\notin B$, or equivalently because $x\in B$ is (vacuously) true for every $B\in\varnothing$.
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Thanks for your reply !! If I understand rightly, the rejection of the conjecture, follows from the very assumption of considering lawful "vacuous truth - reasoning" in mathematics. Is it also equivalent ? – sutorUltraCrepidam Dec 02 '15 at 16:37
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I do not understand what you mean if you ask: "Is it also equivalent?". You must have in mind two statements that are eventually equivalent, but I don't manage to figure out wich statements exactly you mean. – drhab Dec 02 '15 at 17:52
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I'm sorry for my vagueness; what I'm trying to express is:"Let (in a sort of reductio ad absurdum ) $\cap \varnothing \neq \Omega$ . Does exist any statement (of any kind) that, making use of the "vacuous truth - reasoning", you can prove both to be false and true (in the context of ZFC theory) ?" I mention specifically the ZFC theory because I need the AC, that is the Tychonoff's theorem. – sutorUltraCrepidam Dec 02 '15 at 19:35
I worried about this kind of stuff too when I was younger. It turns out that the weirdness you're observing is the result of taking a materialistic view. If we take a structuralist viewpoint, the weirdness goes away, and the empty intersection is indeed $\Omega$, as it should be. I say "as it should be" because from the perspective of lattice theory, the empty meet is always the top element. Note that when $\mathcal{P}(\Omega)$ is viewed a poset, the top element is $\Omega$ and the meet operation is intersection.
For a great book on set theory from a structural standpoint, check out Sets for Mathematics. You'll also want to look into category theory.
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Thanks for your reply !! Luckily, I have Rosebrugh's book at hand, and Bourbaki's "Theory of Sets", also. Bourbaki's "General topology" is one of the books I'm studying, but, at present moment, I'm not skilled enough to appreciate if it fully adheres to $\cap \varnothing = \Omega$ or adopts more exotic points of view. Completely off-topic: Google's translator translates "I have Rosebrugh's book at hand" as "I have NOT Rosebrugh's book at hand" (!?). – sutorUltraCrepidam Dec 02 '15 at 16:59
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@sutorUltraCrepidam, I wouldn't worry too much about the conventions any particular book uses. Just adopt the optimal conventions in your own work, and if a statement in some book ends up being false, or having redundant premises, due to the adoption of bad conventions, well, just modify the statement a little to conform with your own preferences. – goblin GONE Dec 10 '15 at 06:22