Why intersection of nothing is everything?

Question

How do I prove:

$$ \bigcap_{i \in \varnothing} A_i= \Omega $$

Also, another logical explanation will be great.

Suppose that $x \in \Omega$. Is it true that $x \in A_i$ for all $i \in \emptyset$?
In other words, is it true that if $i \in \emptyset$ then $x \in A_i$?

Yes, this is vacuously true. — littleO, Jul 08 '18 at 14:13

score 1 · Accepted Answer · answered Jul 08 '18 at 14:07

1

It ensures that

$$\left ( \bigcap_{i \in I} A_i \right ) \cap \left ( \bigcap_{j \in J} A_j \right ) = \bigcap_{i \in I \cup J} A_i$$

for arbitrary families $A_i$ and index sets $I,J$.

It can also be regarded as a special case of the fact that "$\forall x \in \emptyset \: P(x)$" is true for any predicate $P$: for any $y$, "$\forall i \in \emptyset \: y \in A_i$" is true.

answered Jul 08 '18 at 14:07

Ian

101,645

That is vacuously true – amWhy Jul 08 '18 at 15:17

Why intersection of nothing is everything?

1 Answers1