Question

is this step allowed? if so, why?

Answer 1

I think the reason you can make such a conclusion is because \(B\) is independent of the index \(i_0\).

Is this taken from a proof involving some set \(B\) and the union \(\bigcup A_i\) ?

Answer 2

yes, it came from this exercise. Number 22

Answer 3

Okay, so yes, you can say so because regardless of whichever \(i_0\) you have \(x\in A_{i_0}\), you always have \(x\in B\), so \(x\in B\cap A_{i_0}\), and so \(x\in \bigcup (B\cap A_{i_0})\).

Answer 4

I feel like maybe i'm just thinking too hard, but how does the fact the B is independent of the index i_o allows us to bring the i_o in front of the parentheses. What definition (or maybe theorem?) was used?

Answer 5

@SithsAndGiggles

Answer 6

Think of it in terms of an example.

Suppose \(B\) is the set of all cars in a lot that are green, and \(\bigcup A_i\) is the set of all cars that are BMWs. From the set of BMWs (the whole union), you set up an index that separates the cars by number of wheels. (Forgive me, I'm not an expert on automobiles :P)

Let \(x\) be any one car, but to make it more concrete, let's say this \(x\) represents MY car - a green, two-wheeled BMW. This means that \(x\in B\) due to its color and \(x\in A_2\), where the subscript denotes the particular number of wheels.

The set \(B\cap A_2\) is the set of all green, two-wheeled cars.

In the context of this proof: there's a certain number of wheels \(i_0\) that allows my car to belong to \(A_{i_0}\). Translation: \(\exists~i_0(x\in A_{i_0})\).

My car is green. Translation: \(x\in B\).

There's a certain number of wheels \(i_0\) such that my car has the appropriate number of wheels \(i_0\) to be separated from the rest, AND it happens to be green. This means I can group my car with the other green two-wheelers. Translation: \(\exists~i_0\big[x\in(B\cap A_{i_0})\big]\).

I hope I haven't complicated any of the concepts here. I tried my best to come up with a simple-enough example :)