# Set theory Proof help

## Homework Statement

Suppose {Ai| i $$\in$$ I} is an indexed family of sets and I does
equal an empty set. Prove that $$\bigcap$$ i $$\in$$ I Ai
$$\in$$ $$\bigcap$$ i$$\in$$ I P(Ai ) and P(Ai) is the
power set of Ai

none

## The Attempt at a Solution

Suppose x $$\in$$ {Ai| i $$\in$$ I}. Let i be an arbitrary element of
I where x $$\in$$ Ai . Then let y be an arbitrary element of x. Since x
is an element of Ai and y $$\in$$ x it follows that ...

maybe i want to show that $$\bigcap$$ i $$\in$$ I Ai $$\subseteq$$ $$\bigcap$$ i $$\in$$ I Ai and then
I could say that $$\bigcap$$ i $$\in$$ I Ai $$\in$$ $$\bigcap$$ i$$\in$$ I P(Ai )

Related Precalculus Mathematics Homework Help News on Phys.org
Let $$\left\{ A_{i} \right\}_{i \in I}$$ be your indexed set of family.

Do you mean this $$\bigcap_{i=1} A_i = \left\{ x : \forall i \in I: x \in A_i \right\}$$?

Yes sry about the horrible looking symbols

HallsofIvy
Homework Helper

## Homework Statement

Suppose {Ai| i $$\in$$ I} is an indexed family of sets and I does
equal an empty set.
Did you mean "does not equal and empty set"?

I $$\neq$$ $$\oslash$$ is what I meant